Lecture Notes Math 263 Calculus III (Part II)ee.istanbul.edu.tr/uploads/dersnotlari/5055-Calculus3book2.pdf · Math 263 Calculus III 2 Lecture 10 Sections 16.1 and 16.2 – Integrals

Math 263 Calculus III

Lecture Notes


(Part II)

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Math 263 Calculus III i

Table Of Contents Lecture 10.........................................................................................................................2 Lecture 11 ....................................................................................................................... 17 Lecture 12...................................................................................................................... 26 Lecture 13...................................................................................................................... 40 Lecture 14...................................................................................................................... 55 Lecture 15...................................................................................................................... 68

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Math 263 Calculus III 2

Lecture 10 Sections 16.1 and 16.2 – Integrals In single variable calculus we presented the integral of a continuous function as the ’opposite’ of differentiation and showed how it was related to the area under the curve defined by the graph of the function. The link between area and the integral was illustrated by partitioning or dividing some interval over which the function is defined into n parts and then using the maximum or minimum value of the function on each partition to define the height of a rectangle. The area under the curve defined by the function on the given interval is then approximated by summing the areas of these rectangles. If the maximum value of the function is used to define the height of each rectangle, then the approximated area is greater than the actual area under the curve. If the minimum values are used, the then approximated area is less than the actual area. By choosing finer and finer partitions i.e. larger values of n, the accuracy of this approximation increases. In fact, the actual area under the curve is bounded by the two areas found by using the minimum and maximum values of the function on each partition. If Amin(n) represents the area found by partitioning the interval into n parts and using the minimum values of the function on each partition, and Amax(n) represents the area found by partitioning the interval into n parts and using the maximum values of the function on each partition, and A represents the actual area the we have:

Amin(n) � A � Amax(n) The first known use of this technique of partitioning an interval and using rectangles to approximate the area under a curve was by the Greek Mathematical Archimedes (287BC - 212BC). In fact there is currently a discussion among scholars that Archimedes anticipated both Newton and Leibniz in the discovery/invention of the calculus.

Archimedes (287BC - 212BC)

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Approximately 2000 years ago Archimedes used a similar technique to estimate the area of the circle. Instead of using n-rectangles Archimedes used n-sided polygons however the idea is essentially the same. Figure 1 shows the ‘modern’ version of approximating the area of one quarter of a circle using rectangular approximations.

y

Figure 1. Approximating the area of a circle with rectangles x

In the case of Figure 1, the maximum value of the circle was used on each interval and only n = 6 partitions of the interval were used. Figure 2 shows the same partitioning with the minimum value of the circle on each partition being used. In both cases, the approximation is clearly not very accurate, however by doubling the number of partitions i.e. letting n = 12, it is clear that the approximation is better. The idea of partitioning an interval and taking the minimum and maximum values of the function on the each partition and then summing the product of the partition width and the maximum or minimum value was first presented by Riemann (1826-1866) and is known ad the Riemann sum.

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Georg Friedrich Bernhard Riemann (1826-1866)


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4y

x Figure 2. Area using minimum values.

y

x Figure 3. Partition with n = 12


More formally for a general function, f, we write the Riemann sums as:

� � � �

� � � �

min1

max1

n

i ii

n

i ii

A n f x x

A n f x

x

'

'

¦

¦

�

�

Where ix

� is the place where f takes on the minimum value of the ith partition,

ix� is where f takes on the maximum value on the ith partition, and ix' is the size or width of the ith partition. Note: although we often construct the Riemann sums using the same size partitions, this is not required and is reflected by associating the subscript i with the partition size. We define the definite integral over the interval [a, b] as the limit of either the upper or lower Riemann sum as the partition size tends towards zero or the number of partitions tends towards infinity:

� � � � � �01

lim limb n

i i i in xi ia

f x dx f x x f x xof ' o

§ · § · ' ¨ ¸ ¨ ¸

© ¹© ¹¦ ¦³ '

For functions of two variables we extend this idea of partitioning and summing to define a double Riemann sum and the integral over a region. To see this, we consider the graph of the surface defined by the function f(x, y). In this case, we can partition the area in the xy-pane under the surface into squares (or rectangles) and form boxes whose heights are given by either the minimum or maximum value of the function over the rectangular partitions. Figure 4 illustrates the partitioning for a quarter of a sphere. Like with the circle, we can choose the height of the boxes to like entirely under the surface of the sphere by choosing the minimum value of the sphere over each square in the xy-plane or we can choose the height of the boxes to lie entirely above the surface by choosing the maximum value of sphere over each square in the xy-plane. In both cases, we can compute the volume of each box and then sum these volumes to approximate the volume, V, under the surface.

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Figure 4. Approximating Sphere with boxes

z

y

x

If we partition the x-axis into m-parts and the y-axis into n-parts, and if we let Vmin(m, n) be the volume obtained by summing the volume of the boxes formed by taking the minimum value of the sphere and we let Vmax(m, n) be the volume obtained by taking the maximum value of the sphere on each partition, then we have:

Vmin(m, n) � V � Vmax(m, n)

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And it is clear that the volume of the sphere is bounded by Vmin and Vmax over the specified region of the xy-plane. More formally for a general function, f, we write these Riemann sums as:

� � � �

� � � �

min1 1

max1 1

, ,

, ,

m n

i j i ji j

m n

i j i ji j

V m n f x y x y

V m n f x y x y

' '

' '

¦¦

¦¦

� �

� �

Analogous to what we did for the single variable case, we define the definite integral over the region, R, as the limit of either the upper or lower Riemann sums as the partition sizes tends towards zero or the number of partitions tends towards infinity:

� � � �, , 01 1 ,lim , lim ,

m n

i j i j i j i jm n x yi j i jR

fdA f x y x y f x y x yof ' ' o

§ · § ' ' ¨ ¸ ¨

© ¹ ©¦¦ ¦³

·' ' ¸

¹

The general notation

R

fdA³ implies integration over of an arbitrary shaped

region (area) e.g. triangle, circle, etc. In our presentation above we chose a rectangular shaped region for convenience since it makes the summation limits and partitioning of the xy-pane into squares or rectangles simpler. However, this need not be the case. We can instead picture covering an arbitrary shaped region in the xy-plane with rectangles so that either all the rectangles lie just inside the region or the rectangles extend just outside the region (so that the region is contained inside our rectangles). We can then compute either the minimum or maximum value of the function on each rectangle and compute the volume of the boxes, and sum. We further note that no particular ordering is implied in the summation. We can picture, taking all the boxes and laying them end to end in a line in arbitrary order and summing their volumes. In short, the order of summation is not important since the same volumes of the same boxes are ultimately summed together. This will have an important consequence with regards to the order of integration which we will discuss shortly. Another important point to consider is that if we choose a function f(x, y) = 1 everywhere then our integral becomes:

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R R

fdA dA A ³ ³

for f(x, y) =1 That is, when f(x, y) = 1, the integral gives us the area of the region we are integrating over. Finally we note that, using the definition of continuity and the HG-technique for limits one can formally prove that

� � � �max min,lim , , 0m n

V m n V m nof

�

The formal proof is beyond the scope of the course, but we provide the following brief outline: Since f(x, y) is continuous on the closed region R, then one can show that f(x, y) is uniformly continuous on R. That is for any given an H > 0 we can find G > 0 such that for any two points (x1, y1) and (x2, y2) in R, if

� � � �2 22 1 2 1x x y y G� � � � , then � � � �2 2 1 1, ,f x y f x y H� � . We note that if we

choose our i jx y' ' to lie completely inside R then 1 1

m n

i j Ri j

x y A

' ' d¦¦ where AR is

the area of R. Since f(x, y) is uniformly continuous then if we choose

0RAH

! then we can find G > 0 such that � � � �2 2 1 1, ,R

f x y f x yAH

� � whenever

� � � �2 22 1 2 1x x y y G� � � � . In particular, we choose

2ixG

' � and 2jyG

' � so

that on each square or rectangle of our partition we have

� � � �, ,i j i jR

f x y f x yAH

� �� . Therefore:

� � � � � � � �

� � � �

max min1 1 1 1

1 1

1 1

, , , ,

, ,

m n m n

i j i j i j i ji j i j

m n

i j i j i ji j

m n

i ji jR

V m n V m n f x y x y f x y x y

f x y f x y x y

x yAH H

� ' ' � '

� ' '

� ' ' �

¦¦ ¦¦

¦¦

¦¦

� � � �

� � � �

'

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====================================================================== Integrated Integrals As we mentioned above, the value of the Riemann sum in independent of the order of the summation. This is a direct result of the commutative, associative, and distributive properties of real numbers. As such we are free to sum the terms in which ever way we choose. We can use this ‘freedom’ to show that the integral over a region is equivalent to a double integral over two intervals. To see this we go back to our definition of the integral over a region as the limit of a double Riemann sum:

� � � �

� �

, ,1 1 1 1

1 1

1

lim , lim ,

lim lim ,

lim ( , )

m n m n

i j i j i j jm n m ni j i jR

m n

i i j jm ni j

m d

i icmi

ifdA f x y x y f x y y x

x f x y y

x f x y dy

of of

of of

of

§ · § ' ' ¨ ¸ ¨

© ¹ ©§ ·

' '¨ ¸© ¹

'

¦¦ ¦ ¦³

¦ ¦

¦ ³

·' '¸

¹

Note: we have assumed that xi is ‘constant’ (in the sense we did when taking partial derivative) in going from the second to the last step in the equation above. That is we assumed that f was a function of the form f(xi, y) and could be treated as if it were a single variable function, We then applied the definition of the integral for a single variable function as being a Riemann sum over an interval, in this case [c, d]. We now let

and, substituting into the expression above, repeat this

process:

� � ( , )d

i icF x f x y dy ³

� � � �1

lim ( , )m b b d

i i a a cmiR

fdA F x x F x dx f x y dydxof

' ¦³ ³ ³ ³

Thus we have expressed the integral over a region as the double integral over two intervals. Please note, that we can repeat this argument, reversing the order of the summation so that we sum over i first and j second (i.e. over x first and y second) so the result has the order of integration reversed. That is we can show that the order of integration in the resulting

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double integral can also be reversed. Hence we have the following general result: If f(x, y) is a continuous function on the region R defined by the intervals [a, b] and [c, d] (written R = [a, b] ș [c, d]) then:

� � � �( , ) ( , )b d d b

a c c aR

fdA f x y dy dx f x y dx dy ³ ³ ³ ³ ³

Example 1. Let � �, 124 8x y

f x y � � . Find the integral over the region R = [0, 8]

ș [0, 16]. Using the formula above for the integral:

8 216 8 16

0 0 00

16 216

00

12 124 8 8 8

88 88 12802

R

x y x yfdA dx dy x x dy

yydy y

§ ·§ · � � � �¨ ¸¨ ¸ ¨ ¸© ¹ © ¹

� �

³ ³ ³ ³

³

As noted above, the region of integration need not be rectangular but may be of arbitrary shape. In the formula above where we expressed the integral over the region R as the double integral over two intervals, the formula (does not assume) but is implicitly stated as the integral over a rectangular region. For non-rectangular regions, the formula still applies, but must be stated with more care. In particular, the limits of the inner integral become functions of the variable of the outer integral: If f(x, y) is a continuous function on the region R then:

� �

� ��

� �� 2 2

1 1

( , ) ( , )b g x d h y

a g x c h yR

fdA f x y dy dx f x y dx dy ³ ³ ³ ³ ³

A couple of points are note worthy here. First, the order of integration can still be reversed, however their may be a ‘preferred’ order in the sense that

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the boundary functions (g1(x), g2(x), h1(y), and h2(y)) may be significantly more straight forward to find or compute for one order as opposed to the other. That is, doing the integral over x first say, might be ‘easier’ in the sense that h1(x), and h2(x) are either given, are obvious from the problem, or are relatively simple functions whereas g1(x) and g2(x) might be difficult to determine or be very complex functions. Second, the double integral itself may be significantly simpler to evaluate using one order rather than the other. We shall give examples of these points below. Finally, the limits on the outer integral must always be constants. Thus the chosen order of integration will depend on the problem and is usually determined by the function being integrated and the shape of the region R. The order of integration which results in the ‘simplest’ evaluation of the integrals is the one that is preferred. Example 2. Let f(x, y) = xy. Integrate f(x, y) for the triangular region bounded by the x-axis, the y-axis, and the line y = 2 – 2x. Making a graph of R often helps to visualize the region of integration. Figure 5 show the region of integration for this example. y

R

y = 2 – 2x

2

x 1

Figure 5: Triangular region of integration

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In this case our function is relatively simple and symmetric in the variables, so integrating in either order is equivalent. However, since one of the boundary functions is given in terms of x, i.e. g2(x) = y = 2 – 2x, integrating over y first saves us the effort of inverting this function. Graphically or geometrically integrating over y first is equivalent to moving along the x axis from 0 to 1 and integrating from y = 0 to y = 2 - 2x. That is, summing up the vertical strips as shown in Figure 6.

y

Integrate over these vertical strips

y = 2 – 2x

2

x 1

Figure 6: Integrate over y first corresponding to the vertical strips Hence, the limits of integration are a = 0, b = 1, g1(x) = 0, and g2(x) = 2 – 2x. We therefore have:

� ��

� �

2 2 21 2 2 1

0 0 00

1 12 2

0 0

1 3 41 2 3 2

00

2

1 12 2 4 8 42 2

4 42 2 13 2 3

16

xx yxydy dx x dx

x x dx x x x dx

x xx x x dx x

�� § ·

¨ ¸¨ ¸© ¹

� � �

� � � � � �

³ ³ ³

³ ³

³12

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If we choose to do the integral in the opposite order, then we need to invert the y = 2 – 2x i.e. express x as function of y. In this case this is relatively straightforward and we get: x = 1 - ½y = h2(y). Integrating in this order corresponds to integrating from y = 0 to y = 2 along horizontal strips ranging from x = 0 to x = 1 - ½y, as shown in Figure 7.

y

Integrate over these horizontal strips

x = 1 – ½y

2

x Figure 7: Integrate over x first corresponding to the horizontal strips

1

The limits of integration are c = y = 0, d = y = 2, h1(y) = x = 0, and h2(y) = x = 1 - ½y. Therefore:

111 222 1 22

0 0 00

2 22 2

0 0

23 2 3 42 2

00

2

1 1 11 12 2 2 4

1 1 12 4 2 2 3 1616

yy xxydx dy y dy

yy y dy y y dx

y y y yy y dy

�� § ·§ ·

¨ ¸¨ ¸ ¨ ¸© ¹ © ¹

§ ·§ · � � �¨ ¸¨ ¸© ¹ © ¹

§ · � � � � �¨ ¸

© ¹

³ ³ ³

³ ³

³4 13 2�

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We note that in this example, the order of integration is equally ‘simple’ in both directions. However, this is not always the case, as will be demonstrated below. Example 3. Integrate � �, 1f x y xy � over the region bounded by the line y = x and the parabola y = x2. Figure 8 shows the region of integration. In this case, it’s fairly straightforward to identify the limits of integration as follows: c = 0, d = 1, g1(x) = x2, g2(x) = x.

y

R

y = x 1 y = x2

x

1 Figure 8: Region of Integration defined by y = x and y = x2

We therefore have:

2

2

21 1

0 0

13 5 2 4 3 61 2

00

12

1 1 1 12 2 2 8 3 12 2 8 3

524

xx

xx

yxydy dx y x dx

x x x x x xx x dx

§ ·§ ·� �¨ ¸¨ ¸ ¨ ¸© ¹ © ¹

� � � � � � � � �

³ ³ ³

³ 12

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Example 4. Evaluate

6 2 3

03

1x x y dy d�³ ³ x .

In this case we note that the order of integration is specified by the way the integral is written. However, evaluating the integral in this order is non-trivial since the integral of 3 1y � has no elementary anti-derivative. We would therefore like to reverse the order of integration in the hopes of simplifying the problem. However, we cannot just reverse the order since the lower limit of the inner integral is a function of x. We must therefore determine the region of integration first, so we can express the limits of integration in terms of functions of y for the inner integral. The limits of integration are x = 0, x = 6, y = x/3, and y = 2. Figure 9 shows the region bounded by these lines.

y

x = 6

2

R

x = 0

y = 2

y = x/3

x 6 Figure 9: Region of Integration defined by y = 2, y = x/3, x = 0, x = 6 From the Figure 9 we see the region of integration is the triangular shaped region bounded by the y-axis, and the lines y = 2 and y = x/3. To reverse the order of integration we want to integrate from y = 0 to y = 2 while integrating along horizontal strips from x = 0 to x = 3y. Thus we have:

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� �

6 2 2 33 3

0 0 03

3 22 23 2 3

0 00

2 33 2

0

1 1

91 12 2

1 27 1 26

y

x

y

x y dy dx x y dx dy

xy dy y y d

y

� �

�

� �

³ ³ ³ ³

³ ³ y�

Thus reversing the order of integration allowed us to evaluate the double integral in this case. It is important to note that this required us to determine the region of integration from the integration limits of the original integral.

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Lecture 11 Sections 16.3 and 16.4 – Double & Triple Integrals & Polar Coordinates Triple Integrals In the previous lecture we showed how a function of two variables can be integrated over a region in 2-space i.e. an area and how integration over a region is equivalent to an iterated or double integral over two intervals. This concept and be extended to integration over a solid region or volume of 3-space using triple integrals. In this case, we have a continuous function, f(x, y, z) of three variables that is integrated over a solid region W in 3-space. We define our integral in terms of a triple Riemann sum over a solid region that has been partitioned into boxes having volume . If W is a box-shaped region then

we can let

ijk i j kV x y z' ' ' '

ib a

x xl�

' ' , jd c

y ym�

' ' , and kf e

z zn�

' ' . That is the

solid region W = [a, b] ș [c, d] ș [e, f]. If we let (xi, yj, zk) be a point in ijkth box, then we say: If f is a continuous function of three variables, then we define the integral over the solid region W as:

� �, , 1 1 1lim , ,

l m n

i j k i i kl m ni j kW

fdV f x y z x y zof

'¦¦¦³ ' '

We can show that the integral over the solid region W is equivalent to a triple integral over three intervals. The argument is similar to the argument we used for the double integral and is therefore not repeated here. We state formally that:

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If f(x, y, z) is a continuous function on the solid region W defined by the intervals [a, b], [c, d], [e, f] (written W = [a, b] ș [c, d] ș [e, f]) then:

� �( , , )f d b

e c aW

fdV f x y z dx dy dz§ · ¨ ¸© ¹³ ³ ³ ³

Example 1. Integrate f(x, y, z) = 1 + xyz over the cube of length 4. From above:

� ��

4 24 4 4 4 4

0 0 0 0 00

4 4 4 4 2

0 0 0 0

44 2

0 0

12

4 8 4 4

16 64 16 32 576

W

xfdV xyz dx dy dz x yz dy dz

yz dy dz y y z dz

z dz z z

§ ·§ ·§ · � �¨ ¸¨ ¸¨ ¸ ¨ ¸¨ ¸© ¹ © ¹© ¹

� �

� �

³ ³ ³ ³ ³ ³

³ ³ ³

³

We note that if f(x, y, z) = 1, the triple integral is just the volume of the solid region W. Example 2. Find the volume under the plane z = 12 - ¼x - ƚy over the region [0, 8] ș [0, 16]. We wish to express the double integral of the function z = 12 - ¼x - ƚy over the region [0, 8] ș [0, 16] as a triple integral of the function f(x, y, z) = 1 over a solid region W. We therefore must find W. Figure 10 shows the volume bounded by the region [0, 8] ș [0, 16] and the plane z = 12 - ¼x - ƚy. From Figure 10 we see that that as x goes from 0 to 8 and y from 0 to 16, z ranges from 0 to the plane z = 12 - ¼x - ƚy. Thus the limits of integration are a = 0, b = 8, c = 0, d = 16, e = 0, and f = 12 - ¼x - ƚy.

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z x

16

8

z = 12 - ¼x - ƚy

R

12

y

Figure 10. Volume of integration defined by the region R and a plane We therefore have:

124 88 16 12 8 16

4 80 0 0 0 0

0

16 28 16 8

0 0 00

88 2

0 0

12 124 8 4 16

176 4 176 2 1280

x yx y

V dz dy dx z dy dx

x y x ydy dx y y dx

xdx x x

� �� § ·§ ·§ ·§ ·

¨ ¸¨ ¸¨ ¸¨ ¸¨ ¸ ¨ ¸¨ ¸© ¹© ¹ © ¹© ¹§ ·§ · � � � �¨ ¸¨ ¸ ¨ ¸© ¹ © ¹

� �

³ ³ ³ ³ ³

³ ³ ³

³

Note: In this case we reversed the order of integration so that we integrated over z first and x last. In addition, the limits of integration over z were functions of x and y. This is similar to the results we found for double integrals.

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Example 3. Compute the integral for f(x, y, z) = z over the solid region bounded by the cone 2 2z x y � and the plane z = 3. First we need to determine the solid region of integration based on the information above. Figure 11 shows the cone and the plane specified by the problem. z

x

z = 3

3

y

2 2z x y �

Circle of radius z

-3 3

3

Level Surface i.e. circle of radius z

2 2 9x y� d

Figure 11. Solid region of integration defined by cone and plane z= 3 The intersection of the cone and the plane z = 3 corresponds to the level surface for the cone at z = 3. Thus we have:

23 2x y � or 2 2 9x y� For values of z between 0 and 3 we will get similar level surfaces. We can view the solid region of integration as just built from the areas of the ‘stacked’ circles of radii between 0 and 3. The top circle will have radius of 3 and therefore the limits of integration for x and y are between -3 and 3 corresponding to the circular region in the xy-plane as shown in Figure 11. However for any particular circle, its radius will be z corresponding to the equation of the level surface 2z x y � 2 . To get the area of the circle with

radius z we let y vary from -z to z with x given by 2z y2r � (the plus sign corresponding to the area of the upper half of the circle while the minus

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sign corresponds to the area of the lower half of the circle). Thus as z varies from 0 to 3, y varies from –z to z and x from 2z y2� � to 2 2z y� . Therefore the limits of integration are: z – between 0 and 3 y – between –z and z x – between 2z y� � 2 and 2 2z y� . Hence the triple integral is:

2 2

2 2

3

0

z z y

z z yW

fdV z dx dy dz�

� � �

§ ·§ · ¨ ¸¨ ¸

© ¹© ¹³ ³ ³ ³

We can evaluate the integral by integrating over these limits:

� �

2 22 2

2 22 2

3 3

0 0

3 32 2 2 2 3 1

0 0

3 43 3

00

2 s

814 4

z yz z y z

z z y z z yWz

z

zz

fdV z dx dy dz xz dy dz

yz z y dy dz zy z y z dz

z

zz dzS S S

��

� � � � � �

�

��

§ ·§ · § ·§ · ¨ ¸¨ ¸ ¨ ¸¨ ¸ © ¹© ¹© ¹© ¹§ ·§ · � � �¨ ¸¨ ¸

© ¹© ¹

³ ³ ³ ³ ³ ³

³ ³ ³

³

in

We can also compute this integral in reverse order i.e. with the inner integral over z and the outer integral over x. To do this we note, that for a particular x and y, z can vary between 2z x y � 2 and z = 3. Furthermore since the allowed values of x an y must lie in the circular region in

the xy-plane, for a particular x, y can vary between

2 2 9x y� d29 x� � and 29 x� .

Finally x can vary between -3 and 3. Thus we get:

2

2 2 2

3 9 3

3 9

x

x x yW

fdV zdzdydx�

� � � � ³ ³ ³ ³

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We can integrate to get the value of the integral:

� �

� �

� �

2 2 2

2 2 2 2 22 2

2

2

33 9 3 3 9 3 92 2

3 9 3 9 3 9

393 2

39

33 2 23

32 2 1

3

1 1 92 2

1 92 3

2 932 24345 2 9 sin3 8 8 32 243 813 8 4

x x x

x x y x xx yW

x

x

2fdV zdzdydx z dydx x y dydx

yx y dx

x dx

x xx x

S S

� � �

� � � � � � � � � ��

�

��

�

�

�

�

� �

�

§ · � � � ¨ ¸© ¹

� �

³ ³ ³ ³ ³ ³ ³ ³

³

³

�

Finally we summarize our findings from these examples:

x The limits for the outer integral are constants. x The limits for the middle integral can involve only one variable (that in

the outer integral) x The limits for the inner integral can involve two variables (those on

the two outer integrals). In essence, all the results for the double integral apply to the triple integral with the appropriate generalizations resulting from having three variables as opposed to two variables. ====================================================================== Polar Coordinates Polar coordinates provide and alternate (from Cartesian coordinates) reference-coordinate system for expressing the location of objects and points in the plane. Cartesian coordinates give the location of a point

relative to a defined origin on an equally spaced rectangular grid. We thus write pG

� , �p x y G where x and y are the rectangular (Cartesian) coordinates of

the point. We can also express this same location as the distance r from the

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origin and a direction as given by an angle T measured relative to the x-axis, as shown in Figure 12.

y

(x, y) = (r, T)

x

y

T�

r

x

Figure 12. Cartesian and Polar Coordinates The two coordinates, (x, y) and (r, T) are equivalent and given one pair, the other pair can be computed. Hence, we note from Figure 12 that:

� ��

cos

sin

x r

y r

T

T

Using some basic trigonometry, we have:

2 2

1tan

r x y

yx

T �

�

§ · ¨ ¸© ¹

The choice of coordinate system is a matter of convenience since both systems are equivalent. However, it is often the case, that (depending on the particulars of a problem or situation), one system is preferred over the other due to the resulting simplifications in the problem or because the problem is given or stated using a particular system.

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In the case where the preferred system is the polar system, we would like to be able to work entirely in that system. In particular, if we need to perform an integration we would like to be able to integrate in polar coordinates as opposed to converting to Cartesian coordinates, integrating, and then converting back to polar coordinates. Now in Cartesian coordinates, the elemental area ĄA is given by Ąx Ąy i.e. a small rectangle of length Ąx and height Ąy. In polar coordinates, the elemental area must be expressed in terms of Ąr and ĄǄ. However the shape of this area is not rectangular, but annular, as shown in Figure 13. y

'T�

�U'T�'r�

'r

x

Figure 13. Elemental area in polar coordinates If we consider a small annular area in polar coordinates of extent Ąr and ĄǄ, then if Ąr and ĄǄ are small enough the annulus is approximately rectangular. The width of this rectangle is rĄǄ and its height is Ąr. Therefore the area is:

ĄA = rĄǄĄr Thus in polar coordinates the integral over the region R is:

� �,b

aR

fdA f r r dr dE

DT T ³ ³ ³

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Example 4. Compute the area of a circle of radius 1 using Cartesian and polar coordinates. In Cartesian coordinates the region bounding the circle is given by the equation with both x and y varying (not independently) between -1 and 1. If we allow x to vary between -1 and 1 then y can vary between

2 2 1x y�

21 xr � . Thus we have:

� �

2 2

22

1 1 1 11 2

1 1 1 111

2 1

1

2 1

1 sin

x x

x xA dydx ydx x dx

x x x S

� �

� � � � ��

�

�

� �

³ ³ ³ ³ �

In polar coordinates r varies between 0 and 1 and Ǆ between 0 and 2�. Therefore:

1 2 22 1 2 2

0 0 0 0 00

1 12 2 2r

r dr d d dSS S S

T T T T S ³ ³ ³ ³

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Lecture 12 Sections 16.5 and 16.6 – Integrals in Spherical & Cylindrical Coordinates; Probability Cylindrical and Spherical Coordinates As with polar coordinates (see previous lecture) in 2-space, cylindrical and spherical coordinates provide two alternatives to Cartesian coordinates in 3-space. These coordinate systems are equivalent to Cartesian coordinates but can provide significant simplifications in some problems especially where distance and angle provide the primary or preferred method for describing the location of objects and points in 3-space. Perhaps the best example of this the latitude and longitude coordinate system used for locating aircraft, ships, satellites, and other objects on or above the surface of the earth. Latitude and longitude are essentially part of a spherical coordinate system used for identifying the locations of objects above or on the surface of the earth. Just as with polar coordinates, it is desirable to perform integrations directly in these coordinates as opposed to converting to Cartesian coordinates, integrating, and converting back to either cylindrical or spherical coordinates. In order to do this we must express the elemental volume ĄV in cylindrical or spherical coordinates so that we can integrate directly in these coordinate systems. Before we do this, we describe both these coordinate systems and their relationship to Cartesian coordinates. Figure 14 shows the location of a point in both cylindrical and Cartesian coordinates. In Cartesian coordinates the location of

pG

pG is given by rectangular coordinates (x, y, z)

which specify with distances from the origin along each of the three coordinate axes. In cylindrical coordinates, the location is specified by given the distance ǋ in the xy-plane to the place directly below the point. The direction or angle Ǐ measured from the x-axis to this place is used to identify the location below the point in the xy-plane. Finally the Cartesian z-coordinate is used to identify the ‘height’, h, above the xy-plane where the point is located.

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z

x

y

x

(x, y, z) = (ǋ, Ǐ, h)

z

ǋ h Ǐ

y

Figure 14: Cylindrical Coordinates From Figure 14 we see that the relationship between the cylindrical and Cartesian coordinates is as follows:

� ��

cos

sin

x

y

z h

U \

U \


2 2

1tan

x y

yx

h z

U

\ �

�

§ · ¨ ¸© ¹

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In spherical coordinates the distance r from the origin to the point is used along with the azimuth angle and the elevation angle. The azimuth angle, Ǆ, is measured from the x-axis to the line connecting the origin to the place in the xy-plane directly below where the point is located. The elevation angle, Ĺ, is measured from the z-axis along the line connecting the origin to the point. Figure 15 shows the location of a point in both spherical and Cartesian coordinates. z

x

y

x

(x, y, z) = (r, Ǆ, Ĺ)

z

r Ĺ

Ǆ

y

Figure 15: Spherical Coordinates From Figure 15 we see that the relationship between the spherical and Cartesian coordinates is as follows:

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� � � ��

sin cos

sin sin

cos

x r

y r

z r

M T

M T

M


2 2 2

1

2 21

tan

tan

r x y z

yx

x yz

T

M

�

�

� �

§ · ¨ ¸© ¹§ ·�¨ ¸ ¨ ¸© ¹

With a bit more trigonometry we can show that cylindrical and spherical coordinates are related as follows:

� �

� �

sin

cos

r

h r

U M\ T

M

Or:

2 2

1tan

r h

h

UT \

UM �

�

§ · ¨ ¸© ¹

The elemental volume in cylindrical coordinates is similar to the elemental area in polar coordinates. It resembles a pie-shaped wedge with the ‘point’ removed. Alternatively we can view it as the annulus in polar coordinates with ‘height’ added to make it three-dimensional. Figure 16 shows the elemental cylindrical volume.

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z

x

ĄV = (ǋ ĄǏ) Ąǋ Ąh

Ą ǏĄǋ

ǋ

Ąh

y

Figure 16: Cylindrical Elemental Volume If Ąǋ, Ąh, and Ą Ǐ are sufficiently small we can view the cylindrical elemental volume as a box of height Ąh, length Ąǋ and width ǋ Ą Ǐ. The elemental volume is therefore ĄV = (ǋ Ą Ǐ) Ąǋ Ąh. The integral over the solid region W in cylindrical coordinates is:

� �, ,d b

c aW

fdV f h d d dhE

DU \ U U \ ³ ³ ³ ³

The elemental volume in spherical coordinates is shown in Figure 17. In this case, two angular components define this small solid region. As in the previous cases if Ąr, ĄǄ, and ĄĹ are small enough, then we can regard this volume as approximately box-shaped. The height of the box is rĄĹ, the width rsin(Ĺ)ĄǄ, and the length r. The elemental volume is therefore:

ĄV = r2 sin(Ĺ) Ąr ĄǄ ĄĹ

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The integral over the solid region W in spherical coordinates is:

� � � �2, , sinb

aW

fdV f r r dr d dE \

D JT M M T ³ ³ ³ ³ M

y

z

ĄV = (r sin(Ĺ)ĄǄ) Ąr (rĄĹ)

x

rĄ ĹĄr

r rsin(Ĺ)Ą Ǆ

Ĺ

r sin(Ĺ)

Ǆ

Figure 17: Spherical Elemental Volume Example 1. Find the volume of a hemisphere of radius a. Use cylindrical coordinates to perform the integration. Figure 18 shows the solid region of integration. From Figure 18 we see that the polar radius, ǋ, is a function of h, the distance from the xy-plane. The polar angle Ǐ must sweep out a circle to cover the hemisphere.

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z

x

h

ǋ

y

Figure 18: Hemisphere Volume For Integration Thus the limits of integration are: h – 0 to a Ǐ – 0 to 2� ǋ – 0 to 2 2a h� The volume is therefore:

2 2

2 2

2

0 0 0

22 2 2 2

0 0 0 00

3 32 22

0 00

3 2 3

0

12 2

12 3 3

23 3

a a h

W

a ha a

a

fdV d dhd

dhd a h dhd

h aa h d d

aa

S

S S

S S

S

U U \

U \ \

\ \

S\

�

�

�

³ ³ ³ ³

³ ³ ³ ³

³ ³

�

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Example 2. Find the volume of a hemisphere of radius a. Use spherical coordinates to perform the integration. In this case, spherical coordinates are the ‘natural’ system to use. The limits of integration follow directly from the coordinate system: r – 0 to a Ǆ – 0 to 2� Ĺ – �/2 to � We therefore have:

� �

� � � �

� �

2 2

0 02

3 32 2

0 02 20

3 32 2

0 02

3 2 3

0

sin

sin sin3 3

cos3 3

23 3

a

Wa

fdV r drd d

ad d d d

a ad d

aa

S SS

S S S SS S

SS S

S

S

M M T

U M M T M M T

M T T

ST

�

³ ³ ³ ³

³ ³ ³ ³

³ ³

====================================================================== Probability Probability is the likelihood or chance that a particular event will occur. The word event is used in a broad and general sense to include just about any situation, experience, or incident of interest. The only restriction placed on an event is that it be random. That is, any specific single occurrence of the event is unpredictable. The probability, P, that an event, E, will occur is defined as:

P(E) = N(E)/T where:

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N(E) - is the number of ways event E can occur; T - is the total number of possible outcomes (of all possible events of a

certain type or class and which includes the event E). Classical examples of probabilities involve games of chance: coin-flipping, dice, cards, etc. Consider the following examples:

1. Coin-flipping - the probability of a flipped coin coming up heads. Since there is only one way for this event to occur, N(E) = 1. There are exactly two possible outcomes: heads or tails. Thus P(E) = ½.

2. Dice – the probability of rolling a one or a two. In this case, there are exactly two possible ways for this event to occur i.e. rolling a one or rolling a two. There are six possible outcomes (rolling a one, two, three, four, five, or six). Thus P(E) = 2/6 = ൈ.

3. Cards – the probability of drawing a face card (Jack, Queen, or King). Since there are four suits (heart, diamonds, spades, clubs) and 3 face cards in each suit, there are 12 ways of drawing a face card. The total number of possible outcomes is 52 (the number of cards in a standard deck), therefore P(E) = 12/52 = 3/13.

Although there is great interest in determining the likely outcomes for games of chance (think Las Vegas), probability is used widely across many fields: physics, medicine, insurance, and finance, to name just a few. The medical and insurance industries for example, collect data on disease (cardiovascular, cancer, etc) and mortality (due to accidents, suicide, murder, etc) in the general population and use this information to predict the likely outcomes of these ‘events’ e.g. having a heart attack, being in an automobile accident, etc.

�The medical industry may use this information to concentrate its efforts on development of new drugs, education, or treatment of disease in the general population, while the insurance industry may use this information to set policy rates. Often data is not just collected for the population in general, but is categorized by one or more parameters that characterize the population. For example, the insurance industry may collect data on the number of automobile accidents. These may be categorized by driver age, gender,

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income, and the seriousness of the accident i.e. fatal, serious injuries, minor injuries, no injuries. By doing this, the insurance industry can estimate the likelihood that say, a 16 old male driver will be involved in a fatal accident versus the likelihood that a 35 year old woman will be in such an accident. It can then set rates accordingly. The table below gives accident data for 2002 categorized by driver age and Blood Alcohol Content (BAC).

Alcohol and Accidents*

Driver Age BAC < 0.08 g/dl BAC > 0.08 g/dl

16-20 11.9% 2.4% 21-24 7.4% 3.7% 25-34 14.5% 5.7% 35-44 14.3% 5.0% 45-64 19.9% 4.1% 64 > 10.5% 0.6%

*Data taken from Traffic Safety Facts 2002 Alcohol, US Department of Transportation National Highway Safety Administration, www.nhtsa.dot.gov

�Figure 19 show a histogram plot of the data in the table. It should be noted that the sum of the data in the table is 100% and it therefore includes all fatal accidents for the year 2002. �

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http://www.nhtsa.dot.gov/


Figure 19. Histogram plot of Accident data If we view the boxes of the histogram in Figure 19 as the boxes in a Riemann sum, with Ąx = age difference, Ąy = BAC difference, and p(x, y) = percent accidents for age x and BAC y, then the histogram is an approximation to the volume of accidents in 2002. In fact, we could think of collecting this data with at finer granularity, say partitioning the ages into one year intervals and BAC into units of 0.01 g/dl. If we did this, then the boxes in Figure 19 would have smaller footprints in the xy-plane i.e. ĄA = Ąx Ąy would be smaller and there would be many more boxes in the graph. We could theoretically keep decreasing the granularity by partitioning the ages by months, then days and the BAC at thousandths and then ten-thousandths g/dl. The result would be a closer and closer approximation to a smooth and continuous function p(x, y) giving the percent of accidents for each age and BAC. In the limit where x and y become continuous, p(x, y) would be a continuous and smooth function giving the percent of accidents of a particular age and BAC. However since the total number of accidents must remain at 100%, the integral (instead of the sum, now) of p(x, y) over all x and y must still be 100%. We also note that p(x, y) � 0 for all values of x and y. That is, p(x, y) cannot be negative since it makes no sense to have, say, -12% accidents at a particular age and BAC. Functions that are non-negative everywhere and whose integrals over their domains equal one (or 100%) are called probability density functions. That is, such functions can be used to represent or model the likelihood that an event (defined by one or more parameters i.e. variables) will occur. If probability density is a function of two or more variables, this it is called a joint density function. Formally we make the following definition: A function p(x, y) is called a joint density function of x and y if:

� �, 1p x y dxdyf f

�f �f ³ ³ and p(x, y) � 0 for all x and y

We note that the ±� limits on the integral represent (i.e. are placeholders for) the valid ranges for the variables x and y. These may (or may not) be infinity.

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The product p(x,y)ĄxĄy is the probability that x and y will occur. For example in the table above, the probability that an accident will involve a driver between the ages of 45 to 64 with a BAC greater than 0.08, is 4.1%. In this case, x = 45, y = 0.08, Ąx = 64 – 45 = 19, Ąy = 0.08, and p(x, y) = 12/( 9 ș 0.08) = 16 %/(year-g/dl). In terms of the integral, the limits determine the probability for a range of events. Thus for a generic density function, we write: The probability that x is between a and b and y is between c and d is:

� � �, ,d b

c aP a x b c y d p x y dxdyd d d d ³ ³ �

In the accident data, we would write that the probability that an accident involved a driver between the ages of 45 and 64 with a BAC between 0.08 and 1.0 is:

� �1.0 64

0.08 4545 64,0.08 1.0 ,P x y p x y dxdyd d d d ³ ³ � �

Assuming that p(x, y) could be deduced from the data and represented as a continuous, smooth function we could compute the integral and obtain this probability. In practice, a continuous and smooth function is usually fitted to the data since it is not practical to measure or obtain data at extremely fine granularity. If the data is of sufficient quality, it will typically be well represented by a continuous and smooth function. Example 3. Let p(x, y) = x + y on the region 0 � x � 1, 0 � y � 1. Verify that p is a probability density function. To show that p is a probability density function we must show that

� �, 1p x y dxdyf f

�f �f ³ ³ and p(x, y) � 0 for all x and y. Since x � 0 and y � 0,

then x + y � 0 and hence p(x, y) � 0.

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� � � �1 21 1 1

0 0 00

1 21

00

,2

1 1 12 2 2

xp x y dxdy x y dxdy xydy

yy dy y

f f

�f �f � �

§ · � � ¨ ¸© ¹

³ ³ ³ ³ ³

³

Thus p is a probability density function. Example 4. Find the probability that x � ½ and the probability that x � ½ and y � ½, for the density function of Example 3.

� �11 21 1 22

0 0 00

1 21

00

12 2

1 1 18 2 8 4 8 4 8

xP x x y dxdy xydy

y ydy y

§ ·d � �¨ ¸© ¹

§ · � � � ¨ ¸© ¹

³ ³ ³

³1 3

� �11 1 1 222 2 2

0 0 00

11 222

00

1 1,2 2 2

1 1 18 2 8 4 16 16 8

xP x y x y dxdy xydy

y ydy y

§ ·d d � �¨ ¸© ¹

§ · � � � ¨ ¸© ¹

³ ³ ³

³1 1

Example 5. A machine factory produces components 10 cm long and 5 cm in diameter. To be useable i.e. pass quality control, the length and diameter must be within a 0.1 cm tolerance of these values. The probability density

function � � � � � �2100 10 50 550 2, x yp x y e e2

S� � � � gives the likelihood that a component

of length x and diameter y will result as a variation in the production process. What is the probability that a component is useable? To be useable, the component must have length between 9.9 and 10.1 cm and diameter between 4.9 and 5.1 cm. Therefore:

� � � � � �

� �

2 25.1 10.1 100 10 50 5

4.9 9.9

50 29.9 10.1,4.9 5.1

50 2 0.02556 0.57530

x yP x y e e dxS

dy

S

� � � �d d d d

| |

³ ³

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Note: the integral cannot be solved in closed form. The result was found by using tabulated numerical results. The probability that the component will be useable is approximately 57.5%.

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Lecture 13 Sections 16.7 and 17.1 – Change of Variables & Parameterized Curves Change of Variable In single variable calculus, it is often the case that changing the variable of integration will simplify the evaluation of an integral. In fact, we have already looked at changing variables during integration when we discussed polar, cylindrical, and spherical coordinates. Instead of viewing integration in, say spherical coordinates, as a change in variable we looked at it as a change in coordinates. We can also view this change in coordinates as a change in the variables of the integration. The two views are equivalent, since changing the variable of integration is equivalent to a change in coordinate system and vise versa. Now in single variable calculus, changing the variable of integration involved three steps:

x Substituting the new variable into the function being integrated; x Computing the new differential and making the appropriate

substitution into the integral; x Changing the limits of integration to reflect the equivalent limits in

terms of the new variable. This procedure is essentially identical for integrals involving multivariable functions, except that the second step, computing the new differential, is a bit more involved since there are multiple differentials and associated variables in the integral. Let’s look at a single variable example to illustrate this procedure. Suppose we wish evaluate the following integral:

� �220

cosx x dS

S�³ x

This integral can be readily evaluated by changing variables. Let t = x2 + �. The first step is to substitute the new variable into the integral in the function cos(x2 + �). Thus we have:

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� � � �22 20 0

cos cosx x dx x tS S

S� o³ ³ dx

The next step is to compute the differential dt in terms of dx and substitute. Since t = x2 + �, we have dt = 2xdx. Making the substitution we have:

� � � � � �22 20 0 0

1cos cos cos2

2x x dx x t dx tS S

S� o o³ ³ ³ dtS

Finally we need to change the limits of integration to reflect the new variable. When x = 0, t = � and when x = (�/2)½, t = 3�/2. Thus we have:

� � � �3

2 20 0

1cos cos2

x x dx tSS

S� ³ ³ dt

The integral can now be evaluated and we obtain:

� � � � � �3322 2

0 00

1 1cos cos sin2 2

x x dx t dt tSSS

S� �³ ³12

We can generalize this change of variable for any function as follows. Let f(x) be function we wish to integrate on the interval [a, b] and let x = x(t) be a change in variable we wish to make for x. Then the differential is

dxdx dt

dt and we have:

� � � ��

� �b t x b

a t x a

dxf x dx f x t dt

dt

³ ³

Note that the change of variable introduces the derivative of x with respect to t into the integral as part of the substitution for the differential, dx. In the case of the example above, the function � �x t t S � , takes the interval [0, (�/2)½] and maps it into another interval [0, 3�/2]. However the mapped interval is now curved instead of linear. In essence, the change of variable has deformed the original interval into a new interval and is characterized by the mapping defined by the change of variable, that is in the case of our

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example, the mapping � �x t t S � . We can also view a change of variable as a change in coordinates. If we recall the transformation from Cartesian coordinates to polar coordinates, we encountered exactly this type of deformation as our elemental rectangular area Ä = ¨xÿ was deformed into the annulus r¨Ǆ¨r. Let’s take a closer look at this process. Consider the function f(x, y) = x + y over the region, R, bounded by the x and y axes and the circle x2 + y2 = 16. We are in essence integrating over the quarter circle in the first quadrant of the Cartesian coordinate system, as shown in Figure 20. y

Figure 20: Region of integration in Cartesian plane

x2 + y2 = 16

R

4

4 x

Our integral in Cartesian coordinates is:

� � � �24 16

0 0

x

R

x y da x y dxdy�

� �³ ³ ³

If we decide to use polar coordinates instead, then we get:

� � � � � � � �� 24 16 4

20 0 0 0

cos sinx

R

x y da x y dxdy r r rdrdS

T T T�

� � �³ ³ ³ ³ ³

In the rǄ-plane the region of integration is rectangular, as shown in Figure 21.

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Ǆ

�U��/2�

4

7�

�/2 r

Figure 21: Region of integration in Polar plane Thus by changing coordinates we have deformed the rectangular region of integration into a quarter circular region (see Figure 22). The change of coordinates from Cartesian to polar defines a mapping of a region R in the Cartesian plane to a region T in the polar plane and vise versa. We therefore suspect that we can also view the change in coordinates from Cartesian to polar as a change in variable.

x2 + y2 = 16

y

R

4

Ǆ

r

�U��/2�

x(r, Ǆ) =rcos(Ǆ) y(r, Ǆ) =rsin(Ǆ)

�/2

4

7�

4

x

Figure 22: Mapping from Cartesian to Polar Coordinates

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To confirm our suspicions let’s examine this transformation as a change in variable. We have:

x = x(r, Ǆ) = rcos(Ǆ) and y = y(r, Ǆ) = rsin(Ǆ) For a small region in the rǄ-plane, say a rectangle of width ¨r and height ¨Ǆ, this region will map into the Cartesian plane according the transformations x(r, Ǆ) and y(r, Ǆ). Let us consider the mapping of the four corner of this region: (r, Ǆ) ĺ ( rcos(Ǆ), rsin(Ǆ) ) (r+ ¨r, Ǆ) ĺ ( (r + ¨r) cos(Ǆ), (r + ¨r )sin(Ǆ) ) (r+ ¨r, Ǆ + ¨Ǆ) ĺ ( (r + ¨r) cos(Ǆ+¨Ǆ), (r + ¨r ) sin(Ǆ+¨Ǆ) ) (r, Ǆ + ¨Ǆ) ĺ ( rcos(Ǆ+¨Ǆ), rsin(Ǆ+¨Ǆ) ) The vector along the bottom edge of this region is: (¨r, 0) ĺ (r+ ¨r, Ǆ) - (r, Ǆ) = ( (r + ¨r) cos(Ǆ), (r + ¨r )sin(Ǆ) ) - ( rcos(Ǆ), rsin(Ǆ) ) = ( ¨r cos(Ǆ), ¨r sin(Ǆ) ) And the vector along the left edge of this region is: (0, ¨Ǆ) ĺ (r, Ǆ + ¨Ǆ) - (r, Ǆ) = ( rcos(Ǆ+¨Ǆ), rsin(Ǆ+¨Ǆ) ) - ( rcos(Ǆ), rsin(Ǆ) ) = ( r (cos(Ǆ+¨Ǆ) - cos(Ǆ)), r(sin(Ǆ+¨Ǆ) - sin(Ǆ)) ) = ( r¨Ǆ (cos(Ǆ+¨Ǆ) - cos(Ǆ))/¨Ǆ, r¨Ǆ (sin(Ǆ+¨Ǆ) - sin(Ǆ))/¨Ǆ ) = ( -r¨Ǆ sin(Ǆ), r¨Ǆ cos(Ǆ) ) The area of this region in the xy-plane is:

||(¨r cos(Ǆ), ¨r sin(Ǆ)) ș(-r¨Ǆ sin(Ǆ), r¨Ǆcos(Ǆ))|| = r¨Ǆ¨r = ¨xÿ

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Where we have approximated the region in the xy-plane as a parallelogram with sides defined by the vectors in the cross product above. Thus we have that:

f(x,y) ¨xÿ ĺ f( x(r, Ǆ),y(r, Ǆ) ) r¨Ǆ¨r = f(rcos(Ǆ), rsin(Ǆ) ) ) r¨Ǆ¨r and hence our integral becomes:

� � � �, ,T

R

f x y dxdy f r rdrdT T ³ ³

Thus we can regard the change from Cartesian to polar coordinates as a change in variable from x and y to r and Ǆ. In general, we would like to be able to change the variables x and y to some arbitrary variables, say u and v:

x = x(u, v) and y = y(u, v) and do so using a procedure similar to that for the single variable case. We will do this by repeating the above argument for the general case. Consider a rectangular region in the uv-plane corresponding to some, in general, non-rectangular region in the xy-plane, similar to what we found for polar coordinates. If we consider a small region of the uv-plane, say a rectangle of width ü and height ¨v, then the corners of these regions will be as shown in Figure 23, where the region in the xy-plane is approximated by a parallelogram. The parallelogram will be a good approximation if ü and ü are small enough and our functions are continuous and smooth.

x

( x(u+ü,v), y(u+ü,v) )

( x(u+ü,v+¨v), y(u+ü,v+¨v) )

R ( x(u,v), y(u,v) )

( x(u,v+¨v), y(u,v+¨v) )

aG

bG

y

u

�X�ǻu��Y� �X��Y�

�X�ǻu��Y� ǻv� �X��Y�ǻv�

T

v

x(u, v) y(u,v)

Figure 23: Mapping resulting from general variable change

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Now in the uv-pane the area of the elemental region is just Ä = ü¨v. In the xy-plane, the area of the elemental region is the area of the parallelogram. This area (from previous lectures) is just a bu

GG where a and

are the displaced vectors defining the sides of the elemental parallelogram region. Now:

G

bG

� � � ��

ˆ ˆ, , , ,

, , , ,ˆ ˆ

ˆ ˆ

a x u u v x u v i y u u v y u v j

x u u v x u v y u u v y u vui uj

u ux yui uj

u u

� ' � � � ' �

� ' � � ' � ' �

' 'w w

' � 'w w

G

'

� � � ��

ˆ ˆ, , , ,

, , , ,ˆ ˆ

ˆ ˆ

b x u v v x u v i y u v v y u v j

x u v v x u v y u v v y u vvi vj

v vx yvi vj

v v

� ' � � � ' �

� ' � � ' � ' �

' 'w w

' � 'w w

G

'

Thus:

� ��

,,

x y y xA a b u v u v

u v u v

x y y xu v

u v u v

x xx yu v u v u v

y y u vu v

w w w w§ ·§ · § ·§' u ' ' � ' '¨ ¸¨ ¸ ¨ ¸¨w w w w© ¹© ¹ © ¹©

w w w w§ · � ' '¨ ¸w w w w© ¹

w www w ' ' ' '

w w ww w

GG ·¸¹

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We have introduced some matrix notation here, with

x xu vy yu v

w ww ww ww w

being the

determinant of matrix formed by the partials. In addition, the short-hand

notation for this determinant is � ��

,,x yu v

ww

and is called the Jacobian.

Thus under a change of variable the elemental area Ä is transformed a follows:

� ��

,,

x xx yu vA x y u v u

y y u vu v

w www w' ' ' ' ' ' '

w w ww w

v

Carl Gustav Jacob Jacobi (1804 - 1851) If we make these substitutions into our integral then:

� � � ��

,( , ) , , ,

,R T

x yf x y dxdy f x u v y u v dudv

u vw

w³ ³

To convert an integral from x, y to u, v coordinates/variables we make the following three changes: x Substitute for x and y in the integral in terms of u and v; x Compute the Jacobian and use its absolute value to make the appropriate

change in the differential area i.e. � ��

,,x y

dxdy dudvu v

wo

w

x Change the region of integration to reflect the equivalent limits for the new variables

We note that this procedure is essentially identical to that of the single variable case, except the change in the differential now involves the Jacobian instead of just the derivative of old variable with respect to the new.

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Finally we note that this exact same procedure applies to integrals involving three variables except that the Jacobian is given by:

� ��

, ,, ,

x x xu v w

x y z y y yu v w u v w

z z zu v w

w w ww w w

w w w w

w w w ww w ww w w

Example 1. Compute the Jacobian for polar coordinates.

� ��

� � � �� 2 2cos sin,

cos sinsin cos,

x xrx y r r rry yr

r

T TT T TT TT

T

w w�w w w �

w www w

r

Example 2. Compute the Jacobian for spherical coordinates.

� ��

� � � � � � � � � � � ��

� � � �

� � � � � � � � � � � � � � � � � � � ��

cos sin sin sin cos cos, ,

sin sin cos sin sin cos, ,

cos 0 sin

cos sin sin sin cos cos cos sin sin sinsin sin cos sin sin cos sin sin cos s

x x xr r r

x y z y y yr r

r rr

z z zr

r r rr r r

T M T M T M T MT M T M T M

T M T MM M

T M

T M T M T M T M TT M T M T M T M T

w w ww w w �

w w w w

w w w w�

w w ww w w

� �

M� �

� � � � � ��

� �

2 2 3 2 2 2 2 2 3 2 2 2

2 2 2 2

2

cos sin cos sin sin sin sin cos sin cos

incos 0 sin cos 0

cos sin sin sin

sin

r r r r

r

r r

r

T M M M T T M T M M

MM M M

T M T M

M

� �

�

� �

� �

�

Example 3. Find the volume of the ellipsoid 2 2 2

2 2 2 1x y za b c

� � .

Let x = au, y = bv, and z = cw. Then the Jacobian becomes:

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� ��

0 0, ,

0 0, ,

0 0

x x xu v w a

x y z y y yb a

u v w u v wcz z z

u v w

w w ww w w

w w w w

w w w ww w ww w w

bc

Thus:

W Sdxdydz abcdu dvdw ³ ³

Where W is the solid region defined by the ellipse and S is the solid region defined by the sphere since in uvw-coordinates we have u2 + v2 + w2 = 1. Thus

we have 43W S S

dxdydz abc du dv dw abc du dv dw abcS ³ ³ ³ .

====================================================================== Parameterized Curves Consider a smooth and continuous function of a single variable, f(x). If we consider the graph of this function y = f(x), then we will obtain a smooth and continuous curve in the xy-plane, as shown in Figure 24. y

(x, y)

x Figure 24. Point moving on a curve in the xy-plane

If we consider an arbitrary point (x, y) on the curve, then we can view the curve (graph of f(x, y)) as the path that this point will take if it were to

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move subject to the constraint that it stay on the curve. In essence, the curve is the path this point will follow or trace out if y = f(x). An example of this is car on a road. The car is our point and road is the curve or path that the car takes as it moves. Now with the car, we are not particularly interested in knowing what y-coordinate the car is at for a given x-coordinate, rather we could like to know where the car is i.e. both x and y coordinates at a particular time, t. Mathematically, both x and y are functions of the time t, that is x = x(t) and y = y(t). As the car moves, it traces out a path defined by the road, and at a particular time, t, it is located at (x(t), y(t)). To make our example a little more concrete, say you leave school at noon and start driving home. Five minutes later you stop in front of dry cleaners to pick up some clothes. If we consider t = 0 to correspond to noon, when you left school, then at t = 5 (x(5), y(5)) will be the location on the road where the dry cleaners is located. As you continue on your way, we could pole your location every five minutes to see where you are: passing the grocery store, the park, the public library, etc. Each of these places lies on the curve that defines that path you taking, as shown in Figure 25.

y ( x(25), y(25) )

( x(5), y(5) )

Dry Cleaners

Home Library

( x(20), y(20) )

( x(15), y(15) )

Park Store

School

( x(10), y(10) ) ( x(0), y(0) )

x

Figure 25. Path from school to Home Note that the points all still lie on the curve defined by our function y = f(x) i.e. the road. However, what we are interested in is not so much the relationship between x and y, but rather the path (x(t), y(t)) that a point on this curve is following. In the case of the road example, the time is the independent variable and x and y are the dependant variables. In effect, the time is the parameter we wish to vary or input, and the location (x and y) is the desired output e.g. where you are (or where you will be) at time t.

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A function viewed in this manner as a path dependent on a parameter (or parameters) is called parameterized curve. Of course, we do not need to restrict ourselves to functions of a single variable and paths or curves need not be confined to 2-space. In fact, in our road example, unless the ground under the road is perfectly flat, the road will rise and fall with the terrain, in effect defining a path in 3-space. Think about a road winding though the mountains. In fact, it is often the case, that paths or curves in 3-pace are completely defined by their parameterization. That is, the function relating x, y, and z together is not particularly interesting. Rather the parameterized curve that defines the path in 3-space (or even 2-space) is what is primarily of interest. By way of example, consider the parameterized curve defined by x(t) = cos(t), y(t) = sin(t), and z(t) = t. This curve is known as a helix and is show in Figure 26. A point moving on this curve follows a rising circular path around the z-axis.

Figure 26: Helix Perhaps the simplest curve in both two and three space is a line. What is the parameterized equation for a line in 3-space? Well consider at point

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( , , )p x y z G passing through or starting at a point � �, ,o o o op x y z

G and traveling in a straight line. To travel in a straight line, the point must move in some fixed direction through and way from op

G . Now a fixed direction is given by a vector, v which defines an orientation relative to a coordinate system. A point moving in this direction will move a distance equal to

G

vG in some unit of

time. If it starts at opG then at time t, the point will be located at:

op p v �G G Gt at time t later. Thus we have the following result:

The parametric equation of a line through the point � �, ,o o o op x y z

G and parallel to the vector (or in the direction of the vector) � �, ,v a b c

G is:

� �, ,o o o op p vt x at y bt z ct � � � �G G G

Or

x(t) = xo + at, y(t) = yo + bt, z(t) = zo + ct Example 4. Find the parameterized equation of the line through the points (1, 2, -1) and (3, 3, 4). We use the points to compute the displaced vector giving the direction between the points:

� � � � �3,3,4 1,2, 1 2,1,5v � � G � The parameterized equation of the line is:

x(t) = 2t + 1, y(t) = t + 2, z(t) = 5t – 1 Now let’s consider the general parameterized curve. Any point in the Cartesian plane is defined by its position vector:

ˆˆ ˆr xi yj zk � �G

If x, y, and z are functions of a parameter t, then we have:

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x = f(t), y = g(t), and z = h(t) and we can write these as a single parametric equation in vector form called a parameterization:

� � � � � � � � ˆˆ ˆr t f t i g t j h t k � �G

As the parameter t varies, the vector � �r t

G traces out a path or curve in 3-space. For example the parametric equation of a line is � � or t p vt �

G G G . Thus: The line through the point � �, ,o o o op x y z

G in the direction of the vector � �, ,v a b c

G is:

� � or t p vt �G G G

We can use the parametric equations for a curve to find where a curve intersects a surface or another curve. Consider the following examples. Example 5. Find the places where the line x = t, y = 2t, z = t + 1, intersects the surface of the sphere x2 + y2 + z2 = 100. We are looking for the values of t where t2 + 4t2 + (t+1)2 = 100. Or: t2 + 4t2 + t2 + 2t + 1 = 100 6 t2 + 2t – 99 = 0

� �� 2 4 4 6 99 1 1 6 99 1 59512 6 6

t� r � � r � � r

Thus the points of intersection are t § -4.23 and t § 3.90. Example 6. Do the lines x = t - 1, y = 2t + 1, z = 5 – t and x = 2t + 2, y = t + 4, z = t + 3 intersect? For the lines to intersect we must find t1 and t2 such that:

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t1 - 1 = 2 t2 + 2 2t1 + 1 = t2 + 4 5 – t1 = t2 + 3 Multiplying the first equation by 2 and subtracting from the second we get: 3 = -3 t2 or t2 = -1. If we multiply the second equation by 2 and subtract from the first we get -3t1 - 3 = -6 or t1 = 1. For these lines to intersect, these values must also satisfy the third equation:

5 – t1 = t2 + 3 5 – 1 � -1 + 3 4 � 2

Thus the lines do not intersect.

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Lecture 14 Sections 17.2 and 17.3 – Motion, Velocity, Acceleration & Vecto Fields r Motion, Velocity, and Acceleration In our first lecture we talked about velocity and acceleration and showed that they were vectors. That is, they both had a magnitude and a direction associated with them. In particular the velocity of a moving object e.g. a particle has the following properties: The velocity of a moving object is a vector vG such that: x The magnitude of is the speed of the object; vG

x The direction of v is the direction of motion; G

x The speed of the object is vG , and the velocity is tangent to the object’s path

Physically, the velocity is defined as the derivative of the position with respect to time. Since position (as we also noted in the first lecture) is a vector, then the velocity will also be a vector. Recalling from the previous lecture that the position vector ˆˆ ˆr xi yj zk � �

G we have: The velocity of an object is the derivative of the position vector with respect to time:

ˆ ˆˆ ˆ ˆ ˆx y z

dr dx dy dzv i j k v i v j

dt dt dt dt � � � �GG v k

Where vx, vy, and vz are the components of the velocity vector and are equal to time derivative of the components of the position vector.

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Example 1. The position of a particle moving in a circle of radius R is given by where ǐ is the angular velocity (how fast the

particle is rotating in radians per second). That is, the period (how long it

takes the particle to make one revolution is

� � � � � �ˆ ˆcos sinr t R t i R t jZ Z �G

2SZ

. Find the velocity of the

particle, its speed, and show that it is perpendicular to the position. The velocity is :

� � � �

� � � �

ˆ ˆcos sin

ˆ ˆsin cos

dr d dv R t i R

dt dt dtR t i R t j

Z Z

Z Z Z Z

�

� �

GG t j.

The speed of the particle is:

� � � � � � � �2 2 2 2 2 2ˆ ˆsin cos sin cosv R t i R t j R t R t RZ Z Z Z Z Z Z Z Z � � � G

We note that:

� � � �� 2 2

ˆ ˆ ˆsin cos cos sin

sin cos sin cos 0

v r R t i R t j R t i R t j

R t t R t t

Z Z Z Z Z Z

Z Z Z Z Z Z

� � � � �

� �

G G ˆ

Thus the velocity and the position are perpendicular. As we noted above, the velocity is tangent to the position vector. If the position vector is given or known as a parameterized path or curve, then we can compute the tangent to the path at any point using the velocity. Example 2. Let . Find the tangent line at (1, 1, 2). � � 2 3 ˆˆ ˆ 2r t t i t j tk � �

G

First we note that the point (1, 1, 2) lies on the path defined by . In particular .

� �r tG

� �1 (1, 1, 2)r G

To find the tangent line we first compute the velocity:

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� � � � 2 ˆˆ ˆ2 3 2dv t r t ti t j k

dt � �

G G

At t = 1 we have: . Now the tangent line is of the form: � � ˆˆ ˆ1 2 3 2v i j � �

G k

� � � �ˆ ˆˆ ˆ ˆ ˆ2 3 2o o ox i y j z k i j k t� � � � �

and at t = 1 we want:

� � � �ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 3 2 2o o oˆx i y j z k i j k i j k� � � � � � �

Therefore:

� � � �ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ2 2 3 2 2o o oˆx i y j z k i j k i j k i j� � � � � � � � �

Therefore the equation of the tangent line is:

� � � �ˆˆ ˆ ˆ ˆ2 3 2 2i j k t i j� � � �

The acceleration is defined as the derivative of the velocity with respect to time and is equal to the second derivative of the position with respect to time. Since velocity (as we also noted above) is a vector, then the acceleration is a vector. Thus we have: The acceleration, , of an object is the derivative of the velocity with aG

respect to time:

ˆ ˆˆ ˆ ˆ ˆyx zx y z

dvdv dvdva i j k a i a j

dt dt dt dt � � � �GG a k

Where ax, ay, and az are the components of the acceleration vector and are equal to time derivative of the components of the velocity vector. Since the velocity is the derivative of the position we also have:

2 2 2

2 2 2ˆ ˆˆ ˆ ˆ ˆdv d dx dy dz d x d y d z

a i j k i jdt dt dt dt dt dt dt dt

§ · � � � �¨ ¸© ¹

GGk

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Example 3. Show that the acceleration is parallel to the position and perpendicular to the velocity for a particle moving in a circle. Compute the magnitude of the acceleration. From Example 1 we have:


� � � �ˆ ˆsin cosv R t i R tZ Z Z Z � �

G j The acceleration is:

� � � ��

� � � ��

2 2

2

2

ˆ ˆ ˆsin cos cos sin

ˆ ˆcos sin

dv da R t i R t j R t i R

dt dtR t i R t j

r

Z Z Z Z Z Z Z Z

Z Z Z

Z

� � � �

� �

�

GG

G

ˆt j

Since the acceleration is a multiple of the position, these vectors are parallel. Note the minus sign in the above expression means that the acceleration points in the opposite direction of rH , that is towards the center of the circle. Now: (using the results from Example 1). Thus the acceleration and the velocity are perpendicular.

2 0a v r vZ� � � G G G

The magnitude of the acceleration is:

� � � ��

� �

2 2 2

2 22

ˆ ˆcos sin cos sina R t i R t j R t R

R vR

2 2 t

R R

Z Z Z Z Z

ZZ

� � �

G

GZ

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Combining the results of Examples 1 and 3 we have: A particle moving continually in a circle is said to be in uniform circular motion which is describe by:


The radius of the circle is R and the period of the motion is 2SZ

;

The velocity is tangent to the circle and the speed vG is constant and equal to RZ ;

The acceleration points towards the center of the circle with 2v

aR

G

G ;

The velocity is perpendicular to both the position and the acceleration. In uniform circular motion the magnitude of the velocity is constant but its direction is always changing during the motion. For uniform linear motion (i.e. motion in a straight line), the magnitude of the velocity is changing but its direction is constant. The acceleration is parallel to the velocity vector (versus perpendicular in uniform circular motion) and points in the same direction if the particle is speeding up and in the opposite direction if the particle is slowing down. Example 4. Let � � � � � � � �3 3 ˆˆ ˆ2 2 2 1 3 2r t t t i t t j t t k � � � � � � �

G 3 . Compute velocity, acceleration.

� � � � � � � � � � � ��

2 2 2 2ˆ ˆˆ ˆ ˆ4 3 1 3 3 1 3 1 3 1 4 3

ˆˆ ˆ6 4 3

dv t r t t i t j t k t i j

dtd

a t v t t i j kdt

� � � � � � � �

� �

G G

G G

ˆ k

k

We note that the acceleration and velocity vectors are parallel since they are both multiples of the same vector: ˆˆ ˆ4 3i j� � . For t < 0, the acceleration is negative, so the object is slowing down and for t > 0 the acceleration is positive and the object is speeding up.

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We also note that:

� � � � � � � �

� �� 3 3

3

ˆˆ ˆ2 2 2 1 3 2

ˆˆ ˆ ˆ ˆ4 3 2 6

r t t t i t t j t t k

t t i j k i j

� � � � � � �

� � � � �

G 3

k

So that is also a multiple of � �r t

G ˆˆ ˆ4 3i j� � . Thus the motion is linear. A particle/object whose motion is described by:

� � � �o or t r f t v �G G G

is in uniform linear motion i.e. moving in a straight line. This motion has the following characteristics: The motion is along a straight line through the point or

G and parallel to v ; G

The velocity and acceleration are parallel to each other and the vector vG From the definition of velocity we have:

ˆˆ ˆdr dx dy dzv i j

dt dt dt dt � � kGG

And the speed is therefore:

2 2dx dy dzv

dt dt dt§ · § · § · � �¨ ¸ ¨ ¸ ¨ ¸© ¹ © ¹ © ¹

G 2

The distance, d, of the particle along this curve is the integral of the speed:

� �2

1

t

td v t ³ dtG

Since the distance along he curve is equal to the length of that segment or part of the curve, we use this formula to give us the length of any part of the curve or even the entire curve.

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If the curve C is given parametrically for a � t � b by smooth functions and if the velocity is not zero for a � t � b, then the length L of C is: vG

� �b

aL v t ³ dtG

Example 5. Find the circumference of the ellipse x(t) = 2 cos(t), y(t) = sin(t), 0 � t � 2�. We compute the velocity: vx(t) = -2sin(t) vy(t) = cost(t)

� � � �2 2 24sin cos 9.96o

L t t dtS

�³ �

Note: the integral needs to be evaluated numerically. ====================================================================== Vector Fields Up to this point we have discussed vectors and functions of one, two, and three variables. Although the functions we have presented thus far have contained two or three variables, they have been scalar functions. That is, they have taken a point in 2-space (x, y) or 3-space (x, y, z) and converted or mapped these three values to a single (or scalar) value. Mathematically we write:

:f �o� or 2:f � o� or 3:f � o� where stands for the collection or set of all real numbers and is the collection of all ordered pairs of real numbers (Cartesian plane) and is the collection of all ordered triples of real numbers (Cartesian 3-space).

� 2�3�

The notation means that f is a mapping or function that takes a real number and maps it to another real number i.e. f is a function of a single

:f �o�

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variable. means that f maps a pair of numbers to a single number i.e. f is a function of two variables; and

2:f � o�3:f � o� means f maps a triple of

numbers to a single number i.e. f is a function of three variables. We need not however confine our discussion to scalar functions; instead we can look at functions of the form:

3:f � o�3 or 2 2:f � o� or 3:f �o�

Such functions are known as vector-valued functions. In the case where f assigns a vector to each point in 3-space we call f a vector field. In fact, we have already encountered examples of vector-valued functions and vector fields, although we have not called them as such. The gradient of a scalar function is an example of a

f�vector field. If f is a scalar function of

two variables then:

2 3:f� � o� That is maps a pair of values (x and y) into a vector. Another example of a vector-valued function or vector field is a parameterized curve in 3-space. In this case we have:

f�

3:C �o�

That is C(t) takes a single value, t, and maps it to a vector. An example of this was the helix:

� � � � � � ˆˆ ˆcos sinC t t i t j tk � �G

An example of a physical vector field is the ocean current known as the Gulf Stream (See Figure 27). The Gulf Stream is an example of a velocity vector field. That is, at each point in the ocean, there is a vector giving the velocity of the ocean current at that point. Vector fields are drawn by placing an arrow at each point on grid or map showing the direction and strength of the field at that point. In the case of Figure 27, the strength of the current i.e.

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its speed is indicated by the length of the arrow and its direction by the orientation of the arrow. Figure 27: Gulf Stream. Beginning in the Caribbean and ending in the northern North Atlantic, the Gulf Stream System is one of the world's most intensely studied current systems. This current plays an important role in the poleward transfer of heat and salt and serves to warm the European subcontinent. Image and text taken from: http://oceancurrents.rsmas.miami.edu/atlantic/gulf-stream.html. Another example of a vector field is a Force Field. A force field is a vector field giving the force (magnitude and direction) at every point in space. An example of a force field is the earth’s gravitational field. That is, the force exerted on another body at each point in the space surrounding the earth. It is the gravitational field of the earth that keeps the moon and man-made satellites in orbit about the earth. Figure 28 depicts the earth’s gravitational field which shows that the force lines are directed towards the center of the earth. The length of the arrows get shorter as the distance from the center of the earth increases indicating that the force field gets weaker.

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http://oceancurrents.rsmas.miami.edu/atlantic/gulf-stream.html


Figure 28. Earth’s Gravitational (Force) Field. Another example of a force field is the earth’s magnetic field, shown in Figure 29. In this case, the vectors point around the earth starting at the south-pole towards the north-pole. Figure 29. Earth’s magnetic field. Note: images taken from http://www.unc.edu/depts/oceanweb/turtles/juvenilemap/EarthMF.html With these examples we formally define a vector field as follows:

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http://www.unc.edu/depts/oceanweb/turtles/juvenilemap/EarthMF.html


A vector field in 2-space is a function � �,F x yG

whose value at a given point (x, y) is a 3-dimensional vector. Similarly, a vector field in 3-space is a function � , ,F x y zG

�whose values are 3-dimesional vectors.

The arrow over the function F

Gindicates that its value is a vector, not a

scalar. We can also represent the point (x, y) or (x, y, z) as a vector and write .

rG

� �F rG G

Vector fields can be visualized by sampling the vector function at various points and drawing arrows in the appropriate directions. The length of the arrows is proportional to the magnitude of the vector function at the point. Example 6. Sketch the vector field � � ˆ ˆ,F x y yi xj � �

G.

In this case the magnitude of the vector field is � � 2 2,F x y x y �

Gand we

note that is perpendicular to the position vector � ,F x yG

� ˆˆr xi yj �G i.e.

. The vector field will therefore

look like a spiral of arrows centered around the origin with the length of the arrows increasing as the distance from the origin increases, as shown in Figure 30.

� � � �� ˆ ˆ ˆ ˆ, 0F x y r yi xj xi yj yx xy� � � � � � G G

y

x

Figure 30: Vector field � � ˆ ˆ,F x y yi xj � �G

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Example 7. Sketch the vector fields � � ˆ,F x y xj G

and � � ˆ,G x y xi G

. The vector ˆxj is parallel to the y-direction, pointing up when x is positive and down when x is negative. The larger the x is the longer the vector. The vectors in the field are constant along vertical lines since the vector field does not depend on y. See Figure 31.

Figure 31: Vector field � � ˆ,F x y xj G

y

x

The vector ˆxi is parallel to the x-direction, pointing right when x is positive and left when x is negative. The larger the x is the longer the vector. The magnitudes of vectors in the field are constant along horizontal lines since the vector field does not depend on y. See Figure 32. y

Figure 32: Vector field � � ˆ,G x y xi G

x

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Newton’s Law of Gravitation states that the magnitude of the gravitational force exerted by an object of mass M on an object of mass m is proportional to M and m and inversely proportional to the square of the distance between them. The direction of the force is from m to M along the line connecting them:

� � � � 2ˆ, , GMm r GMm

F x y z F r rr r r r

� ��

GG G GG G G

where 22r r r r � G G G and ˆ r

rr

GG . Note that the minus sign means that the

force is attractive.

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Lecture 15 Sections 17.4 and 17.5 – Flow and Parameterized Surfaces Flow of a vector Field Let’s re-examine to notion of a vector field as a velocity field. If we take a look at a velocity field, say the Gulf Stream (as shown in Figure 33), then each point in the field gives the speed and direction of the flow of water in the Atlantic Ocean. Figure 33: Gulf Stream. Beginning in the Caribbean and ending in the northern North Atlantic, the Gulf Stream System is one of the world's most intensely studied current systems. This current plays an important role in the pole-ward transfer of heat and salt and serves to warm the European subcontinent. Image and text taken from: http://oceancurrents.rsmas.miami.edu/atlantic/gulf-stream.html. If we look at the white area of Figure 33, it shows the flow off the east cost of the United States, extending from Florida up through Maine and out into the North Atlantic. We note that the arrows in this zone of the map are relatively long, indicating a strong or fast current and pointing in generally

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http://oceancurrents.rsmas.miami.edu/atlantic/gulf-stream.html


the same direction: north-east. Consider an object placed in the water off the coast of Florida say an empty bottle or a raft. What can we say about the path of this object? Well from the magnitude and direction of the velocity field our bottle or raft will be swept up the east coast of the United States and out into the north Atlantic. We expect the speed of the bottle or raft to be approximately the speed of the current which is given by the velocity field. If is the Gulf Stream velocity field, and v is the velocity of our bottle or raft, then:

� ,G x yG

� G

� �,v G x yGG �

Where x and y are the position of the bottle or raft in the ocean. Since the bottle or raft will move with the flow of the ocean water, x, y, and are functions of time. That is we could think of setting our bottle adrift in the water on Monday morning and then asking where is it on Tuesday morning, Wednesday morning, etc. Because the bottle will be carried along by the currents of the Gulf Stream, it position will change over time. In the case where the bottle offers no resistance to the water, its speed will match that of the water exactly and we will have:

vG

� � � � � �� ,v t G x t y t

GG But the velocity is just time derivative of the position, rG , with respect to time, so we have:

� � � � � �� ,drt G x t y t G r t

dt

G G G G

Hence the velocity of the bottle is determined by its position in the velocity field and it will follow a path determined by the field. That is, it will follow a flow line or streamline. The white area in Figure 33 is an area of the Gulf Stream with a clear streamline or flow line where water is moving from the south near Florida to the north-east. We generalize this concept with the following definition: A flow line of a vector field � � � �� v t G r t

GG G is a path � �r tG whose velocity

equals . Thus: � �v tG

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� � � �� dr t

G r tdt

G G G

The flow of a vector field is the family of all its flow lines. Flow lines are also called streamlines or integral curves. The equation

� � � ��dr tG r t

dt

G G G � is known as a differential equation (Math 275) since it

relates the derivates of its variables to a function of those variables. That is it is an equation that contains both variables and their derivatives. For vector fields in 2-space, � � � � � �ˆ, , ,x yG x y G x y i G x y j �

G ˆ , we have:

� � � � � ��

,

,

x

y

dx tG x t y t

dtdy t

G x t y tdt

In general, differential equations are difficult to solve and numerical methods must be used. However if � �,xG x y and � �,yG x y are not too complicated, closed form solutions can be found. We consider some examples. Example 1. Let , that is a constant vector field. Find x and y if (x, y) = (1, 2) at t= 0.

� � ˆ, 3 4G x y i j �G ˆ

In this case we have:

� �

� �

3

4

dx tdtdy tdt

So that:

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x(t) = 3t + xo and y(t) = 4t + yo. Since x(0) = xo = 1 and y(0) = yo = 2, we have:

x(t) = 3t + 1 and y(t) = 4t + 2. or � � � � � �ˆ ˆ3 1 4 2r t t i t j � � �G

Example 2. Let and let x(0) = -2, and y(0) = 2. Find . � � ˆ,G x y i xj �

G ˆ � �r tG

In this case we have:

� �

� �

1

( )

dx tdtdy t

x tdt

From the first equation we have: x(t) = t + xo. And since x(0) = xo = -2, we have: x(t) = t - 2. Putting this into the second equation we get:

� � 2dy t

tdt

�

Or y(t) = ½t2 – 2t + yo. Since y(0) = yo = 2, we get: y(t) = ½t2 – 2t +2. Thus:

� � � � 21ˆ ˆ2 22

r t t i t t j§ · � � � �¨ ¸© ¹

G 2

Example 3. Let � � ˆ,G x y yi xj � �

G ˆ . Find � �r tG .

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We have:

� � � �

� �( )

dx ty t

dtdy t

x tdt

�

In this case we note that if we let x(t) = acos(t) and y(t) = asin(t), then these equations will be satisfied. Hence � � � � �ˆ ˆcos sinr t a t i a t j � �G . As we noted above, in general, the differential equation for the flow line will not have a closed form solution and will require numerical methods to determine a solution. In such cases, Euler’s method is often used in a first attempt at finding a solution. Euler’s method makes used of a local linearization to approximation to using its first derivative. That is � �r t

G

� � � � � �dr tr t t r t t

dt� ' � '

GG G� . In this case we note that since

� � � ��dr tG r t

dt

G G G � we have: � � � � � �� r t t r t tG r t� ' � 'GG G G� . Starting

at t = 0 we can construct a series of values for � �r tG using

this approximation. Hence we have:

Leonhard Euler 1707-1783

� ��

1

2 1 1

0o

o o

r r

r r tG r

r r tG r

� '

� '

G GGG G GGG G G

And so on. In general we can write the nth term as r r �1 1n n n �tG r� � � '

GG G G which will give us the position at time n¨t. By graphing the points we can plot the flow line. As usual, as we make ¨t smaller and smaller, our approximation well be better and better. Although this method will work in principle, if the vector field G

1 2, , ,o nr r r rG G G G!

G is a complicated function, more sophisticated

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numerical techniques (beyond the scope of this course) are required. For examples and a more comprehensive discussion see, for example: Numerical Recipes in C++: The Art of Scientific Computing by William H. Press, Saul A. Teukolsky, William T. Vetterling, Brian P. Flannery, Cambridge University Press, 2002, ISBN: 0521750334. ====================================================================== Parameterized Surfaces Consider the helix again, as shown in Figure 34. We recall that the helix is a parameterized curve given by vector � � � � � � ˆˆ ˆcos sinr t t i t j tk � �

G . Although the helix exists in 3-space, it is a curve and is therefore one-dimensional. That is, we an think of ‘unwinding’ or ‘un-twisting’ the helix and ‘straightening’ it out so that it is just a straight line, which is of course a one-dimensional object.

Figure 34: The Helix

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Surfaces on the other hand are two-dimensional objects. Just like the helix, they may exist in three dimensions but we can think of ‘pounding’ them flat or ‘stretching’ them out again so that they lie flat or in a plane, which is two-dimensional. Unlike curves however, surfaces depend on two variables or parameters. Thus consider the curve:

x(t) = cos(t) y(t) = sin(t) z(t) = 0 This is just a circle in the xy-plane. If we set z to some other constant value, say 10, then we still have a circle except that this circle will be in a plane parallel to the xy-plane located 10 units above it. However if we were to allow z to vary freely i.e. make it a parameter independent of t, say s, then we will no longer have curve but rather a surface and this surface will be a cylinder. Thus:

x(s, t) = cos(t) y(s, t) = sin(t) z(s, t) = s are the parametric equations that define a cylinder. The parameters s and t are independent of each other and as such can vary independently over their allowed ranges. In the case of the cylinder we have: 0 � t � 2� and -� � s � �. In general we can express the points on a surface in terms of two parameters, s and t as follows:

x(s, t) = f1(s, t) y(s, t) = f2(s, t) z(s, t) = f3(s, t) As the values of s and t vary over their allowed ranges, the point (x, y, z) will ‘sweep’ out the surface. The functions f1(s, t), f2(s, t), and f3(s, t) are called the parameterization of the surface. As with the curve, we can use the position vector rG to define the location of all the points on the surface:

� � � � � � � �1 2 3ˆˆ ˆ, , , ,r s t f s t i f s t j f s t k � �

G We note that the graphs of functions of two variables are surfaces in 3-space. That is we let z = f(x, y). In this case we can parameterize these

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surfaces by letting x(s, t) = f1(s, t) = t, y(s, t) = f2(s, t) = s, and z(s, t) = f3(s, t) = f(s, t). Example 4. Give the parameterization of the lower hemisphere x2 + y2 + z2 = 1. In this case we have 21z x � � � 2y . We let x = t, y = s and 2 21z t s � � � where . 2 2 1s t� d Consider a plane containing two non-parallel vectors 1v

G and 2vG and a point op

G . We can get to any point on the plane by starting at op

G and moving parallel to and then to . Since is parallel to 1vG

2vG

1svG

1vG and 2tv

G is parallel to we have: 2vG

The parametric equations of a plane through the point op

G and containing the non-parallel vectors and is: 1v

G2vG

� � 1 2, or s t p sv tv � �G G G G

If ˆˆ ˆ

o o o op x i y j z k � �G , , and 1 1 2 3

ˆˆ ˆv a i a j a k � �G

2 1 2 3ˆˆ ˆv b i b j b k � �

G , then

x(s, t) = xo + sa1 + tb1 y(s, t) = yo + sa2 + tb2 z(s, t) = zo + sa3 + tb3

Note that x, y, z are linear functions of s and t. Example 5. Write the parametric equations for the plane through the point (2, -1, 3) and containing the vectors 1

ˆˆ ˆ2 3v i j k � �G and 2

ˆˆ ˆ4 5v i j k � �G .

Using the results above we have:

x(s, t) = 2 + 2s + t y(s, t) = -1 + 3s - 4t z(s, t) = 3 – s + 5t

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******************************************************************* In our discussion of spherical coordinates we showed that the distance r from the origin to a point along with the azimuth angle and the elevation angle were an alternative to Cartesian coordinates for locating points in 3-space. Figure 35 shows these angles and their relationship to an arbitrary point (x, y, z) in Cartesian 3-space. z

x

y

x

(x, y, z) = (r, Ǆ, Ĺ)

z

r

Ǆ

Ĺ

y

Figure 35: Spherical Coordinates Figure 35 shows that the relationship between the spherical and Cartesian coordinates is as follows:

� � � ��

sin cos

sin sin

cos

x r

y r

z r

M T

M T

M

We can use the angles Ǆ and Ĺ as alternative parameters to the Cartesian-like parameters s and t presented above when developing parametric equations for surfaces. The angle Ǆ is the azimuth angle and can be

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identified with the longitude angle used for locating points on or near the surface of the earth. This angle ranges either between 0 and 2� or equivalently between –� and � (the later is used for longitude). The elevation angle Ĺ can be identified with the latitude angle. This angle ranges between 0 and �. We thus have:

0 � Ǆ � 2� and 0 � Ĺ � �. Points in the Northern hemisphere range between 0 and �/2 i.e. 0 � Ĺ � �/2. Points in the Southern hemisphere range between �/2 and � i.e. �/2� Ĺ � �. If a point is located at (Ǆo, Ĺo) then the point located on the other side of the sphere on the line through this point and the center is called the antipodal point. In azimuth this point is located at Ǆo + �. In elevation, the antipodal point is located at � -Ĺo. Example 6. Find the parametric equations for the sphere with center at the point (2, -1, 3) and having radius 2. The sphere with radius 2 has the following parametric equations:

x = 2cos(Ǆ)sin(Ĺ) y = 2sin(Ǆ)sin(Ĺ) z = 2cos(Ĺ) The center of this sphere is at (0, 0, 0). To shift it to (2, -1, 3) we add this point to the parameterization above:

x = 2cos(Ǆ)sin(Ĺ) + 2 y = 2sin(Ǆ)sin(Ĺ) - 1 z = 2cos(Ĺ) + 3 ******************************************************************* The many cases, surfaces have an axis of rotational symmetry with circular cross sections perpendicular to that axis. These surfaces are referred to as surfaces of revolution. Consider the following case: what is the parameterization of the cone whose base is the circle of radius a in the xy-plane and whose vertex is height h above the xy-plane? In this case we note that the radius of the cone is a function of the height, s, above the xy-plane, as shown in Figure 36.

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Figure 36: Cross section for the cone

s

a

r

h

In the xy-plane the radius of the cone is a and at height h above the xy-plane, the radius is zero since we are at the vertex of the cone. The triangle in Figure 36 is a cross section of half the cone. If we were to rotate this triangle about its vertical leg, we would sweep out the cone. The vertical or z-axis is therefore the rotational axis of symmetry. If we find the radius r as a function of the height parameter s and then rotate this radius as it moves between the xy-plane and the vertex we will sweep out the cone. Using similar triangles we have:

a rh h s

�

So that � � 1 sr s a

h§ �¨© ¹

·¸ . Rotating about this z-axis is parametrically

equivalent to computing the x and y components of the parameterization by taking r(s) and multiplying by the cosine and sine of the rotation angle, t. Thus we have:

x(s, t) = r(s)cos(t) y(s, t) = r(s)sin(t) Finally the z-component of the parameterization is just s, the height above the xy-plane. Thus we get:

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x(s, t) = a(1 – s/h) cos(t) y(s, t) = a(1 – s/h) sin(t) z(s, t) = s where 0 � s � h and 0 � t � 2�. This is the parameterization of the cone. Note the parameterization essentially uses cylindrical coordinates. ******************************************************************* Finally we talk about parameter curves. These are essentially parameterized curves on a surface. That is, unlike a helix which is a parameterized curve in 3-space, a parameter curve exists in 3-space like the helix, but it is contained entirely on surface. On a parameterized surface, such a curve is obtained by setting one of the parameters equal to a constant and letting the other vary. Thus if the surface is parameterized by:

� � � � � � � �1 2 3ˆˆ ˆ, , , ,r s t f s t i f s t j f s t k � �

G then there are two families of parameter curves on the surface, one family with t constant and the other with s constant. Example 7. For the vertical cylinder x(s, t) = cos(t) y(s, t) = sin(t) z(s, t) = s describe the parameter curves with t constant and s constant. If t is constant, then x(s, t) = xo = constant and y(s, t) = yo = constant. Thus x and y are fixed. We note that the constants xo and yo are such that

and therefore this point lies on the surface of the cylinder. Since z(s, t) = s, then z will vary up and down above the point (x

2 2 1o ox y�

o , yo) on the surface of the cylinder. This curve is therefore a vertical line on the surface, as are all the curves in this family, as shown in Figure 37. If s is constant then we have:

x(s, t) = cos(t) y(s, t) = sin(t) z(s, t) = constant. In this case, x(s, t) and y(s, t) define a circle on the surface of the cylinder that is in a plane parallel the xy-plane and located at a height equal to the

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constant value of s above or below the xy-pane. This family of curves in shown in Figure 38. z

t = constant family

x

y

Figure 37: The constant t parameter curve family on the cylinder z

s = constant family

x

y

Figure 38: The constant s parameter curve family on the cylinder

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Example 8. Describe the families of parameter curves for the sphere with parameterized surface given by:

x(Ǆ, Ĺ) = cos(Ǆ)sin(Ĺ) y(Ǆ, Ĺ) = sin(Ǆ)sin(Ĺ) z(Ǆ, Ĺ) = cos(Ĺ). In this case our parameters are Ǆ and Ĺ. If we let Ǆ = constant then we have: x(Ǆ, Ĺ) = Ĵsin(Ĺ) y(Ǆ, Ĺ) = ǀsin(Ĺ) z(Ǆ, Ĺ) = cos(Ĺ) where Ĵ and ǀ are constants such that Ĵ = cos(Ǆ) and ǀ = sin(Ǆ). We note that

� �2 2 sinx y M� so that 2 2x y� and z define a circle running from pole to pole in the plane that contains the z-axis and is defined by the constant angle Ǆ. Figure 39 shows this family of curves. On the surface of the earth, these would be lines of longitude. z

y

Ĳ = constant family

x

Figure 39: The constant Ǆ parameter curve family on the sphere If we now let Ĺ = constant we get:

x(Ǆ, Ĺ) = acos(Ǆ) y(Ǆ, Ĺ) = asin(Ǆ) z(Ǆ, Ĺ) = b

where a and be are constants. In this case x an y define a circle of radius a = sin(Ĺ) in a plane parallel to the xy-plane and at height b above or below the xy-plane. These are lines of constant latitude on the surface of the earth and are shown in Figure 40.

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z

Ĺ = constant family

y

x

Figure 40: The constant Ĺ parameter curve family on the sphere Finally we note that the cross sections of a surface z = f(x, y) are examples of parameter curves. So are the grid lines on a computer plot of the surface. The small regions shaped like parallelograms surrounded by nearby pairs of parameter curves are called parameter rectangles. See Figure 41.

Parameter Rectangle

Parameter curves i.e. cross-sections

Figure 41: Parameter Rectangles

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