Surface Area with Parametric Equations — Question 9

Question 9

Problem

Rotate the astroid quadrant $x=\cos^3t,y=\sin^3t$, $0\le t\le\pi/2$, about the $x$-axis. Set up a simplified integral.

Question 9 – Solution

Solution

1. Differentiate: $$x'(t)=-3\cos^2t\sin t, \qquad y'(t)=3\sin^2t\cos t.$$

2. As in the astroid arc-length calculation, $$ds=3|\sin t\cos t|\,dt.$$ Both factors are nonnegative on $0\le t\le\pi/2$, so $$ds=3\sin t\cos t\,dt.$$

3. The radius about the $x$-axis is $y=\sin^3t$. Therefore, $$\begin{aligned} S&=2\pi\int_0^{\pi/2}y\,ds\\ &=6\pi\int_0^{\pi/2}\sin^4t\cos t\,dt. \end{aligned}$$

4. Let $u=\sin t$, so $du=\cos t\,dt$ and $0\le u\le1$: $$S=6\pi\int_0^1u^4\,du =6\pi\left[\frac{u^5}{5}\right]_0^1 =\boxed{\frac{6\pi}{5}}.$$