Pertemuan

April 22, 2023 Kalkulus II 1

PertemPertemuan uan

1717Integral lipatIntegral lipat


Double Riemann Sums

In first year calculus, the definite integral was defined as a Riemann sum that gave

the area under a curve. There is a similar definition for the volume of a region

below a function of two variables. Let f(x,y) be a positive function of two variables

and consider the solid that is bounded below by f(x,y) and above a region R in the

xy-plane.


For a two dimensional region, we approximated the area by adding up the areas of

many approximating rectangles. For the volume of a three dimensional solid, we

take a similar approach. Instead of rectangles, we use rectangular solids for the

approximation. We cut the region R into rectangles by drawing vertical and

horizontal lines in the xy-plane. Rectangles will be formed. We let the rectangles

be the base of the solid, while the height is the z-coordinate of the lower left

vertex. One such rectangular solid is shown in the figure.


Dengan cara yang sama integral dapat juga dinytakan sebagai berikut :

Atau


Taking the limit as the rectangle size approaches zero (and the number of rectangles

approaches infinity) will give the volume of the solid. If we fix a value of x and look at

the rectangular solids that contain this x, the union of the solids will be a solid with

constant width Dx. The face will be approximately equal to the area in the yz-plane

of the (one variable since x is held constant) function z = f(x,y).

This area is equal to

Untuk mendekati 0 integral menjadi :


Volume

Cari volume yang dibatasi oleh g(x,y) dan fungsi

f (x,y) seperti pada gambar berikut. Click gambar

untuk melihat animasi

http://www2.scc-fl.edu/lvosbury/CalculusIII_Folder/VolumeWorksheet.html

http://www2.scc-fl.edu/lvosbury/CalculusIII_Folder/VolumeWorksheet.html


Volume ( seperti soal sebelumnya )

Click gambar untuk melihat animasinya

http://www2.scc-fl.edu/lvosbury/images/V3animation.gif


Volume Seperti soal sebelumnya

Click disini animation


Solution

The cone is sketched below

We can see that the region R is the blue circle in the xy-plane. We can find the

equation by setting z = 0.

Solving for y (by moving the square root to the left hand side, squaring both sides, etc) gives


Volume benda di Oktant pertama

Yang dibatasi oleh :

Click disini DPGraph Picture

Find the volume of the solid in the first octant

bounded by the graphs of http://www2.scc-fl.edu/lvosbury/images/Mpic132no32.gif


Luas PermukaanCari luas permukaan yang ditentukan oleh

f(x,y) seperti pada gambar Click here untuk melihat animasi


Fubini's Theorem

Let f, g1, g2, h1, and h2 be defined and continuous on a region R. Then the double

integral equals

Example

Set up the integral to find the volume of the solid that lies below the cone

and above the xy-plane. Solution

The cone is sketched below


The "-" gives the lower limit and the "+" gives the upper limit. For the outer limits,

we can see that

-4 < x < 4

Putting this all together gives

Either by hand or by machine we can obtain the result

Volume = 64 /3

Notice that this agrees with the formula

Volume = r2h/3


Example

Set up the double integral that gives the volume of the solid that lies below the sphere x2 + y2 + z2 = 6

and above the paraboloid z = x2 + y2 Solution

We substitute x2 + y2 + (x2 + y2)2 = 6or x2 + y2 + (x2 + y2)2 - 6 = 0


Now factor with x2 + y2 as the variable to get (x2 + y2 - 2)(x2 + y2 + 3) = 0

The second factor has no solution, while the first is x2 + y2 = 2

Solving for y gives and - < x <


Just as we did in one variable calculus, the volume between two surfaces is the

double integral of the top surface minus the bottom surface. We have

Again we can perform this integral by hand or by machine and get Volume = 7.74

Polar Double Integration FormulaBentuk integral dapat dibawa dalam bebntk koordinat polar terutama yang

berbentuk lingkatan yaitu x2 + y2. Untuk ini digunakan koordinat polar.

Sebagai contoh lihat berikut ini.


Theorem: Double Integration in Polar Coordinates

Let f(x,y) be a continuous function defined over a region R bounded in polar

coordinates by r1() < r < r2() 1 < < 2

Then


Even if r and are very small the area is not the product (r)(). This

comes from the definition of radians. An arc that extends radians a

distance r out from the origin has length r. If both r and are very

small then the polar rectangle has area

Area = r r


The equation of the parabola becomes z = 9 - r2

We find the integral

This integral is a matter of routine. It evaluates to 28.


Example

Find the volume of the part of the sphere of radius 3 that is left after

drilling a cylindrical hole of radius 2 through the center.

Solution

The picture is shown below

The region this time is the annulus (washer)

between the circles r = 2 and r = 3 as shown below.


Example

Find the volume to the part of the paraboloid

z = 9 - x2 - y2

that lies inside the cylinder

x2 + y2 = 4

Solution

The surfaces are shown below.

The region R is the part of the xy-plane that is inside the cylinder. In polar coordinates, the

cylinder has equation r2 = 4

Taking square roots and recalling that r is positive gives

r = 2The inside of the cylinder is thus the polar

rectangle 0 < r < 2 0 < q < 2p


The sphere has equation x2 + y2 + z2 = 9

In polar coordinates this reduces to

r2 + z2 = 9

Solving for z by subtracting r2 and taking a square root we

get top and bottom surfaces of

We get the double integral


We get the double integral

This integral can be solved by letting u = 9 - r2 du = -2rdr

After substituting we get


http://stores.ebay.de/comic4you

Documents

Pertemuan