Section 7.2
Solids of Revolution
1st Day
Solids with Known Cross Sections
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.
Examples
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
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
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
Side of equilateral triangle =
f(x) – g(x)
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
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.
Diameter of semicircle = 2x – x2
21radius of semicircle 2
2x x
2
21 12
2 2A x x x
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
2nd Day
Disk Method
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.
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
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.
Where do these flat cylinders (disks) come from?
They are created by rotating an infinitely many rectangles around the axis of revolution.
y x
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
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
Examples
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
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
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
sin x
2
sin sinx x
radius of a disk
Area of a disk
Volume of the solid 0
sin x dx
0cos x
cos cos0 32 u
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
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
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
3rd Day
Washer Method
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
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
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
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
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
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
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
Examples
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
outer radius
inner radius
x
2x
Volume of the solid 21 22
0x x dx
1 4
0x x dx
12 5
02 5
x x
33 u
10
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
outer radius
inner radius
26 1y
2
21x y
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