I'm stuck on this problem. I've looked up a bunch of answers and watched youtube videos on solving similar problems, but when I try to apply what I learn to the below problem I get a weird answer ( $0$).
The problem I'm solving asks:
Evaluate the surface integral $\int\int_S{F{\rm d}S}$ for the vector field $F(x,y,z)=xz\hat{i}+x\hat{j}+y\hat{k}$ where $S$ is the hemisphere $x^2+y^2+z^2=25, y\geq0$ , oriented in the direction of the positive $y$-axis.
Here is where I am so far:
- Parameterize $F(x,y,z)$ in spherical coordinates and find $n\cdot {\rm d}S$: $$r(\phi,\theta)=\begin{pmatrix}5\sin\phi \cos\theta\\ 5\sin\phi \sin\theta\\ 5\cos\phi\end{pmatrix}$$ $$r_\phi =\begin{pmatrix}5\cos\phi \cos\theta\\ 5\cos\phi \sin\theta\\ -5\sin\phi\end{pmatrix}$$ $$r_\theta =\begin{pmatrix}-5\sin\phi \sin\theta\\ 5\sin\phi \cos\theta\\0\end{pmatrix}$$ $$r_\phi \times r_\theta = \begin{pmatrix}25\sin^2\phi \cos\theta\\ 25\sin^2\phi \sin\theta\\ 25\sin\phi \cos\phi\end{pmatrix}$$ $$0\leq \phi \leq \pi$$ $$0\leq \theta \leq \pi$$
- Plug into equation for the surface integral:
$$\int_0^\pi\int_0^\pi\begin{pmatrix}25\sin\phi \cos\phi \cos\theta\\ 5\sin\phi \cos\theta\\ 5\sin\phi \sin\theta\end{pmatrix}\cdot\begin{pmatrix}25\sin^2\phi \cos\theta\\ 25\sin^2\phi \sin\theta\\ 25\sin\phi \cos\phi\end{pmatrix}{\rm d}\phi {\rm d}\theta$$ $$=\int_0^\pi\int_0^\pi \left(625\sin^3\phi \cos\phi \cos^2\theta + 125\sin^3\phi \cos\theta \sin\theta + 125\sin^2\phi \cos\phi \sin\theta \right){\rm d}\phi {\rm d}\theta$$
- Evaluate:
inner integral:
$$= \frac{625}{4}\sin^4\phi \cos^2\theta + 125\left(\frac{\cos^3\phi}{3}-\cos\phi\right)\cos\theta \sin\theta + \frac{125}{3}\sin^3\phi \sin\theta \vert_0^\pi$$ $$=\frac{500}{3}\left(\cos\theta \sin\theta\right)$$
outer integral:
$$= \frac{500}{6}\sin^2\theta \vert_0^\pi$$ $$=0$$
This definitely seems wrong. :/