🌌 Calculus Kingdom · Calculus

Volumes: Cross-Sections & Solids of Revolution

Add up thin slices to measure a solid: discs and washers spun about an axis or a line, solids built on a region from square, semicircular or triangular cross-sections, and the cone and sphere recovered from an integral.

In short

Every volume here is one sentence: slice the solid perpendicular to the axis, write the area of one slice, and integrate that area across the interval.
A disc has area $pi*r 2$ and a washer has area pi*((outer)^2 - (inner)^2) — the squares are subtracted, never the radii.
Slices are cut perpendicular to the axis of rotation, so a horizontal axis gives an integral in x and a vertical axis gives an integral in y, with the limits read off that same axis.
A radius is a distance from the axis of rotation, so spinning about the line $y = k$ turns every radius into a subtraction that must stay positive across the region.
A cross-section solid is not spun at all: the base supplies the side s, the named shape supplies a constant, and the volume is that constant times the integral of $s 2$ .

One idea: slice it, then add the slices up

Keeper Leiba keeps a potter's wheel at the back of the Integral Vault. Put a curve on the wheel, spin it, and the curve sweeps out a solid. The question this craft answers is always the same: how much space is inside it?

The whole method is one sentence. Cut the solid into thin slices, work out the area of one slice, and integrate.

$V =$ the integral from a to b of A(x) dx

That is it. A(x) is the area of the slice at position x, and dx is its thickness. Everything else in this lesson is just different pictures of what one slice looks like.

The first picture is a disc. Take the region under $y =$ f(x), between $x = a$ and $x = b$ and above the x-axis, and spin it about the x-axis. Slice the solid straight down at position x, perpendicular to the axis it was spun about. What you see is a circle, of radius equal to the height of the curve there:

$r =$ f(x) A(x) $= pi*r 2 = pi*(f(x)) 2$

$V = pi$ times the integral from a to b of $(f(x)) 2 dx$

Square the radius before integrating, not after. $pi*(f(x)) 2$ is not the same as $pi*f(x)$ , and it is not the same as (pi times the integral of $f(x)) 2$ either. The pi is a constant, so it can sit outside the integral sign and wait; the squaring cannot.

Two habits make every question in this craft easier.

Always slice perpendicular to the axis of rotation. Spinning about a horizontal axis means vertical slices and dx. Spinning about a vertical axis means horizontal slices and dy.
A radius is a distance from the axis, never a height above the x-axis, unless the x-axis happens to be the axis of rotation.

Washers: a disc with a hole in it

Spin the region between two curves and the solid comes out hollow. Slice it and you get a washer: a big circle with a small circle removed. Its area is the difference of the two circles.

A(x) $=$ pi*(outer)^2 - pi*(inner)^2 $=$ pi*((outer)^2 - (inner)^2)

$V = pi$ times the integral from a to b of ((outer)^2 - (inner)^2) dx

The outer radius belongs to the curve further from the axis of rotation; the inner radius belongs to the curve nearer it. Test one x in the middle of the interval if you are not sure which is which.

Subtract the squares. Never square the difference. This is the single most common slip in the topic, so it is worth seeing why it is wrong:

(outer)^2 - (inner)^2 is what a washer's area really is (outer - inner)^2 is the square of the gap between the curves

Those are different numbers. With outer $= 3$ and inner $= 1$ : the first gives 9 - $1 = 8$ , the second gives $22 = 4$ . The gap between the curves is the *width* of the washer, and a washer's area is not the square of its width — it is the big circle take away the small one.

When the region touches the axis of rotation somewhere, the inner radius there is 0 and the washer closes up into a solid disc. That is why a disc is really just a washer whose hole has shrunk to nothing, and why one formula covers both.

Choosing the variable, and spinning about a line

Two things change the setup, and neither one changes the method.

A vertical axis means an integral in y. Slices are always cut perpendicular to the axis of rotation, so spinning about the y-axis gives horizontal slices of thickness dy. Every radius then has to be written as a function of y, which usually means rearranging the boundary so x is on its own:

$y = x 2$ becomes $x =$ sqrt(y) $y =$ sqrt(x) becomes $x = y 2 y = 3x$ becomes $x = y/3$

The limits change too: they are now the lowest and highest y of the region, read off the vertical axis. Reaching for the x-values out of habit is the classic error here.

An axis that is a line $y = k$ means every radius is a subtraction. The radius is the distance from the line to the curve, so:

for a region below the line $y = k, a$ radius is k - (the curve)
for a region above the line $y =$ -k, a radius is (the curve) $+ k$

Check the sign by testing one $x: a$ radius must come out positive everywhere on the region. If your expression is negative somewhere, you subtracted the wrong way round.

One useful consequence: if the region does not touch the axis of rotation, the solid has a hole, so the slice is a washer even when the region is bounded by a single curve. The far edge of the region gives the outer radius, the near edge gives the inner one.

Solids built from cross-sections

Not every solid comes from a wheel. A solid can also be built on a flat base, with a shape of your choosing standing on every slice.

Here the base only supplies one number: the length s of the slice at position x, which is the height of the region there — top boundary minus bottom boundary. The named shape then turns that length into an area.

squares, side s on the base $A = s 2$ isosceles right triangles, a leg on $s A = (1/2)s 2$ equilateral triangles, side $s A =$ (sqrt(3)/4)s^2 semicircles, DIAMETER s on the base $A = (pi/8)s 2$ rectangles of height $2s A = 2s 2$

Every one of them is a constant multiplied by $s 2$ , so the work is always the same: integrate $s 2$ across the base and multiply by the shape's own constant at the end.

$V =$ (the shape's constant) times the integral from a to b of $s 2 dx$

The semicircle deserves a second look, because the wording matters. The base of a slice is the diameter, so the radius is $s/2$ , and half a disc of that radius has area $(1/2)pi*(s/2) 2 = (pi/8)s 2$ . Using s itself as the radius gives four times too much.

Watch where the pi is allowed to go. Nothing is being spun in this section, so a square, triangular or rectangular cross-section produces an answer with no pi in it at all — an ordinary fraction. A semicircular cross-section is the one shape here that is round, and it is the one answer that carries $a pi$ . An equilateral triangle carries a sqrt(3) instead, and the answer is left exactly like that.

The cone and the sphere, and reading an integral backwards

The formulas you already know are not separate facts. They are what this integral gives.

The cone. Take the straight line through the origin and the point $(h, r): y = (r/h)x$ . Spin it about the x-axis from $x = 0$ to $x = h$ .

$V = pi$ times the integral from 0 to h of $((r/h)x) 2 dx = pi*(r 2 /h 2)*(h 3 /3) = pi*r 2 *h/3$

The sphere. Take the upper half of the circle $x 2 + y 2 = R 2$ , which is $y =$ sqrt(R^2 - $x 2)$ , and spin it about the x-axis from -R to R. The square root disappears the moment it is squared:

$V = pi$ times the integral from -R to R of $(R 2$ - $x 2) dx = pi*(2R 3$ - $2R 3 /3) = 4*pi*R 3 /3$

The cylinder is the flattest case of all: $y = R$ is a constant, so the integral is $pi*R 2 *h$ . Stopping the sphere's integral at 0 instead of -R gives the hemisphere, $2*pi*R 3 /3$ , and a slanted line that does not pass through the origin gives a truncated cone.

These are worth doing once, because they turn the formula sheet into something you could rebuild from scratch — and because they are a free check. Whenever a solid of revolution happens to be a familiar shape, the school formula and the integral must agree.

Reading an integral backwards. Given a volume integral, you can usually say what it came from:

one squared bracket means a disc, so the region reaches all the way to the axis
two squared brackets subtracted means a washer, so the region stands clear of the axis
dx means slices stacked along the x-axis, so the axis of rotation is horizontal; dy means it is vertical
take the square root of the integrand and you are looking at the radius itself

So pi times the integral from 0 to 2 of (4 - $x 2) 2 dx$ has one bracket, $a dx$ and a radius of 4 - $x 2$ measured from the x-axis: it is the region under $y = 4$ - $x 2$ from $x = 0$ to $x = 2$ , spun about the x-axis. If it had been spun about the line $y = 4$ instead, the radius would have been 4 - (4 - $x 2) = x 2$ , and the integrand would have read $(x 2) 2$ .

Worked examples

Example 1

The region under $y =$ sqrt(x) from $x = 0$ to $x = 4$ , above the x-axis, is revolved about the x-axis. Find the volume. Give an exact answer in terms of pi.

The axis is horizontal, so slice vertically. Each slice is a disc of radius $r =$ sqrt(x).
Area of one slice $= pi*r 2 =$ pi*(sqrt(x))^2 $= pi*x$ . Squaring has removed the square root, which is why disc problems like square roots.
$V = pi$ times the integral from 0 to 4 of $x dx$ .
An antiderivative of x is $x 2 /2$ , and between the limits that is $16/2$ - $0 = 8$ .
$V = 8*pi. (A$ student who forgot the pi would hand in 8; the pi belongs to the area of the disc.)

Example 2

The region enclosed by $y = x$ and $y = x 2$ is revolved about the x-axis. Find the volume. Give an exact answer in terms of pi.

First the limits: $x = x 2$ gives x(1 - $x) = 0$ , so the curves meet at $x = 0$ and $x = 1$ .
Between those, test $x = 1/2$ : the line gives 0.5 and the parabola gives 0.25, so $y = x$ is the outer radius and $y = x 2$ is the inner one.
Area of a washer $= pi*((x) 2$ - $(x 2) 2) = pi*(x 2$ - $x 4)$ . Subtract the SQUARES, not the difference.
$V = pi$ times the integral from 0 to 1 of $(x 2$ - $x 4) dx$ , whose antiderivative is $x 3 /3$ - $x 5 /5$ .
At $x = 1$ that is $1/3$ - $1/5 = 2/15$ , and at $x = 0$ it is 0.
$V = 2*pi/15$ . Squaring the gap instead would have given pi times the integral of (x - $x 2) 2$ , which is $pi/30$ — a different solid altogether.

Example 3

The region under $y = x 2$ from $x = 0$ to $x = 2$ and above the x-axis is revolved about the y-axis. Find the volume. Give an exact answer in terms of pi.

The axis is vertical, so the slices are horizontal and the integral is in y with thickness dy.
The region runs from $y = 0$ up to $y = 4$ (the height of the curve at $x = 2)$ , so those are the limits.
At height y the slice runs from the parabola across to the line $x = 2$ . Rearranging $y = x 2$ gives $x =$ sqrt(y), so the outer radius is 2 and the inner radius is sqrt(y).
Area of the washer $= pi*((2) 2$ - (sqrt(y))^2) $= pi*(4$ - y).
$V = pi$ times the integral from 0 to 4 of (4 - $y) dy = pi*(16$ - $8) = 8*pi$ .
Check the shape: a cylinder of radius 2 and height 4 has volume $16*pi$ , and the solid is exactly half of it, which is what a parabola should carve away.

Example 4

The base of a solid is the region between $y = x$ and $y = x 2$ , from $x = 0$ to $x = 1$ . Every cross-section perpendicular to the x-axis is an equilateral triangle with one side on the base. Find the volume. Give an exact answer in terms of sqrt(3).

Nothing is being spun, so there is no pi anywhere in this answer.
The side sitting on the base at x is $s =$ (top) - (bottom) $= x$ - $x 2$ .
An equilateral triangle of side s has area (sqrt(3)/4)s^2, so first square the side: $s 2 = x 2$ - $2x 3 + x 4$ .
The integral from 0 to 1 of $(x 2$ - $2x 3 + x 4) dx = 1/3$ - $1/2 + 1/5 = 1/30$ .
$V =$ (sqrt(3)/4)*(1/30) $=$ sqrt(3)/120.
If the cross-sections had been squares instead, the answer would be the bare $1/30$ ; the shape of the slice only ever changes the constant in front.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

The region under $y =$ sqrt(x) from $x = 0$ to $x = 1$ , above the x-axis, is revolved about the x-axis. Find the volume of the solid. Give an exact answer in terms of pi.

Answer: $pi/2$

A slice at x is a disc of radius sqrt(x), so its area is pi*(sqrt(x))^2 $= pi*(x)$ .
$V = pi$ times the integral from 0 to 1 of $(x) dx$ .
An antiderivative of x is $(1/2)x 2$ .
Evaluating between the limits gives $1/2$ , so $V = pi/2$ .

Problem 2

Difficulty 3 of 5

The region enclosed by $y = x$ and $y = x 2 /2$ is revolved about the x-axis. They meet at $x = 0$ and $x = 2$ . Find the volume of the solid. Give an exact answer in terms of pi.

Answer: $16*pi/15$

On this stretch $y = x$ is the outer radius and $y = x 2 /2$ is the inner one.
Area of a washer at $x = pi*((x) 2$ - $(x 2 /2) 2) = pi*(-(1/4)x 4 + x 2)$ .
$V = pi$ times the integral from 0 to 2 of $(-(1/4)x 4 + x 2) dx$ .
An antiderivative is $-(1/20)x 5 + (1/3)x 3$ , and between the limits it gives $16/15$ .
$V = 16*pi/15$ .

Problem 3

Difficulty 4 of 5

The region under $y = x 2$ from $x = 0$ to $x = 3$ and above the x-axis is revolved about the y-axis. Find the volume of the solid. Give an exact answer in terms of pi.

Answer: $81*pi/2$

Slice horizontally, because the axis is vertical: $y = x 2$ rearranges to $x =$ sqrt(y), so a slice at height y is a washer with outer radius 3 and inner radius sqrt(y).
Area of a slice at height $y = pi*((3) 2$ - (sqrt(y))^2) $= pi*(-y + 9)$ .
$V = pi$ times the integral from 0 to 9 of (-y $+ 9) dy$ .
An antiderivative is $-(1/2)y 2 + 9y$ , and between the limits it gives $81/2$ .
$V = 81*pi/2$ .

Common mistakes

Squaring the gap instead of subtracting the squares: writing (outer - inner)^2 in place of (outer)^2 - (inner)^2. Those are different numbers, and only the second one is the area of a washer.
Leaving the pi off a solid of revolution, or putting one into a solid with square, triangular or rectangular cross-sections. Pi arrives with a circle and with nothing else — and in a semicircular cross-section the base is the diameter, so the area is $(pi/8)s 2$ rather than $(pi/2)s 2$ .
Integrating the radius without squaring it. A slice is an area, so the radius is squared before the integral sign ever appears.
Using the x-limits when the axis of rotation is vertical. Horizontal slices are stacked up the y-axis, so the limits are heights and every radius has to be rewritten in y.
Measuring a radius from the x-axis when the solid was spun about the line $y = k$ . The radius is the distance from that line, so it is k - (the curve), not the curve itself.

What you should be able to do

Set up and evaluate a disc-method volume about the x-axis or y-axis.
Use washers for the region between two curves, including revolution about a line $y = k$ .
Find the volume of a solid with square, semicircular or equilateral cross-sections on a given base.
Read a volume integral back to the region and axis that produced it.

Where this fits in the curriculum

Common Core

CHA-5.B
AP Calculus AB, Unit 8 — Calculate volumes of solids with known cross sections using definite integrals.
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.
CHA-5.C
AP Calculus AB, Unit 8 — Calculate volumes of solids of revolution using definite integrals: the disc and washer methods.
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.