How do I know whether to integrate in x or in y?

Match the differential to the axis. If you revolve around a horizontal line (y=k), the washers are stacked horizontally, so integrate in x with vertical radii. If you revolve around a vertical line (x=h), integrate in y with horizontal radii. Cross sections must always be perpendicular to the axis.

What offset do I add when the axis is y = k?

Measure the vertical distance from each boundary curve to the line. If the axis is below the region, radius =curve-k; if the axis is above, radius =k-curve. Pick whichever makes the distance positive, then square it.

Why can't I just use (R - r) squared?

Because the area of a washer is the outer disc area minus the inner disc area: R^2- r^2, not (R-r)^2. Squaring the difference ignores the ring geometry and gives the wrong volume. Always square R and r separately before subtracting.

What if the axis of revolution cuts through the region?

Then measuring a single outer and inner radius fails, because part of the region is on each side. You typically split the integral or reconsider the geometry; on the AP exam, axes are usually chosen so the region lies entirely on one side, giving clean washers or discs.

AP-CALCBC-8.12

U8.12 Washer Method: Revolving Around Other Axes

Master the washer method when revolving a region around a horizontal or vertical line other than an axis. Learn to write R and r with offsets for AP Calculus BC.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U8.12 Washer Method: Revolving Around Other Axes, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U8.12 Washer Method: Revolving Around Other Axes free

What this lesson covers

In U8.9 you learned the disc and washer methods for revolving a region around the x- or y-axis. But the AP exam loves to shift the axis to a line like

y=-2

x=5

. The mechanics of

V=\pi\int (R^2-r^2)\,dx

don't change — what changes is how you measure the radii.

The entire skill in 8.12 is turning "distance from a curve to a shifted axis" into an algebraic expression. Once you can write the outer radius

R

and inner radius

r

as distances with the correct offset, the integral is routine. This lesson shows you how to set up those radii carefully, avoid sign traps, and integrate confidently.

Radii as Distances From a Shifted Axis

The washer method computes volume as

V=\pi\int_a^b \left(R^2-r^2\right)\,dx

where

R

is the distance from the axis of revolution to the farther boundary of the region, and

r

is the distance to the nearer boundary. When the axis is a coordinate axis, these distances are just function values. When the axis moves, you must build the distance with an offset.

For a horizontal axis

y=k

, distances are measured vertically. The distance from a point on curve

y=f(x)

to the line is

|f(x)-k|

. So each radius has the form

\text{(curve)}-k

k-\text{(curve)}

, whichever is positive.

For a vertical axis

x=h

, distances are measured horizontally, giving radii of the form

|g(y)-h|

.

The most common mistake is to plug the bare function into

R

r

and forget the offset. Always ask: how far is this boundary from the line I'm spinning around? That distance — never the raw function value — is the radius.

A Reliable Setup Procedure

Use this sequence on every problem to keep the offsets straight.

Step	What to do
1	Sketch the region and the axis of revolution.
2	Decide if slices are vertical ( $dx$ , horizontal axis) or horizontal ( $dy$ , vertical axis). Slices must be perpendicular to the axis.
3	Identify which boundary is farther from the axis ( $R$ ) and which is nearer ( $r$ ).
4	Write each radius as a positive distance: axis-minus-curve or curve-minus-axis.
5	Integrate $\pi\int (R^2-r^2)$ over the correct bounds.

Key orientation rule: if you revolve around a horizontal line, your washers are stacked in the

x

-direction, so integrate with respect to

x

and radii are vertical distances. If you revolve around a vertical line, integrate with respect to

y

and radii are horizontal distances. Matching the differential to the axis prevents the single most common error on this topic.

Choosing R and r When the Axis Is Below, Above, or Beside the Region

The offset formula depends on where the axis sits relative to the region.

Suppose the region lies between

y=f(x)

(top) and

y=g(x)

(bottom), with

f\ge g

.

If the axis

y=k

is below the whole region (

k\le g

), the top curve is farther, so

R=f(x)-k,\qquad r=g(x)-k.

If the axis

y=k

is above the whole region (

k\ge f

), the bottom curve is now farther from the axis, so

R=k-g(x),\qquad r=k-f(x).

Notice the roles of

f

and

g

swap depending on which side the axis is on. This is why sketching matters — the curve closer to the axis always gives the smaller radius

r

.

A useful check: the difference

R-r

should equal the width of the region,

f(x)-g(x)

, regardless of the offset

k

. If your

R-r

doesn't simplify to the gap between the curves, you set up a radius incorrectly. This invariance is a quick error-catcher on exam day.

How the Exam Tests This

On the AP exam, 8.12 shows up in two ways. On the calculator-active section, you may be asked to set up and numerically evaluate a volume where the axis is

y=-1

x=3

; you get credit for a correct integrand and bounds even before evaluating, so write the integral cleanly.

On the no-calculator section, expect friendly functions (lines, simple parabolas) so the algebra of

(R^2-r^2)

is doable by hand. Graders look for: correct radii with offsets, correct bounds, the factor of

\pi

, and correct handling of the

R^2-r^2

expansion.

A frequent trap is

(R-r)^2 \ne R^2 - r^2

. You must square each radius separately, then subtract. Another trap: forgetting that when the axis passes through or touches the region, you may not need a washer at all (it becomes discs), or the setup may require splitting the integral. Reading whether the axis is outside the region tells you whether you truly have a hole.

Key terms

Washer method.: A technique for volumes of revolution that uses cross sections shaped like washers (rings), with volume $\pi\int (R^2-r^2)\,dx$ or $dy$ .
Outer radius R.: The distance from the axis of revolution to the boundary of the region that is farther away.
Inner radius r.: The distance from the axis of revolution to the boundary of the region that is nearer, creating the hole in the washer.
Offset.: The constant added or subtracted when the axis is a line like $y=k$ or $x=h$ , converting a function value into a true distance.
Axis of revolution.: The fixed line about which the region is rotated to generate the solid; here it is a horizontal or vertical line other than a coordinate axis.
Perpendicular slicing.: The rule that washer cross sections must be taken perpendicular to the axis of revolution, which determines whether you integrate in $x$ or $y$ .

Worked example

Let

R

be the region bounded by

y=x^2

and

y=2x

. Find the volume of the solid formed when

R

is revolved about the line

y=-1

First find intersections:

x^2=2x\Rightarrow x^2-2x=0\Rightarrow x=0

x=2

. On

[0,2]

, the line

y=2x

is above the parabola

y=x^2

(test

x=1

2>1

).

The axis

y=-1

lies below the region, so the top curve

y=2x

is farther from the axis and the bottom curve

y=x^2

is nearer. Add the offset of

1

to each distance:

R=2x-(-1)=2x+1,\qquad r=x^2-(-1)=x^2+1.

Check:

R-r=(2x+1)-(x^2+1)=2x-x^2

, which equals the region's width. Good.

Set up the volume:

V=\pi\int_0^2\left[(2x+1)^2-(x^2+1)^2\right]dx.

Expand:

(2x+1)^2=4x^2+4x+1

and

(x^2+1)^2=x^4+2x^2+1

. Subtract:

4x^2+4x+1-x^4-2x^2-1=-x^4+2x^2+4x.

Integrate:

\pi\int_0^2(-x^4+2x^2+4x)\,dx=\pi\left[-\frac{x^5}{5}+\frac{2x^3}{3}+2x^2\right]_0^2.

Evaluate at

2

-\frac{32}{5}+\frac{16}{3}+8=-6.4+5.333\ldots+8=6.9333\ldots=\frac{104}{15}.

V=\frac{104\pi}{15}\approx 21.78.

Practice questions

The region bounded by

y=\sqrt{x}

y=0

, and

x=4

is revolved about the line

y=3

. Which integral gives the volume of the resulting solid?

$\pi\int_0^4\left[(3-\sqrt{x})^2-(3)^2\right]dx$
$\pi\int_0^4\left[(3)^2-(3-\sqrt{x})^2\right]dx$
$\pi\int_0^4\left[(3+\sqrt{x})^2-(3)^2\right]dx$
$\pi\int_0^4\left[(\sqrt{x})^2-(3)^2\right]dx$

Answer: $\pi\int_0^4\left[(3)^2-(3-\sqrt{x})^2\right]dx$

The axis

y=3

is above the region (which sits between

y=0

and

y=\sqrt{x}

, and

\sqrt{x}\le 2<3

[0,4]

). The farther boundary is

y=0

, at distance

3-0=3

, so

R=3

. The nearer boundary is

y=\sqrt{x}

, at distance

3-\sqrt{x}

, so

r=3-\sqrt{x}

. Thus

V=\pi\int_0^4[(3)^2-(3-\sqrt{x})^2]\,dx

The region bounded by

x=y^2

and

x=4

is revolved about the vertical line

x=6

. Set up, but do not evaluate, an integral for the volume.

Answer: $V=\pi\int_{-2}^{2}\left[(6-y^2)^2-(6-4)^2\right]dy$

Because the axis is vertical, slice horizontally and integrate in

y

. The curves meet where

y^2=4

, i.e.

y=\pm 2

. The region lies between

x=y^2

(left) and

x=4

(right). The axis

x=6

is to the right of the region, so the farther boundary is the left curve

x=y^2

, giving

R=6-y^2

, and the nearer boundary is

x=4

, giving

r=6-4=2

. Hence

V=\pi\int_{-2}^{2}[(6-y^2)^2-2^2]\,dy

A region has top curve

f(x)

and bottom curve

g(x)

with

f\ge g\ge 0

, and is revolved about

y=-c

for some

c>0

. Explain why

R-r

does not depend on

c

, and why that is a useful check.

Answer: $R-r=f(x)-g(x)$ , independent of $c$ .

With the axis below the region,

R=f(x)+c

and

r=g(x)+c

. Subtracting, the offsets cancel:

R-r=(f+c)-(g+c)=f-g

. This equals the vertical width of the region regardless of how far the axis is shifted. It is a useful check because if your computed

R-r

does not reduce to the gap between the two curves, you have made a sign or offset error in one of the radii.

FAQ

How do I know whether to integrate in x or in y?: Match the differential to the axis. If you revolve around a horizontal line ( $y=k$ ), the washers are stacked horizontally, so integrate in $x$ with vertical radii. If you revolve around a vertical line ( $x=h$ ), integrate in $y$ with horizontal radii. Cross sections must always be perpendicular to the axis.
What offset do I add when the axis is y = k?: Measure the vertical distance from each boundary curve to the line. If the axis is below the region, radius $=\text{curve}-k$ ; if the axis is above, radius $=k-\text{curve}$ . Pick whichever makes the distance positive, then square it.
Why can't I just use (R - r) squared?: Because the area of a washer is the outer disc area minus the inner disc area: $\pi R^2-\pi r^2$ , not $\pi(R-r)^2$ . Squaring the difference ignores the ring geometry and gives the wrong volume. Always square $R$ and $r$ separately before subtracting.
What if the axis of revolution cuts through the region?: Then measuring a single outer and inner radius fails, because part of the region is on each side. You typically split the integral or reconsider the geometry; on the AP exam, axes are usually chosen so the region lies entirely on one side, giving clean washers or discs.

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The Crimsora tutor teaches U8.12 Washer Method: Revolving Around Other Axes live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

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