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Calculus AB · Unit 8 · 8.10
AP Calculus · Applications of Integration
Disc Method: Revolving Around Other Axes
Same coin, same integral. Only the measurement of R changes — by one subtraction.
key idea$R=\big\lvert\, \text{boundary}-\text{axis}\,\big\rvert\ \Rightarrow\ V=\pi\int R^{2}$
Picture the region
Nothing about the method changes when the axis moves off the coordinate axes. The disc is still a coin of area $\pi R^{2}$. The only new work is measuring $R$ from the new line instead of from zero — a subtraction.
The region between $y=x^{2}$ and the line $y=4$, revolved about that line. The rose dashes mark the axis; the faint copy above it is where the region sweeps. The radius of the amber disc runs from the axis down to the parabola, so $R=4-x^{2}$ — not $x^{2}$.[Figure: Revolving about a line that is not an axis. The radius is the distance from that line to the curve, which is why the integrand carries a subtraction the x-axis case does not.]
Revolving about a line that is not an axis. The radius is the distance from that line to the curve, which is why the integrand carries a subtraction the $x$-axis case does not.Setting up the integral
Measuring from a shifted axis
- Draw the axis of revolution as an actual line on your sketch. Most errors here are errors of picture, not of algebra.
- Decide whether the region lies above or below the axis. That decides the order of the subtraction.
- Write $R$ as the positive gap: $R=f(x)-k$ if the curve is above the line $y=k$, and $R=k-f(x)$ if it is below.
- Check the region touches the axis along the whole interval. If a gap opens up anywhere, you need two radii, not one.
- Integrate $\pi R^{2}$ with the differential matching the slice direction — $dx$ for a horizontal axis, $dy$ for a vertical one.
| Axis | Region relative to it | Radius | Differential |
|---|
| $y=k$ | curve above the line | $f(x)-k$ | $dx$ |
| $y=k$ | curve below the line | $k-f(x)$ | $dx$ |
| $x=k$ | curve to the right | $g(y)-k$ | $dy$ |
| $x=k$ | curve to the left | $k-g(y)$ | $dy$ |
| $y=0$ or $x=0$ | either | $f$ or $g$ alone | matching |
Every row is $\lvert\text{boundary}-\text{axis}\rvert$. Writing the subtraction so the result is positive is not strictly necessary — squaring will fix the sign — but it keeps the algebra readable and the set-up checkable.
Axis of revolutionThe fixed line the region spins about. Every radius is measured perpendicular to it.
Radius $R$The distance from the axis to the outer boundary. Always non-negative.
DiscA slice with no hole — possible only when the region touches the axis.
WasherA slice with a hole, needed when a gap separates the region from the axis.
Slice directionAlways perpendicular to the axis: vertical for $y=k$, horizontal for $x=k$.
Worked examples
EX 1
About a horizontal line above the region
$y=4$The region between $y=x^{2}$ and $y=4$ is revolved about the line $y=4$. Find the volume.
1
Find the limits
$x^{2}=4$ gives $x=\pm2$.
2
Locate the region
The parabola lies below the line on $(-2,2)$, and the region reaches the line at both ends.
3
Write the radius
$R=4-x^{2}$, measured downward from the axis.
4
Set up
$V=\pi\displaystyle\int_{-2}^{2}\left(4-x^{2}\right)^{2}dx=\pi\int_{-2}^{2}\left(16-8x^{2}+x^{4}\right)dx$.
5
Evaluate using symmetry
$2\pi\left[16x-\dfrac{8x^{3}}{3}+\dfrac{x^{5}}{5}\right]_{0}^{2}=2\pi\left(32-\dfrac{64}{3}+\dfrac{32}{5}\right)=\dfrac{512\pi}{15}$.
EX 2
About a horizontal line below the region
$y=-1$The region between $y=\sqrt{x}$ and $y=-1$, for $0\le x\le4$, is revolved about $y=-1$. Find the volume.
1
Locate the region
The curve lies above the axis throughout, and the region extends down to it.
2
Write the radius
$R=\sqrt{x}-(-1)=\sqrt{x}+1$.
3
Square it
$R^{2}=x+2\sqrt{x}+1$ — the cross term is the part that is easiest to lose.
4
Set up
$V=\pi\displaystyle\int_{0}^{4}\left(x+2x^{1/2}+1\right)dx$.
5
Evaluate
$\pi\left[\dfrac{x^{2}}{2}+\dfrac43x^{3/2}+x\right]_{0}^{4}=\pi\left(8+\dfrac{32}{3}+4\right)=\dfrac{68\pi}{3}$.
EX 3
About a vertical line
$x=3$The region bounded by $x=y^{2}$ and $x=3$, for $-\sqrt3\le y\le\sqrt3$, is revolved about the line $x=3$. Find the volume.
1
Choose the slice
A vertical axis needs horizontal slices, so the integral is in $dy$.
2
Find the limits
$y^{2}=3$ gives $y=\pm\sqrt3$.
3
Write the radius
The parabola is to the left of $x=3$, so $R=3-y^{2}$.
4
Set up
$V=\pi\displaystyle\int_{-\sqrt3}^{\sqrt3}\left(3-y^{2}\right)^{2}dy=2\pi\int_{0}^{\sqrt3}\left(9-6y^{2}+y^{4}\right)dy$.
5
Evaluate
$2\pi\left[9y-2y^{3}+\dfrac{y^{5}}{5}\right]_{0}^{\sqrt3}=2\pi\left(9\sqrt3-6\sqrt3+\dfrac{9\sqrt3}{5}\right)=\dfrac{48\sqrt3\,\pi}{5}$.
EX 4
Why the shift cannot be ignored
compare two axesCompare revolving the region under $y=\sqrt{x}$ on $[0,4]$ about $y=0$ with revolving it about $y=-1$.
1
About $y=0$
$R=\sqrt{x}$, so $V=\pi\displaystyle\int_{0}^{4}x\,dx=8\pi$.
2
About $y=-1$
$R=\sqrt{x}+1$, so $R^{2}=x+2\sqrt{x}+1$.
3
Integrate
$V=\pi\left(8+\dfrac{32}{3}+4\right)=\dfrac{68\pi}{3}\approx71.2$.
4
Compare
The second is more than eight times larger, from a shift of one unit.
5
Explain
Squaring amplifies the shift, and the whole solid is now swung around a farther line. This is why $R$ must be rewritten rather than adjusted at the end.
Watch out
Rewrite $R$; do not patch the answer
Shifting the axis changes the integrand, not the answer at the end. There is no correction factor you can apply to a volume computed about the wrong line.
Expand the square fully
$\left(\sqrt{x}+1\right)^{2}=x+2\sqrt{x}+1$. Dropping the middle term is the most frequent arithmetic error on shifted-axis problems.
Match the differential to the axis
A vertical axis $x=k$ needs horizontal slices and $dy$, with every boundary rewritten as $x=g(y)$.
Confirm the region reaches the axis
The disc method assumes no hole. If the region stops short of the line anywhere on the interval, the slice is a washer instead.