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Section 7.2
Solids of Revolution
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1st Day
Solids with Known Cross Sections
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Method of Slicing
1
Find a formula for A(x) = the area of a cross section.(When you multiply A(x) by dx, you will have the third length necessary to find the volume)
Sketch the solid and a typical cross section.
2
3 Find the limits of integration.
4 Integrate A(x) to find volume.
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Examples
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1. Find the volume of the solid whose base is bounded by the circle x2 + y2 = 9 and each cross section perpendicular to the x-axis is a square.
x
y
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x
y
side of each square = 2y
Area = (2y)2 = 4y2
29y x
224 9A x x
23 2
02 4 9V x dx
3 2
08 9V x dx
33
0
8 93
xV x
3144 uV
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2. Find the volume of the solid whose base is bounded by the graphs f(x) = x + 1 and g(x) = x2 – 1 whose cross sections perpendicular to the x-axis are equilateral triangles. Area of an equilateral triangle is given by 2 3
4
sA
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Side of equilateral triangle =
f(x) – g(x)
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2side 1 1x x 22 x x
2232
4A x x x
4 3 232 3 4 4
4x x x x
2 4 3 2
1
32 3 4 4
4V x x x x dx
381 3 u
40
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3. The region bounded by the graph of y = 2x – x2 and the x-axis is the base of a solid. For this solid each cross section perpendicular to the x-axis is a semicircle.
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Diameter of semicircle = 2x – x2
21radius of semicircle 2
2x x
2
21 12
2 2A x x x
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2
21 12
2 2A x x x
2212
8A x x x
2 2 3 4
0
14 4
8V x x x dx
32 u
15
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2nd Day
Disk Method
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y x Suppose I start with this curve.
How did we first find the approximation for area under this curve?
Drawing rectangles and finding the area of each rectangle and adding them together.
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If I take the graph and rotate it about the x-axis, I will get a 3-dimensional solid that is cone shaped.
y x
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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.
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Where do these flat cylinders (disks) come from?
They are created by rotating an infinitely many rectangles around the axis of revolution.
y x
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Volume of a Cylinder = r2hWhat is the radius of each cylinder(disk) in the cone?What is the height of each cylinder(disk) in the cone?
The radius of each cylinder(disk) is y coordinate of its rectangle.The height of each cylinder(disk) is Δx.Volume of each cylinder(disk) = (y2)Δ x
y x
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How do we find the volume of the solid?
Add the volume of all the cylinders(disks).
y x
24
0πV x dx 4
0π x dx
42
0
π
2V x 8π cubic units
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Examples
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The region between the curve , and the
y-axis is revolved about the y-axis. Find the volume.
1x
y 1 4y
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
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24
1
1π V dy
y
volume of disk
4
1
1π dy
y
4
1π ln y
π ln 4 ln1
2π ln 232π ln 2 u
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Find the volume of the solid formed by rotating the region bounded by the x-axis (0 ≤ x ≤ ) and the graph of
about the x-axis.
sinf x x
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sin x
2
sin sinx x
radius of a disk
Area of a disk
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Volume of the solid 0
sin x dx
0cos x
cos cos0 32 u
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Find the volume of the solid by revolving by
f(x) = 2 – x2 and g(x) = 1 about the line y = 1.
radius of a disk 2 22 1 1x x
Area of a disk 221 x
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Volume of the solid 1 22
11 x dx
1 2 4
11 2x x dx
153
1
2
3 5
xx x
316
u15
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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.
Solid of Revolution Formula with a vertical axis:
2π b
aV y dx
2π b
aV x dy
Solid of Revolution Formula with a horizontal axis:
End of 2nd Day
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3rd Day
Washer Method
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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 r
2 2R r dy
outerradius
innerradius
2y x
2y x
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The volume of the washer is:
2 2 thicknessR r
2 2R r dy
outerradius
innerradius
2y x
2
yx
2y x
y x
2y x
2y x
2
24
0 2
yV y dy
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4 2
0
1
4 V y y dy
4 2
0
1
4V y y dy
4
2 3
0
1 1
2 12
y y
168 0
3
38 u
3
2
24
0 2
yV y dy
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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 dx
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2y xIf the same region is rotated about the line x = 2:
2y x
The outer radius is:
22
y
RR
The inner radius is:
2 r y
r
2y x
2
yx
2y x
y x
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4 2 2
0 V R r dy
2
24
02 2
2
yy dy
2
4
04 2 4 4
4
yy y y dy
24
04 2 4 4
4
yy y y dy
14 2 2
0
13 4
4 y y y dy
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432 3 2
0
3 1 8
2 12 3
y y y
16 6424 0
3 3
38 u
3
14 2 2
0
13 4
4 V y y y dy
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Examples
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Find the volume of the solid by revolving the region bounded by the graphs of
about the x-axis.
2 and y x y x
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outer radius
inner radius
x
2x
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Volume of the solid 21 22
0x x dx
1 4
0x x dx
12 5
02 5
x x
33 u
10
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Find the volume of the solid formed by the equations:
rotated about the line x = 6.
, 1, a1 nd 4yx xy
1y x
1x y
21x y
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outer radius
inner radius
26 1y
2
21x y
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Volume of the solid 23 2 2
16 1 2y dy
3 4 3 2
14 6 20 21y y y y dy
54 3 22 10 21
5
yy y y y
1
3
3192 u
5