Section 5.2

Finding Volumes of SolidsWe have used integrals to find the area 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 section and thenext, we will develop several techniques for doing so.

Technique 1: Volumes by SlicingThe first technique we will learn is helpful for evaluating the volume of a solid for which we can

determine the area of a typical cross-section. We note that a cylindrical solid with base area a andlength h has volume V = ah:

We will think about the slicing technique by considering the volume of the figure below:

We begin by approximating the volume of the figure. We could make an (admittedly poor)approximation by replacing the figure with a cylinder, whose volume we know:

1

Section 5.2

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

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:

2

Section 5.2

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:

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 a function giving the areaof 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 and we can integrate A(x) from x = a to x = b. However, if the slicesare perpendicular to the y-axis, then the slices will be pushed along the y-axis; we will need tointegrate an area formula in terms of y with bounds of integration in terms of y as well.

3

Section 5.2

Examples: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 the base.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:

4

Section 5.2

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

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

5

Section 5.2

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.

∫ 1

0

π

2(1− 2x+ x2)dx =

π

2

∫ 1

01− 2x+ x2dx

=π

2(x− x2 +

1

3x3

∣∣∣∣10

)

=π

2(1− 1 +

1

3)

=π

6.

So the volume of the figure is π6 .

A round column with base radius r = 1 is sliced in half vertically so that its base is a semi-circle.A wedge is then sliced from the half-column so that the blade of the circular saw makes an angleof π

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

6

Section 5.2

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

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

Then a slice of the solid is a right triangle with one angle of π3 whose base is perpendicular to

the y-axis:

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

2(base)(height).

The formula for the circle is x =√

1− y2, which is the length of the base of the triangle:

7

Section 5.2

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− y2. We can use the Pythagorean identityto see that the height of the triangle must be

h =

√(2√

1− y2)2 − (√

1− y2)2

=√

4(1− y2)− (1− y2)

=√

3(1− y2).

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

1− y2)(√

3(1− y2)) =√32 (1 − y2). The slices

8

Section 5.2

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

V =

∫ 1

−1

√3

2(1− y2)dy

=

√3

2(y − 1

3y3)

∣∣∣∣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:

To find the volume of such a figure, we note that we can always cut it so that slices look likediscs:

9

Section 5.2

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

In this case, since slices are shaped like discs, a typical slice has area A = π(r(x))2, where r(x)is the radius of the disc. So the volume of the solid is

V =

∫ b

aπ(r(x)2)dx.

It is possible that, when we revolve the shape to create the desired solid, we end up with afigure that has a hole in the middle, such as the one graphed below:

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

10

Section 5.2

So a typical slice has area A = π(R(x))2 − π(r(x))2, where R(x) is the outside radius of thewasher and r(x) is the inside radius.

Again, we can push the washers along the axis of revolution to generate the desired figure:

So the volume of such a figure is given by

V =

∫ b

aπ((R(x))2 − (r(x))2)dx.

To evaluate the volume of a figure obtained by rotating a shape about a line, follow the stepsbelow:

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).

11

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 along the axis whosevariable 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.

Examples: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:

A typical disc is graphed below:

12

Section 5.2

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.

We will be using the volume formula

V =

∫ b

aπ(R(y))2dy.

The radius of the disc is given by y = x3, which we rewrite as x = 3√y:

Then the volume of the region is given by

13

Section 5.2

V = π

∫ 8

0( 3√y)2dy

= π

∫ 8

0y

23dy

= π(3

5y

53 )

∣∣∣∣80

=3π

5· 32

=96π

5.

Find the volume of the solid obtained by rotating the region enclosed by y = x and y = x2

about the line y = 2.

The two functions intersect when x = x2, that is when 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:

14

Section 5.2

A typical slice is graphed below:

The slices will be pushed from x = 0 to x = 1 to create the desired shape, so we will beintegrating 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 centered

15

Section 5.2

at 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.

16

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.

17