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. 
 A: 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 $u=1+\cos\phi$.
A: 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.
A: 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
$$\omega(S)=-2\pi\sqrt{2}\int_0^\pi \bigl(r(\phi)\bigr)^{3/2}r'(\phi)\ d\phi =
-2\pi\sqrt{2}{2\over5}\bigl(r(\phi)\bigr)^{5/2}\Biggr|_0^\pi ={32\pi\over5}\ .$$
