Volumes of revolution from parametric curves

Sketch, identify sections, and substitute

For a region bounded by a parametric curve, first understand the curve and its branches. Next, cut the solid perpendicular to its axis of revolution and write its cross-sectional area. Finally, substitute the parameter into the resulting integral.

For rotation about the $x$-axis, a section may be a disk of area $\pi R(x)^2$ or a washer of area $\pi\{R(x)^2-r(x)^2\}$. A hole occurs when the original section does not meet the axis. Do not automatically use a disk for every $x$.

Parameter substitution uses $dx=x'(t)\,dt$, including its sign. Branch integrals sometimes join into one parameter integral, but this depends on the geometry and orientation. There is no general guarantee that arbitrary sections can always be merged.

See parametric sketches and parametric areas for the supporting calculations.

Example 1: revolving a cardioid

Find the volume swept out by rotating the region enclosed by

$$x=\cos t(1+\cos t),\qquad y=\sin t(1+\cos t),\qquad0\le t\le2\pi$$

once about the $x$-axis.

The region is symmetric about that axis. Its upper half alone generates the whole solid, so do not double the volume after rotating it.

On $0\le t\le\pi$,

$$x'=-\sin t(1+2\cos t),\qquad y'=\cos t+\cos2t.$$

The point runs from $(2,0)$ through $(3/4,3\sqrt3/4)$ at $t=\pi/3$, then to the leftmost point $(-1/4,\sqrt3/4)$ at $t=2\pi/3$, and finally to $(0,0)$.

Call the height on $0\le t\le2\pi/3$ $y_+$, and that on $2\pi/3\le t\le\pi$ $y_-$. For $-1/4<x<0$ the section is a washer; for $0\le x\le2$ it is a disk. Thus

$$V=\pi\int_{-1/4}^{2}y_+^2\,dx -\pi\int_{-1/4}^{0}y_-^2\,dx.$$

Notice that the inner-radius integral ends at zero, not at two. Correctly orienting each parameter interval gives

$$\begin{aligned} V&=-\pi\int_0^\pi y(t)^2x'(t)\,dt\\ &=\pi\int_0^\pi\sin^3t(1+\cos t)^2(1+2\cos t)\,dt. \end{aligned}$$

Retain one factor $\sin t$ and write the rest using $u=\cos t$. Since $du=-\sin t\,dt$,

$$\begin{aligned} V&=\pi\int_{-1}^{1}(1-u^2)(1+u)^2(1+2u)\,du\\ &=\pi\int_{-1}^{1}(-2u^5-5u^4-2u^3+4u^2+4u+1)\,du\\ &=\pi\left(-2+\frac83+2\right) =\boxed{\frac{8\pi}{3}}. \end{aligned}$$

Odd powers integrate to zero on the symmetric interval. This also avoids carrying unnecessary terms in the antiderivative.

Example 2: revolving an astroid

Find the volume swept out by rotating the region enclosed by

$$x=\cos^3t,\qquad y=\sin^3t,\qquad0\le t\le2\pi$$

once about the $x$-axis.

The curve is symmetric about both coordinate axes. Revolving the first-quadrant region gives the half of the solid with $x\ge0$, so double that volume. There is no further factor for reflection across the axis of rotation.

On $0\le t\le\pi/2$,

$$x'=-3\cos^2t\sin t,\qquad y'=3\sin^2t\cos t.$$

Each perpendicular section is a disk. The half-volume is

$$\begin{aligned} \frac V2&=\pi\int_0^1y^2\,dx =3\pi\int_0^{\pi/2}\sin^7t\cos^2t\,dt\\ &=3\pi\int_0^{\pi/2}\sin t(1-\cos^2t)^3\cos^2t\,dt. \end{aligned}$$

Again let $u=\cos t$:

$$\begin{aligned} \frac V2&=3\pi\int_0^1u^2(1-u^2)^3\,du\\ &=3\pi\left[\frac{u^3}{3}-\frac{3u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{9}\right]_0^1\\ &=\frac{16\pi}{105}. \end{aligned}$$

Therefore

$$\boxed{V=\frac{32\pi}{105}.}$$

Useful checks

A volume must be nonnegative, but a negative result should prompt a review of bounds and orientation rather than an automatic sign change. Check whether sections are disks or washers, whether any part was swept more than once, and which symmetry factors remain after rotation.

For powers of sine and cosine, an odd power often allows you to reserve one factor for substitution and express the remaining even powers using $\sin^2t+\cos^2t=1$. That is why both volume integrals above become polynomial integrals.