Section 5.2

Finding Volumes of Solids

We have used integrals to find the areas of regions under curves; it may not seem obvious at first,but we can actually use similar methods to find volumes of certain types of solids. In this sectionand the next, we will develop several techniques for doing so.

Technique 1: Volumes by Slicing

The first technique we will learn is helpful for evaluating the volume of a solid for which we candetermine the area of a typical cross-section. We note that a cylindrical solid with base area A andheight H has volume V = AH:

Assuming that we know the values for A and H above, calculating the volume of the solid inthe graphic is quite simple.

Of course, there are many solids whose volumes are not so easy to understand. Consider thesolid below:

1

Section 5.2

Finding the volume of this solid is much more challenging than it would be in the last examplebecause the solid has “curved” sides, as opposed to the straight sides of the original cylindricalsolid.

Instead of trying to find the exact volume of the solid, we can start by approximating the volumeof the figure. We could make an (admittedly poor) approximation by replacing the figure with acylinder, whose volume we can easily calculate by multiplying the area of its base by its height:

We could get a better approximation by using two cylinders in our estimations instead of justone:

2

Section 5.2

In particular, the sum of the volumes of the two regions above would be a decent approximationfor the volume of the original solid.

Obviously the more cylinders or “slices” we use, the better the approximations become. As theheight of the cylinders decreases, they begin to look more like “slices” of the figure:

The volume of the ith slice is its area A(xi) multiplied by its height h = ∆x, that is A(xi)∆x.We approximate the volume of the figure by adding up the volumes of each of the slices,

V ≈n∑i=1

A(xi)∆x.

We find the actual volume by letting the number of slices increase without bound:

3

Section 5.2

V = limi→∞

n∑i=1

A(xi)∆x.

Since this is simply a Riemann sum, we can accomplish the same thing by evaluating

V =

∫ b

aA(x) dx,

where a is the “beginning” and b is the “end” of the solid, and A(x) is athe function giving thearea of a slice of the solid at value x.

To use slicing to evaluate the volume of a figure, follow the steps below:

1. Sketch the base of the figure and draw a typical slice.

2. Determine a formula A(x) for the area of a typical slice.

3. Determine where the slices begin (x = a) and end (x = b), i.e. the bounds of integration.

4. Evaluate

∫ b

aA(x) dx.

One minor note: depending on the shape of the solid, it may be convenient to switch the orderof integration. If the slices of the solid are perpendicular to the x-axis, then the slices will be“pushed” along the x-axis to create the solid, and we should integrate A(x) from x = a to x = b.However, if the slices are perpendicular to the y-axis, so that pushing the slices along the y-axiscreates the solid, then we will need to integrate an area formula in terms of y with bounds ofintegration in terms of y as well.

For example, in the figure below, we can slide slices of the figure along the x-axis to create theentire solid:

4

Section 5.2

In this example, we should choose to integrate with respect to x.On the other hand, sliding slices of the figure along the y-axis in the following figure will create

the entire solid:

In this example, we should choose to integrate with respect to y.

Example. The base of a solid is bounded between the y-axis and the lines y = x and y = 2 − x.Cross-sections of the solid perpendicular to the x-axis are semi-circles whose diameters lie on thebase. Find the volume of the solid.

The lines y = x and y = 2 − x intersect at x = 1, so the base of the solid is the region drawnbelow:

We know that one typical slice of the figure is a semicircle:

5

Section 5.2

The semicircle gets pushed from one end of the base to the other to form the solid whose volumewe would like to determine:

We need to determine the area of a typical slice. We can use the formula for the area of asemicircle, A = 1

2πr2, but how do we determine the radius r? We know that the diameter d runs

from y = x to y = 2− x, that isd = 2− x− x = 2− 2x.

Since the radius is half of the diameter, r = 1− x. So the area of a typical slice is

A =1

2π(1− x)2 =

π

2(1− 2x+ x2).

Since the slices start at x = 0 and go to x = 1, the volume of the solid is described by

V =

∫ 1

0

π

2(1− 2x+ x2) dx.

6

Section 5.2

Let’s evaluate the integral:∫ 1

0

π

2(1− 2x+ x2) dx =

π

2

∫ 1

01− 2x+ x2 dx

=π

2(x− x2 +

1

3x3∣∣∣∣10

)

=π

2(1− 1 +

1

3)

=π

6.

So the volume of the figure is π6 .

Example. A round column with base radius r = 1 is sliced in half vertically so that its base isa semi-circle. A wedge is then sliced from the half-column so that the blade of the circular sawmakes an angle of π

3 with the base of the column. Find the volume of the wedge.

The wedge will be cut from the column as shown in the following diagrams:

7

Section 5.2

We can situate the base of the column on the xy coordinate plane:

Inspecting the original solid, we see that slices perpendicular to the x-axis are right triangles,with one angle of π

3 :

8

Section 5.2

We need to determine the area of a typical slice. Since slices are triangles, we can use theformula A = 1

2(base)(height).Fortunately, we know that the base of each triangle lies on the circle, whose formula is y =√

1− x2. Thusbase =

√1− x2.

We will need to use trigonometry to determine the heights of the triangles. If we “double” thetriangle, we end up with a triangle all of whose angles are equal; thus all of the sides are equal aswell:

So the hypotenuse of the original triangle is 2√

1− x2. We can use the Pythagorean identity torelate the side lengths:

h2 = a2 + b2

9

Section 5.2

Thus we see that the height of the triangle must be√(2√

1− x2)2 − (√

1− x2)2 =√

4(1− x2)− (1− x2)

=√

3(1− x2).

Thus the area of a typical triangle is A = 12(√

1− x2)(√

3(1− x2)) =√32 (1 − x2). The slices

run from x = −1 to x = 1, so the volume of the wedge is

V =

∫ 1

−1

√3

2(1− x2) dx

=

√3

2(x− 1

3x3)

∣∣∣∣1−1

=

√3

2(1− 1

3+ 1− 1

3)

=

√3

2(2− 2

3)

=

√3

2(6− 2

3)

=

√3

2(4

3)

=2√

3

3.

Volumes of Solids of RevolutionWe can create a solid by revolving a shape around a line. For instance, we can create a sphere

by revolving a semi-circle around the line y = 0:

10

Section 5.2

In the example below, we have revolved a two-dimensional shape about the y-axis:

To find the volume of such a figure, we note that we can always cut it perpendicular to the axisof rotation so that slices look like discs:

We can use the ideas from the previous section to find the volume of the figure; find the areaof a typical slice, then “add up” all of the areas by integrating the area formula. In a sense, theintegral pushes the disc along the axis of revolution:

11

Section 5.2

Since slices are shaped like discs, a typical slice has area A = πr2; in the case of the figureabove, the radius of a typical disc depends on the height of the disc as compared with the y-axis.That is, the radius of the disc is a function r(y) of y, so we write

A = πr(y)2.

So the volume of the solid is

V =

∫ b

aπ(r(y)2) dy,

where the slices start at y = a and end at y = b.If we instead revolve the shape around the x-axis, the volume of the solid created is given by

V =

∫ b

aπ(r(x)2) dx,

where the slices start at x = a and end at x = b.It is possible that, when we revolve the shape to create the desired solid, we end up with a

figure that has a hole in the middle:

This time, a typical slice has a hole in the middle and looks like a washer:

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Section 5.2

Thus the area of a typical slice is given by

A = π(R(x))2 − π(r(x))2,

where R(x) is the outside radius of the washer at x and r(x) is the inside radius.Again, we can push the washers along the axis of revolution to generate the desired figure:

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Section 5.2

So if the x-axis is the axis of revolution, then the volume of the figure is given by

V =

∫ b

aπ((R(x))2 − (r(x))2) dx,

where the slices start at x = a and end at x = b. If the axis of revolution is the y-axis, then thevolume is given by

V =

∫ b

aπ((R(y))2 − (r(y))2) dy,

where the slices start at y = a and end at y = b.To evaluate the volume of a figure obtained by rotating a shape about a line parallel to the

x-axis, follow the steps below. If the axis of revolution is parallel to the y-axis instead, just changeall of the xs below to ys.

1. Sketch the shape to be rotated, as well as a typical disc (or washer).

2. Determine the radius R(x) of the disc (or the outer radius R(x) and inner radius r(x) of thewasher).

3. Determine where the discs (or washer) begin (x = a) and end (x = b), i.e. the bounds ofintegration.

4. Evaluate π

∫ b

a(R(x))2dx (or π

∫ b

a((R(x))2 − (r(x))2)dx).

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Section 5.2

Again, depending on the shape we have created, it may be desirable to integrate a formula interms of y instead of in terms of x. The discs or washers always get pushed parallel to the axiswhose variable we should use for integration.

One minor note: students often have difficulty determining how to write a formula for the radiusof the disc. Keep in mind that the radius is always measured from right to left (or top to bottom,in terms of y) between the center of the disc and the function that is being used to create the disc.

Example. Find the volume of the solid obtained by rotating the region bounded by y = x3, y = 8,and x = 0 about the y-axis.

The region given in the problem is graphed below:

We rotate the region around the y-axis to create the desired solid:

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Section 5.2

A typical disc is graphed below:

Notice that discs can be pushed from y = 0 to y = 8 along the y-axis to generate the solid, sowe will need the formula for the area of a typical disc to be in terms of y.

16

Section 5.2

We will be using the volume formula

V =

∫ b

aπ(R(y))2dy.

The radius of a typical disc is included on the graph below:

Notice that the radius has ends on the y and on the curve y = x3; since we plan to integratewith respect to y, we rewrite this last equation as x = 3

√y, so that

R(y) = 3√y.

Then the volume of the region is given by

V = π

∫ 8

0( 3√y)2 dy

= π

∫ 8

0y

23 dy

= π(3

5y

53 )

∣∣∣∣80

=3π

5· 32

=96π

5.

Example. Find the volume of the solid obtained by rotating the region enclosed by y = x andy = x2 about the line y = 2.

17

Section 5.2

The two functions intersect when x = x2: solving this equation, we have

x2 − x = 0

x(x− 1) = 0,

so x = 0 or x = 1. Accordingly, the ”base” of the region is graphed below:

We rotate the shape around y = 2 to generate the desired solid:

A typical slice is graphed below:

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Section 5.2

The slices will be pushed along the x-axis from x = 0 to x = 1 to create the desired shape, sowe will be integrating with respect to x.

Notice that the slice is a washer this time instead of a disc, so we use the volume formula

V =

∫ b

aπ((R(x))2 − (r(x))2)dx.

We need to determine the outer radius R(x) and the inner radius r(x). The washer is centeredat the line y = 2, so the outer radius is measured from y = 2 to the outer edge of the washer:

Keep in mind that we measure the radius from the center of the disc to the function; soR(x) = 2− x2.

Similarly, the inner radius r(x) is measured from y = 2 to the inner edge of the washer:

So r(x) = 2− x.

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Section 5.2

The volume formula is

V = π

∫ 1

0((2− x2)2 − (2− x)2)dx

= π

∫ 1

0(4− 4x2 + x4 − 4 + 4x− x2)dx

= π

∫ 1

0(x4 − 5x2 + 4x)dx

= π(1

5x5 − 5

3x3 + 2x2)

∣∣∣∣10

= π(1

5− 5

3+ 2)

= π

(3− 25 + 30

15

)=

8π

15.

20