Find the area of the surface of revolution generated by revolving about the $x$-axis the hypocycloid $x=a\cos^3\theta$, $y=a\sin^3\theta$ Find the area of the surface of revolution generated by revolving about the $x$-axis the hypocycloid $x=a\cos^3\theta$, $y=a\sin^3\theta$ ($0 \leq \theta \leq \pi$)
I know you have to integrate $2\pi y ds$ for the limits $0$ to $\pi$ and I know that $ds$ is the square root of the sum of each derivative squared but I'm stuck on the integration, how do we do it? Please help!
 A: The standard formula for rotation about the $x$-axis, through an angle of $2\pi$-radians, when $x$ and $y$ are given as functions of $\theta$ is
$$A = 2\pi \int_{\theta_1}^{\theta_2} y\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^{\!\!2} + \left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^{\!\!2}}~\mathrm{d}\theta$$
In your case $x=a\cos^3\theta$ and $y=a\sin^3\theta$. It follows that
\begin{eqnarray*}
\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^{\!\!2} + \left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^{\!\!2} &=& \left(-3a\cos^2\theta\sin\theta\right)^{\!2}+\left(3a\sin^2\theta\cos\theta\right) \\ \\
&=&9a^2\cos^4\theta\sin^2\theta+9a^2\sin^4\theta\cos^2\theta \\ \\
&\equiv& 9a^2\cos^2\theta\sin^2\theta
\end{eqnarray*}
Since $0 \le \theta \le \pi$ we need to take care because $\cos\theta$ changes sign. We have
\begin{eqnarray*}
A &=& 2\pi\int_0^{\pi} a\sin^3\theta\sqrt{9a^2\cos^2\theta\sin^2\theta}~\mathrm{d}\theta \\ \\
&=& 6\pi a^2 \int_0^{\pi} \sin^3\theta \cdot |\cos\theta| \cdot |\sin\theta|~\mathrm{d}\theta
\end{eqnarray*}
Since $\sin \theta \ge 0$ for all $0 \le \theta \le \pi$ we have $|\sin\theta| \equiv \sin\theta$. However, $\cos\theta \ge 0$ for all $0 \le \theta \le \frac{\pi}{2}$ and $\cos\theta \le 0$ for all $\frac{\pi}{2} \le \theta \le \pi$. Hence $|\cos\theta| \equiv \cos\theta$ for all $0 \le \theta \le \frac{\pi}{2}$, while $|\cos\theta| \equiv -\cos\theta$ for all $\frac{\pi}{2} \le \theta \le \pi$. It follows that
\begin{eqnarray*}
A &=& 6\pi a^2 \int_0^{\pi/2} \sin^4\theta\cos\theta~\mathrm{d}\theta - 6\pi a^2\int_{\pi/2}^{\pi} \sin^4\theta\cos\theta~\mathrm{d}\theta \\ \\
&=& 6\pi a^2\left[\frac{1}{5}\sin^5\theta\right]_0^{\pi/2}-6 \pi a^2\left[\frac{1}{5}\sin^5\theta\right]_{\pi/2}^{\pi} \\ \\
&=& \frac{12}{5}\pi a^2
\end{eqnarray*}
A: Hint: We assume $a\gt 0$. Minor modification can be made if $a\lt 0$.
We have $\frac{dx}{d\theta}=-3a\cos^2\theta\sin\theta$ and $\frac{dy}{d\theta}=3a\sin^2\theta\cos\theta$. 
Square, add, simplify a bit. We get $9a^2\cos^2\theta\sin^2\theta$.  Now take the square root. Here we have an opportunity to make a major error. One should get $3a|\sin\theta\cos\theta|$.  
Multiply by $2\pi a\sin^3\theta$, and integrate. You will need to split into two parts. For each part, make the substitution $u=\sin\theta$.
Remark: I would rather exploit symmetry, by finding the surface area from $0$ to $\frac{\pi}{2}$, and doubling the result. 
