MISSION · Calculus
Volumes of Revolution
Spin a region round a line and integrate the circles it sweeps out
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
Slice the solid into coins, note that each coin is a cylinder of area \(\pi r^2\) and thickness \(dx\), and integrate.
Building the solid
Take the region under \(y = \sqrt{x}\) from \(0\) to \(4\) and spin it a full turn about the \(x\)-axis. It sweeps out a solid — something like a horn lying on its side.
Now cut it at some value of \(x\), perpendicular to the axis. The cross-section is a circle, and its radius is exactly the height of the original curve there: \(r = \sqrt{x}\).
A slice of thickness \(dx\) is a coin of volume \(\pi r^2\,dx\). Adding the coins gives
\[V = \int_a^b \pi \left[f(x)\right]^2 dx\]For our horn:
\[V = \int_0^4 \pi (\sqrt{x})^2 dx = \pi \int_0^4 x\,dx = \pi \left[\tfrac{x^2}{2}\right]_0^4 = 8\pi\]This is the disk method, and it is one idea repeated: find the radius as a function of position, square it, multiply by \(\pi\), integrate.
Washers, when there is a hole
If the region does not touch the axis, spinning it leaves a tunnel down the middle. Each slice is then a washer — a disk with a smaller disk removed:
\[V = \int_a^b \pi \left( R_{\text{outer}}^2 - R_{\text{inner}}^2 \right) dx\]The trap is subtracting in the wrong place. You subtract the areas, so both radii get squared first. \(R^2 - r^2\) is not \((R - r)^2\), and mixing them up is the single most common error in this topic.
Spinning the region between \(y = x\) and \(y = x^2\) about the \(x\)-axis from 0 to 1, \(y = x\) is the outer curve and \(y = x^2\) the inner, so \(V = \pi\int_0^1 (x^2 - x^4)\,dx = \frac{2\pi}{15}\).
Spinning about a different line
Nothing changes except the radius. The radius is always the distance from the axis of rotation to the curve, so read it off the picture rather than reaching for a formula.
- About the \(x\)-axis: \(r = f(x)\).
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About \(y = 2\): $$r = f(x) - 2 $$. - About the \(y\)-axis: rewrite the curve as \(x = g(y)\) and slice horizontally, so \(V = \int \pi [g(y)]^2 dy\).
The slices must be perpendicular to the axis of rotation. Spinning about a horizontal line means vertical slices and \(dx\); spinning about a vertical line means horizontal slices and \(dy\).
Shells, when slicing the other way is awkward
Sometimes rewriting \(y = f(x)\) as \(x = g(y)\) is unpleasant or impossible. The shell method avoids it by slicing parallel to the axis instead, peeling the solid like an onion. Each shell is a thin cylinder unrolled into a rectangle:
\[V = \int_a^b 2\pi \, (\text{radius}) (\text{height}) \, dx\]Both methods give the same answer whenever both apply. Choose whichever keeps the algebra clean: if the region is easy to describe in \(x\) but you are spinning about a vertical axis, shells save you the rearrangement.
A sanity check that catches real errors
Before integrating anything, picture the solid and estimate it as a cylinder. Our horn sits inside a cylinder of radius 2 and length 4, volume \(16\pi\), and it visibly fills about half of it. Getting \(8\pi\) is therefore believable; getting \(100\pi\) or \(0.3\pi\) would not be.
A cone of radius \(r\) and height \(h\) is a third of its cylinder, and a sphere is two thirds of its cylinder. Both facts fall straight out of this method, and both are worth recomputing once as practice.
Common mix-ups
- Forgetting to square the radius. The integrand is \(\pi r^2\), not \(\pi r\).
- Subtracting radii instead of areas in the washer method.
- Using \(dx\) when spinning about a vertical line. Slices must cut the axis at right angles.
- Losing the \(\pi\). It multiplies the whole integral, so factor it out at the start.
- Wrong limits after switching to \(dy\). The limits become \(y\) values, not \(x\) ones.
Check yourself
Spin \(y = \sqrt{x}\) on \([0,4]\) about the \(x\)-axis, say what the disk-method integrand is \(\pi\) times, and spin the line \(y = 2\) on \([0,5]\).
\(\pi\int_0^4 x\,dx = 8\pi\). The integrand is \(\pi\) times the radius squared. And a constant radius 2 over a length 5 is a cylinder: \(\pi(2^2)(5) = 20\pi\).
LEVEL 3 · REFLECT
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