# Would this not take a ridiculously long calculation? (Surface area of parametric surface)

One of the question in my homework asks to verify that the surface are of

$\mathbf{r} = a(1+\cos\phi)\sin\phi \cos\theta \mathbf{i} + a(1+\cos\phi)\sin\phi \sin\theta \mathbf{j} + a(1+\cos\phi)\cos\phi \mathbf{k}$

is $\frac{32}{5}\pi a^2$

I started with the equation to the area of the surface being the double integral of $|\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}|$... but MAN this is so long it's almost unworkable. Am I doing this the stupid way? Can this be simplified?

All I can tell is that $\mathbf{r} = a\sin\phi \cos\theta \mathbf{i} + a\sin\phi \sin\theta \mathbf{j} + a\cos\phi \mathbf{k}$ is te equation of a sphere of radius $a$, but you can't really factor out areas, that I know of... so I don't see how this helps.

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I'll ignore the common scaling factor of $a$; we can multiply by $a^2$ at the end.

The surface is a surface of revolution about the symmetry axis of a cardioid: $$\mathbf r = (f(\phi)\cos\theta, f(\phi)\sin\theta, g(\phi))$$ with $$f(\phi) = (1+\cos\phi)\sin\phi = \sin\phi + \sin(2\phi)$$ $$g(\phi) = (1+\cos\phi)\cos\phi = \cos\phi + \cos(2\phi) + 1/2$$ Therefore we can use the formula for the area of a surface of revolution: $$A = 2\pi \int_0^\pi f(\phi)\sqrt{f'(\phi)^2+g'(\phi)^2} \;d\phi$$ where only the range $[0,\pi]$ for $\phi$ is relevant because $f$ is odd and $g$ is even. The expression under the square root is \begin{align} &\big(-\sin\phi - \sin(2\phi)\big)^2 + \big(\cos\phi - \cos(2\phi)\big)^2\\ =&\sin^2\phi+\cos^2\phi + \sin^2(2\phi)+\cos^2(2\phi) + 2\Bigl(\sin\phi\sin(2\phi)+\cos\phi\cos(2\phi)\Bigr) \\ =&2+2\Bigl(\sin\phi(2\cos\phi\sin\phi)+\cos\phi(2\cos^2\phi-1)\Bigr) \\ =&2+2(2\cos\phi-\cos\phi) = 2(1+\cos\phi) \end{align} so we get $$A = 2\pi \int_0^\pi \sqrt 2 (1+\cos\phi)^{3/2} \sin \phi\; d\phi$$ which is easily evaluated by substituting $u=1+\cos\phi$.

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Thanks, I don't remember seeing the area for surface of revolution for $f$ and $g$ in class or in my textbook (Stewart's), but I looked into it now and it seems clear. –  iDontKnowBetter Oct 30 '11 at 18:50
@Henning_Makholm I'm not sure what you mean by "only the range $[0,\pi]$ for $\phi$ is relevant because $f$ is odd and $g$ is even." — doesn't the function go up to $2\pi$? –  iDontKnowBetter Oct 30 '11 at 19:27
Yes, but points with $\phi\in(\pi,2\pi)$ can all be hit with $\theta\in(\pi,2\pi)$ instead. Either of $\phi\mapsto 2\pi-\phi$ and $\theta\mapsto\theta+\pi$ will negate the first two coordinates and leave the third one intact, so if we allow both, we'll count everything twice. –  Henning Makholm Oct 30 '11 at 21:11

I don't really see anything that can significantly help cut down on your calculations.

This is the sort of problem I like to throw to Maple.

Maple finds that

$${\bf r}_\theta \times {\bf r}_\phi = \begin{array}{l} (-a^2\sin^2(\phi)\cos(\theta)(3\cos(\phi)+1+2\cos^2(\phi))){\bf i}\\ (-a^2\sin^2(\phi)\sin(\theta)(3\cos(\phi)+1+2\cos^2(\phi))){\bf j}\\ (-a^2\sin(\phi)(-1+3\cos^2(\phi)+2\cos^3(\phi))){\bf k}\end{array}$$

Then $|{\bf r}_\theta \times {\bf r}_\phi| = a^2\sqrt{2}\sin(\phi)(1+\cos(\phi))^{3/2}$

Integrating over $0 \leq \phi \leq \pi$ and $0 \leq \theta \leq 2\pi$ gives $\frac{32}{5}a^2\pi$ as predicted.

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The surface is a surface of revolution around the $z$-axis. The meridian curve $\gamma$ in the $(\rho,z)$-halfplane is given by $$\gamma:\quad \phi\mapsto\bigl((1+\cos\phi)\sin\phi,(1+\cos\phi)\cos\phi\bigr)\qquad(0\leq\phi\leq\pi)\ ;$$ whence it has the simple polar representation $$\gamma:\qquad r(\phi):=1+\cos\phi\quad(0\leq\phi\leq\pi)\ .$$ It follows that its line element is given by $$ds=\sqrt{r^2+r'^2}\ d\phi=\sqrt{2(1+\cos\phi)}\ d\phi\ .$$ To this line element belongs the annular area element $$d\omega=2\pi\rho(\phi)ds=2\pi\sqrt{2}(1+\cos\phi)^{3/2}\sin\phi\ d\phi=-2\pi\sqrt{2}\bigl(r(\phi)\bigr)^{3/2}r'(\phi)\ d\phi\ .$$ Therefore we get by integration $$\omega(S)=-2\pi\sqrt{2}\int_0^\pi \bigl(r(\phi)\bigr)^{3/2}r'(\phi)\ d\phi = -2\pi\sqrt{2}{2\over5}\bigl(r(\phi)\bigr)^{5/2}\Biggr|_0^\pi ={32\pi\over5}\ .$$

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Nice observation on the polar representation of $\gamma$. –  Henning Makholm Oct 30 '11 at 21:16