I'm trying to evaluate the surface integral
$\iint_S (x^2z + y^2z)dS$, where S is the hemisphere $x^2+y^2+z^2 =4, z\geq0$.
To start, I parametrize S through spherical coordinates to end up with
$r(\phi, \theta) =
4sin(\phi)cos(\theta)i+4sin(\phi)sin(\theta)k+4cos(\phi)$
Evaluating the cross product between $r_\phi$ and $r_\theta$ results in the expression $4sin(\phi)$. This then results in the integral
$\int\int z(x^2+y^2)\cdot4sin(\phi) d\phi d\theta$
which can be rewritten as
$32\int_{0}^{2\pi}\int_{0}^{\pi/2} sin^3(\phi)cos(\theta) d\phi d\theta$.
This, however, ends up evaluating to $0$ (as the inner integral evaluates to $\frac{cos^3(x)}{3} – cos(x)|_{0}^{\pi/2} = \frac{2}{3}$, which then leads to the integral of
$32\int_{0}^{2\pi}\frac{2}{3}cos(\theta)d\theta$
which evaluates to $0$. Does anyone know where/if I'm going wrong?
Best Answer
List below all the pieces in the surface integral, $$z=r\cos\phi\\ x^2+y^2=r^2\sin^2\phi\\ dS=r^2 \sin\phi d\phi d\theta $$ Plug them into the integral, $$ \iint_S z(x^2 + y^2)dS = r^5\int_0^{\pi/2}\cos\phi \sin^3\phi d\phi \int_0^{2\pi}d\theta$$ which does not integrate to zero.
The problem is that the term $\cos\theta$ in your integral should be $\cos\phi$, which stems from $z=2\cos\phi$.