I am trying to do the following exercise:
Let $B=\{(x,y,z) \in \mathbb{R}^3: x^2+y^2+z^2=1\}$ be the unit ball.
Calculate the surface integral of $f(x,y,z)=(x^3,y^3,z^3)$ over the surface of the unit ball.
Since f is a vector field, I need to use the "surface Integral for vector fields" definition.
But first I need a parametrization of the unit ball.
I decided to use the parametrization:
$\phi (u,v)= \left(\begin{array}{c} \cos(u)\cos(v) \\ \sin(u)\cos(v) \\ \sin(v) \end{array}\right)$
for $0 \leq u\leq 2 \pi $ and $-\pi/2 \leq v \leq \pi/2$
to keep it clean let $K:=\{(u,v) \in \mathbb{R}^2:0 \leq u\leq 2 \pi ,\, -\pi/2 \leq v \leq \pi/2\}$
Now the surface integral is:
$$\int_{\phi} \langle f,n\rangle=\int_K \langle f(\phi(u,v)), \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \rangle d(u,v)$$
In this case:
$ \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} =\cos (v) \phi (u,v)$
and
$f(\phi(u,v))= \left(\begin{array}{c} \cos^3(u)\cos^3(v) \\ \sin^3(u)\cos^3(v) \\ \sin^3(v) \end{array}\right)$
Thus,
$$\langle f(\phi(u,v)), \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \rangle=\cos^4(u)\cos^5(v)+\sin^4(u)\cos^5(v)+\sin^4(v)\cos(v)$$
The last step seems to be to calculate the integral:
$$\int_K \cos^4(u)\cos^5(v)+\sin^4(u)\cos^5(v)+\sin^4(v)\cos(v) d(u,v)=
\int_{-\pi/2} ^{\pi/2} [\int_0 ^{2 \pi} \cos^4(u)\cos^5(v)+\sin^4(u)\cos^5(v)+\sin^4(v)\cos(v)du] dv =\frac{12 \pi}{5}.$$
Questions:
Are my calculations correct (the way I calculate it)?
Another way to do a parametrization of the unit ball would be to solve $x^2+y^2+z^2=1$ (locally) for $z$ and use $\phi_1 (x,y) =\left(\begin{array}{c} x \\ y \\ \sqrt{1-x^2-y^2} \end{array}\right)$ and $\phi_2 (x,y) =\left(\begin{array}{c} x \\ y \\ -\sqrt{1-x^2-y^2} \end{array}\right)$ as a parameterization.
In this case, do I need to do the surface Integral for both parameterizations and sum them up? (if not, how?)
Best Answer
Your approach and the final result are both correct. By the divergence theorem, this surface integral, which is the flux of the vector field $f$ through the surface of the unit sphere, can also be obtained as $$\begin{align} \iiint_{\|x\|\leq 1} \text{div}(f)\,dxdydz&=\iiint_{\|x\|\leq 1} 3(x^2+y^2+z^2)\,dxdydz\\&=3\int_{0}^{2\pi}d\theta\int_{0}^{\pi}\sin(\phi)d\phi\int_{0}^1\rho^4 d\rho=\frac{12\pi}{5}. \end{align}$$