Volume with Washer Method: Revolving Around Other Axes

AP Calculus · Applications of Integration

Washer Method: Revolving Around Other Axes

Two radii, one shift, applied to both. That is the entire difference.

key idea$R=\lvert\text{far}-\text{axis}\rvert,\quad r=\lvert\text{near}-\text{axis}\rvert$

Picture the region

Both radii are now measured from a shifted line. Everything else — the $\pi$, the squaring, the subtraction, the differential — is unchanged. Moving the axis is a bookkeeping problem, and the whole difficulty is doing that bookkeeping on two radii at once.

The region between $y=\sqrt{x}$ and $y=x$ on $[0,1]$, revolved about $y=-1$. Both radii now start at the rose line one unit below the axis: $R=\sqrt{x}+1$ reaches the far boundary and $r=x+1$ reaches the near one. Neither is the curve's own $y$-value any more.

Setting up the integral

Shifting both radii together

Draw the axis of revolution and both boundary curves. Label which curve is farther from the axis — that is $R$.
Write each radius as a distance from the axis, not from zero: $R=f-k$ and $r=g-k$ for an axis $y=k$ below the region, or $k-f$ and $k-g$ for one above it.
Apply the shift to both radii. Adjusting only one is the characteristic error of this topic.
Expand $R^{2}-r^{2}$ carefully. Each square produces a cross term, and the two cross terms rarely cancel.
Integrate with the differential matching the slice direction, over the interval where the two curves bound the region.

Axis	Far curve $f$, near curve $g$	$R$	$r$
$y=0$	both above the axis	$f$	$g$
$y=k$, below the region	both above the line	$f-k$	$g-k$
$y=k$, above the region	both below the line	$k-f$	$k-g$
$x=k$, left of the region	both to the right	$f(y)-k$	$g(y)-k$
$x=k$, right of the region	both to the left	$k-f(y)$	$k-g(y)$

Notice that row 1 is just row 2 with $k=0$. There is only one formula here — distance from the axis — and the coordinate axes are the special case where the subtraction is invisible.

Worked examples

EX 1

About a line below the region

$y=-1$

The region between $y=\sqrt{x}$ and $y=x$ on $[0,1]$ is revolved about $y=-1$. Find the volume.

Sort the boundaries

For $0\lt x\lt1$, $\sqrt{x}\gt x$, so $\sqrt{x}$ is the far curve from an axis below.

Shift both radii

$R=\sqrt{x}+1$ and $r=x+1$.

Square each

$R^{2}=x+2\sqrt{x}+1$ and $r^{2}=x^{2}+2x+1$.

Subtract

$R^{2}-r^{2}=x+2\sqrt{x}+1-x^{2}-2x-1=2\sqrt{x}-x-x^{2}$.

Integrate

$\pi\left[\dfrac43x^{3/2}-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{0}^{1}=\pi\left(\dfrac43-\dfrac12-\dfrac13\right)=\dfrac{\pi}{2}$.

EX 2

About a line above the region

$y=2$

The same region is revolved about $y=2$. Find the volume.

Re-sort the boundaries

From above, the farther curve is now $y=x$, since it sits lower. The roles swap.

Write the radii

$R=2-x$ and $r=2-\sqrt{x}$.

Square each

$R^{2}=4-4x+x^{2}$ and $r^{2}=4-4\sqrt{x}+x$.

Subtract

$R^{2}-r^{2}=-4x+x^{2}+4\sqrt{x}-x=x^{2}-5x+4\sqrt{x}$.

Integrate

$\pi\left[\dfrac{x^{3}}{3}-\dfrac{5x^{2}}{2}+\dfrac83x^{3/2}\right]_{0}^{1}=\pi\left(\dfrac13-\dfrac52+\dfrac83\right)=\dfrac{\pi}{2}$. Equal to Example 1 here by symmetry of the two distances — but check, never assume.

EX 3

About a vertical line

$x=2$

The region bounded by $y=x^{2}$, $y=0$ and $x=1$ is revolved about the line $x=2$. Set up the volume.

Choose the slice

A vertical axis needs horizontal slices, so integrate in $dy$.

Rewrite the boundaries

$y=x^{2}$ becomes $x=\sqrt{y}$; the other boundary is the vertical line $x=1$, and $y$ runs from $0$ to $1$.

Sort by distance from $x=2$

$x=\sqrt{y}$ is farther from $2$ than $x=1$ is, for $0\lt y\lt1$.

Write both radii

$R=2-\sqrt{y}$ and $r=2-1=1$.

Set up

$V=\pi\displaystyle\int_{0}^{1}\left[\left(2-\sqrt{y}\right)^{2}-1\right]dy=\pi\int_{0}^{1}\left(3-4\sqrt{y}+y\right)dy=\pi\left(3-\dfrac83+\dfrac12\right)=\dfrac{5\pi}{6}$.

EX 4

Shifting only one radius

the diagnostic error

A student revolving the Example 1 region about $y=-1$ writes $\pi\displaystyle\int_{0}^{1}\left[\left(\sqrt{x}+1\right)^{2}-x^{2}\right]dx$. Identify the mistake.

Read what was written

$R$ was shifted to $\sqrt{x}+1$, but $r$ was left as $x$.

Name the inconsistency

The two radii are being measured from different lines — one from $y=-1$, the other from $y=0$.

Compute the wrong value

$\pi\displaystyle\int_{0}^{1}\left(x+2\sqrt{x}+1-x^{2}\right)dx=\pi\left(\dfrac12+\dfrac43+1-\dfrac13\right)=\dfrac{5\pi}{2}$.

Compare with the correct $\dfrac{\pi}{2}$

Five times too large — the phantom hole is far too small, so the solid is far too solid.

State the check

Both radii must contain the same $k$. If only one of them mentions the shift, the set-up is wrong.

Watch out

four checks before integrating

Shift both radii by the same $k$. One shifted and one not is the signature error here.
Which curve is ‘outer’ can swap when the axis moves from below the region to above it.
Expand each square separately, then subtract. Cross terms survive and matter.
A vertical axis means $dy$ and every boundary rewritten as $x=g(y)$.

Both radii carry the shift

$R=f-k$ and $r=g-k$. Applying $k$ to only one measures the two boundaries from different lines and inflates the volume badly.

The outer curve can change when the axis moves

From below, the higher curve is farther; from above, the lower one is. Re-sort the boundaries whenever the axis moves across the region.

Expand before subtracting

$\left(f-k\right)^{2}-\left(g-k\right)^{2}$ has cross terms in both squares. Cancelling them by inspection is where sign errors creep in.

A sanity check on size

Moving the axis farther away should make the volume larger. If your answer shrinks when the axis retreats, re-examine the radii.

Course map. Topic 8.12 · LO CHA-5.C — calculate volumes of solids of revolution using definite integrals.