Volume with Washer Method: Revolving Around the x- or y-Axis

AP Calculus · Applications of Integration

Volume with the Washer Method

The gap between the region and the axis becomes the hole in every slice.

key idea$V=\pi\int_a^b\Big[R^{2}-r^{2}\Big]dx$

Picture the region

When the region does not touch the axis, each slice sweeps out a washer — a disc with a bite taken out of the middle. Its area is the big circle minus the small one, and the gap between the region and the axis is exactly what becomes the hole.

the gap becomes the hole

one washer: area $\pi R^{2}-\pi r^{2}$

The region between $y=\sqrt{x}$ and $y=x$ on $[0,1]$, revolved about the $x$-axis. The outer radius $R$ reaches the far boundary and the inner radius $r$ reaches the near one. Both are measured from the axis, and the ring between them is the slice.

[Figure: Revolving a region that does not touch the axis leaves a hole. Each slice is a washer, and its area is the outer disc minus the inner one — never (R-r)2.]

Revolving a region that does not touch the axis leaves a hole. Each slice is a washer, and its area is the outer disc minus the inner one — never $(R-r)^2$.

Setting up the integral

Two radii instead of one

Draw the axis and the region. If any daylight shows between them, you need washers rather than discs.
Identify the outer boundary — the curve farther from the axis — and call its distance $R$.
Identify the inner boundary, the curve nearer the axis, and call its distance $r$.
Write $\pi\left(R^{2}-r^{2}\right)$. Square each radius separately, then subtract. $(R-r)^{2}$ is a different and wrong quantity.
Integrate over the interval where the two boundaries enclose the region — usually between their intersections.

$R$distance to the farther curve

$r$distance to the nearer curve

Ordersquare, then subtract

Expression	Value at $R=3$, $r=1$	Correct?
$\pi\left(R^{2}-r^{2}\right)$	$8\pi$	yes
$\pi(R-r)^{2}$	$4\pi$	no
$\pi R^{2}-\pi r^{2}$	$8\pi$	yes — the same thing
$\pi\displaystyle\int R^{2}-\pi\int r^{2}$	—	yes — two integrals, same result

Row 2 is the error this topic is built to catch: the area of a ring is not the area of a circle whose radius is the gap. Squaring must happen before subtracting, every time.

Worked examples

EX 1

About the $x$-axis

$y=\sqrt{x}$ and $y=x$

The region between $y=\sqrt{x}$ and $y=x$ is revolved about the $x$-axis. Find the volume.

Find the limits

$\sqrt{x}=x$ gives $x=x^{2}$, so $x=0$ and $x=1$.

Sort the radii

At $x=\tfrac14$: $\sqrt{x}=0.5$ and $x=0.25$. So $R=\sqrt{x}$ (outer) and $r=x$ (inner).

Write the integrand

$\pi\left[\left(\sqrt{x}\right)^{2}-x^{2}\right]=\pi\left(x-x^{2}\right)$.

Integrate

$\pi\left[\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{0}^{1}$.

Evaluate

$\pi\left(\dfrac12-\dfrac13\right)=\dfrac{\pi}{6}$.

EX 2

About the $y$-axis

rewrite both boundaries

The same region is revolved about the $y$-axis instead. Find the volume.

Switch the slice direction

A vertical axis means horizontal slices and a $dy$ integral.

Rewrite both curves

$y=\sqrt{x}$ becomes $x=y^{2}$; $y=x$ becomes $x=y$.

Sort the radii

At $y=\tfrac12$: $x=y=0.5$ and $x=y^{2}=0.25$. So $R=y$ and $r=y^{2}$.

Set up

$V=\pi\displaystyle\int_{0}^{1}\left(y^{2}-y^{4}\right)dy$.

Evaluate

$\pi\left[\dfrac{y^{3}}{3}-\dfrac{y^{5}}{5}\right]_{0}^{1}=\pi\left(\dfrac13-\dfrac15\right)=\dfrac{2\pi}{15}$. A different axis gives a different solid, so a different number is expected.

EX 3

The classic wrong turn

$(R-r)^{2}$

A student writes $V=\pi\displaystyle\int_{0}^{1}\left(\sqrt{x}-x\right)^{2}dx$ for Example 1. Show what this computes and why it is wrong.

Expand what was written

$\left(\sqrt{x}-x\right)^{2}=x-2x^{3/2}+x^{2}$.

Integrate it

$\pi\left[\dfrac{x^{2}}{2}-\dfrac45x^{5/2}+\dfrac{x^{3}}{3}\right]_{0}^{1}=\pi\left(\dfrac12-\dfrac45+\dfrac13\right)=\dfrac{\pi}{30}$.

Compare

$\dfrac{\pi}{30}$ against the correct $\dfrac{\pi}{6}$ — five times too small.

Say what it actually computes

The volume of a solid whose slices are discs of radius equal to the gap — a genuinely different solid.

State the rule

$R^{2}-r^{2}\neq(R-r)^{2}$. Squaring is not distributive over subtraction, and the difference is not small.

EX 4

When a gap appears partway

check the whole interval

The region between $y=x^{2}$ and $y=2$ on $[0,\sqrt2]$ is revolved about the $x$-axis. Set up the volume.

Locate the region

It sits between the parabola below and the horizontal line above.

Check the gap

At every $x$ in $(0,\sqrt2)$ the parabola is strictly above the axis, so there is always a hole.

Assign the radii

The line $y=2$ is farther from the axis, so $R=2$; the parabola is nearer, so $r=x^{2}$.

Write the integral

$V=\pi\displaystyle\int_{0}^{\sqrt2}\left(4-x^{4}\right)dx$.

Evaluate

$\pi\left[4x-\dfrac{x^{5}}{5}\right]_{0}^{\sqrt2}=\pi\left(4\sqrt2-\dfrac{4\sqrt2}{5}\right)=\dfrac{16\sqrt2\,\pi}{5}$.

Watch out

Square each radius, then subtract

$R^{2}-r^{2}$, never $(R-r)^{2}$. This one error accounts for more lost points on volume questions than anything else.

$R$ is the farther curve, not the upper one

Above and below are irrelevant. Sort the two boundaries by distance from the axis, testing an interior point if the picture is unclear.

Both radii switch when the axis does

Revolving about the $y$-axis means rewriting $R$ and $r$ as functions of $y$ and integrating $dy$ — a full change of set-up.

If the region touches the axis, use discs

A washer with $r=0$ is a disc. Forcing an inner radius where there is no hole adds a term that should not be there.

Course map. Topic 8.11 · LO CHA-5.C — calculate volumes of solids of revolution using definite integrals.