3
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ This is impossible over the real numbers. Since $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$, for all $ x \in \mathbb{R}$, the inverse function $\arcsin$ and $\arccos$ are only defined on the closed interval $[-1,1]$. $\endgroup$ May 12, 2014 at 19:33
  • $\begingroup$ Yes I see now ~ Whoops. I imagine $0 < \theta < 1$ was intended? $\endgroup$
    – Ayesha
    May 12, 2014 at 19:36
  • $\begingroup$ It was the function $\endgroup$ May 12, 2014 at 19:44
  • $\begingroup$ @Ayesha It was just that $\arcsin$ should have been $\sin$ and $\arccos$ should have been $\cos$. The letter $\theta$ is, in this context, usually reserved for an angle. The restriction $0 < \theta < 1$ is quite odd for an angle in radians. $\endgroup$ May 12, 2014 at 19:46

2 Answers 2

4
$\begingroup$

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*}

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ You are welcome. I forgot to write what I almost always write, "Draw a picture." $\endgroup$ May 12, 2014 at 20:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .