Upload
eleanore-pierce
View
228
Download
0
Embed Size (px)
Citation preview
7.3 Day One: Volumes by Slicing7.3 Day One: Volumes by Slicing
x2
π x2
2
π x2
8
34
⎛⎝⎜
⎞⎠⎟ x2
154
⎛⎝⎜
⎞⎠⎟x2
33
3
Find the volume of the pyramid:
Consider a horizontal slice through the pyramid.
sdh
The volume of the slice is s2dh.
If we put zero at the top of the pyramid and make down the positive direction, then s=h.
0
3
h
2sliceV h dh=
3 2
0V h dh=∫
33
0
1
3h= 9=
This correlates with the formula:1
3V Bh=
19 3
3= ⋅ ⋅ 9=
→
Method of Slicing:
1
Find a formula for V(x).(Note that I used V(x) instead of A(x).)
Sketch the solid and a typical cross section.
2
3 Find the limits of integration.
4 Integrate V(x) to find volume.
→
xy
A 45o wedge is cut from a cylinder of radius 3 as shown.
Find the volume of the wedge.
You could slice this wedge shape several ways, but the simplest cross section is a rectangle.
If we let h equal the height of the slice then the volume of the slice is: ( ) 2V x y h dx= ⋅ ⋅
Since the wedge is cut at a 45o angle:x
h45o h x=
Since2 2 9x y+ = 29y x= −
→
xy
( ) 2V x y h dx= ⋅ ⋅h x=
29y x= −
( ) 22 9V x x x dx= − ⋅ ⋅
3 2
02 9V x x dx= −∫
29u x= −2 du x dx=−
( )0 9u =( )3 0u =
10
2
9V u du=−∫
93
2
0
2
3u=
227
3= ⋅ 18=
Even though we started with a cylinder, π does not enter the calculation!
→
Cavalieri’s Theorem:
Two solids with equal altitudes and identical parallel cross sections have the same volume.
Identical Cross Sections
π
Cavalieri’s Theorem: Volume of a Sphere
π
( )
( )( )
[ ]
2
0
0
0
2
V sin2
11 cos 2x
2 2
sin
4 2
4
4
= x dx
= dx
2x = x +
=
=
π
π
π
π
π−
π ⎡ ⎤⎢ ⎥⎣ ⎦
ππ
π
∫
∫
7.3 Disk and Washer Methods
0
1
2
1 2 3 4
y x= Suppose I start with this curve.
My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape.
So I put a piece of wood in a lathe and turn it to a shape to match the curve.
→
0
1
2
1 2 3 4
y x=How could we find the volume of the cone?
One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes.
The volume of each flat cylinder (disk) is:
2 the thicknessrπ ⋅
In this case:
r= the y value of the function
thickness = a small change
in x = dx
π ( )2x dx
→
0
1
2
1 2 3 4
y x=The volume of each flat cylinder (disk) is:
2 the thicknessrπ ⋅
If we add the volumes, we get:
( )24
0x dxπ∫
4
0 x dxπ=∫
42
02x
π= 8π=
π ( )2x dx
→
This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk.
If the shape is rotated about the x-axis, then the formula is:
2 b
aV y dxπ= ∫
2 b
aV x dyπ= ∫A shape rotated about the y-axis would be:
→
The region between the curve , and the
y-axis is revolved about the y-axis. Find the volume.
1x
y= 1 4y≤ ≤
0
1
2
3
4
1
y x
1 1
2
3
4
1.707
2=
1.577
3=
1
2
We use a horizontal disk.
dy
The thickness is dy.
The radius is the x value of the function .1
y=
24
1
1 V dy
yπ⎛ ⎞
= ⎜ ⎟⎜ ⎟⎝ ⎠
∫
volume of disk
4
1
1 dy
yπ=∫
4
1ln yπ= ( )ln 4 ln1π= −
02ln 2π= 2 ln 2π=
→
The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis:
2.000574 .439 185x y y= − +x
y
500 ft
( )500 22
0.000574 .439 185 y y dyπ − +∫
The volume can be calculated using the disk method with a horizontal disk.
324,700,000 ft≈→
0
1
2
3
4
1 2
The region bounded by and is revolved about the y-axis.Find the volume.
2y x= 2y x=
The “disk” now has a hole in it, making it a “washer”.
If we use a horizontal slice:
The volume of the washer is: ( )2 2 thicknessR rπ π− ⋅
( )2 2R r dyπ −
outerradius
innerradius
2y x=
2
yx=
2y x=
y x=
2y x=
2y x=
( )2
24
0 2
yV y dyπ
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫
4 2
0
1
4V y y dyπ ⎛ ⎞= −⎜ ⎟
⎝ ⎠∫4 2
0
1
4V y y dyπ= −∫
42 3
0
1 1
2 12y yπ ⎡ ⎤= −⎢ ⎥⎣ ⎦
168
3π ⎡ ⎤= −⎢ ⎥⎣ ⎦
8
3
π=
→
This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.
The washer method formula is: 2 2 b
aV R r dxπ= −∫
→
0
1
2
3
4
1 2
2y x=If the same region is rotated about the line x=2:
2y x=
The outer radius is:
22
yR = −
R
The inner radius is:
2r y= −
r
2y x=
2
yx=
2y x=
y x=
4 2 2
0V R r dyπ= −∫
( )2
24
02 2
2
yy dyπ ⎛ ⎞= − − −⎜ ⎟
⎝ ⎠∫
( )24
04 2 4 4
4
yy y y dyπ
⎛ ⎞= − + − − +⎜ ⎟
⎝ ⎠∫
24
04 2 4 4
4
yy y y dyπ= − + − + −∫
14 2 2
0
13 4
4y y y dyπ= − + +∫
432 3 2
0
3 1 8
2 12 3y y yπ
⎡ ⎤= ⋅ − + +⎢ ⎥
⎣ ⎦16 64
243 3
π ⎡ ⎤= ⋅ − + +⎢ ⎥⎣ ⎦
8
3
π=
π
Washer Cross Section
The region in the first quadrant enclosed by the y-axis and the graphs of y = cos x and y = sin x is revolved about the x-axis to form a solid. Find its volume.
Washer Cross Section
The region in the first quadrant enclosed by the y-axis and the graphs of y = cos x and y = sin x is revolved about the x-axis to form a solid. Find its volume.
( )4 4 2 2
0 0
4
0
43
0
V = A(x) dx = cos x - sin x dx
= cos 2x dx
sin 2x = = units
2 2
π π
π
π
π
π
π⎡ ⎤π ⎢ ⎥⎣ ⎦
∫ ∫
∫
g
7.3 The Shell Method
Find the volume of the region bounded by , , and revolved about the y-axis.
2 1y x= + 2x =0y =
0
1
2
3
4
5
1 2
2 1y x= +
We can use the washer method if we split it into two parts:
( )25 2 2
12 1 2 1y dyπ π− − + ⋅ ⋅∫
21y x− = 1x y= −
outerradius
innerradius
thicknessof slice
cylinder
( )5
14 1 4y dyπ π− − +∫
5
15 4y dyπ π− +∫
52
1
15 4
2y yπ π⎡ ⎤− +⎢ ⎥⎣ ⎦
25 125 5 4
2 2π π⎡ ⎤⎛ ⎞ ⎛ ⎞− − − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
25 94
2 2π π⎡ ⎤− +⎢ ⎥⎣ ⎦
164
2π π⋅ +
8 4π π+ 12π=→
Japanese Spider CrabGeorgia Aquarium, Atlanta
0
1
2
3
4
5
1 2
If we take a vertical slice and revolve it about the y-axis
we get a cylinder.
cross section
If we add all of the cylinders together, we can reconstruct the original object.
2 1y x= +
Here is another way we could approach this problem:
→
0
1
2
3
4
5
1 2
cross section
The volume of a thin, hollow cylinder is given by:
Lateral surface area of cylinder thickness⋅
=2 thicknessr hπ ⋅ ⋅ r is the x value of the function.
circumference height thickness= ⋅ ⋅
h is the y value of the function.
thickness is dx.( )2=2 1 x x dxπ +
r hthicknesscircumference
2 1y x= +
→
0
1
2
3
4
5
1 2
cross section
=2 thicknessr hπ ⋅ ⋅
( )2=2 1 x x dxπ +
r hthicknesscircumference
If we add all the cylinders from the smallest to the largest:
( )2 2
02 1 x x dxπ +∫
2 3
02 x x dxπ +∫
24 2
0
1 12
4 2x xπ ⎡ ⎤+⎢ ⎥⎣ ⎦
[ ]2 4 2π +
12π
This is called the shell method because we use cylindrical shells.
2 1y x= +
→
0
1
2
3
4
1 2 3 4 5 6 7 8
( )2410 16
9y x x= − − +
Find the volume generated when this shape is revolved about the y axis.
We can’t solve for x, so we can’t use a horizontal slice directly.
→
0
1
2
3
4
1 2 3 4 5 6 7 8
( )2410 16
9y x x= − − +
Shell method:
Lateral surface area of cylinder
=circumference height⋅
=2 r hπ ⋅Volume of thin cylinder 2 r h dxπ= ⋅ ⋅
If we take a vertical sliceand revolve it about the y-axiswe get a cylinder.
→
0
1
2
3
4
1 2 3 4 5 6 7 8
( )2410 16
9y x x= − − +Volume of thin cylinder 2 r h dxπ= ⋅ ⋅
( )8 2
2
42 10 16
9x x x dxπ⎡ ⎤− − +⎢ ⎥⎣ ⎦∫
rh thickness
160π=
3502.655 cm≈
Note: When entering this into the calculator, be sure to enter the multiplication symbol before the parenthesis.
circumference
→
When the strip is parallel to the axis of rotation, use the shell method.
When the strip is perpendicular to the axis of rotation, use the washer method.
π
Find the volume of the solid when the region bounded by the curve y = , the x-axis, and the line x = 4 is revolved about the x-axis. Find the volume of the solid using cylindrical shells.
x
( ) ( )2 2
0V = 2 y 4 - y dy
= 8
π
π
∫
2
2
Radius = y
x = y
Shell height = 4 - y
Find the volume of the solid when the region bounded by the curve y = , the x-axis, and the line x = 4 is revolved about the x-axis. Find the volume of the solid using cylindrical shells.
x
Find the volume of the solid of revolution formed by revolving the region bounded by the graph of and the y axis, 0 ≤ y ≤ 1, about the x-axis. Use the Shell Method.
x =e−y2
2
1
-1
2
Find the volume of the solid of revolution formed by revolving the region bounded by the graph of and the y axis, 0 ≤ y ≤ 1, about the x-axis. Use the Shell Method.
x =e−y2
2
1
-1
2
V =2π ye−y2
0
1
∫ dy
=−π e−y2⎡
⎣⎤⎦0
1
=π 1−1e
⎛
⎝⎜⎞
⎠⎟
=1.986
Find the volume of the solid formed by revolving the region bounded by the graphs y = x3 + x + 1, y = 1, and x = 1 about the line x = 2.
3
2
1
-1
2 4
axis ofrevolution
Find the volume of the solid formed by revolving the region bounded by the graphs y = x3 + x + 1, y = 1, and x = 1 about the line x = 2.
3
2
1
-1
2 4
axis ofrevolution
V =2π 2−x( )0
1
∫ x3 + x+1−1( ) dx
=2π −x4 + 2x3 −x2 + 2x( ) 0
2
∫ dx
=2π −x5
5+x4
4−x3
3+ x2⎡
⎣⎢
⎤
⎦⎥0
1
=2π −15+12−13+1
⎡
⎣⎢
⎤
⎦⎥
=29π15