Find the flux of $F(x,y,z)=(xy^2,yz^2+xze^{\sin(z^2)},zx^2+e^{x^2})$ through $S=\{(x,y,z):x^2+y^2+z^2=9,x\geq 0\}$ with a non negative $x$ normal.
My attempt:
We want to use gauss. First we need to close $S$, so we define $S_2=\{(x,y,z):y^2+z^2\leq3^2,x=0\}$.
Now $S_1=S \cup S_2$ is a closed surface. $F\in C^1$ so using gauss:
$$\iint\limits_{S_1}F\cdot dr=\iiint\limits_{V}div(F)dxdydz=\iiint\limits_{V}x^2+y^2+z^2dxdydz=_{\text {spherical coordinates}}\int_0^\pi \int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^3r^4\sin\phi drd\phi d\theta=$$
$$48.6\pi\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sin\phi d\phi=0$$
Now using additive of integrals:
$$\iint\limits_{S}F\cdot dr=-\iint\limits_{S_2}F\cdot dr$$
We know the normal of $S_2\rightarrow \vec n_2=(1,0,0)$
$$-\iint\limits_{S_2}F\cdot \vec n_2 ds=-\iint\limits_{S_2}xy^2$$
Im stuck here. Maybee i've misplayed all of this. Anyway, any help will be appreciated.
Best Answer
You have a mistake in the order of your integral because we have
$x \geq 0$ i.e. $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.
$0 \leq \phi \leq \pi$.
So the integral should be,
$\displaystyle \int_0^{\pi} \int_{-\pi/2}^{\pi/2}\int_0^3 r^4 \sin\phi \ dr \ d\theta \ d\phi = \frac{486 \pi}{5}$
Now to apply divergence theorem, we closed the surface with disk $y^2+z^2 = 9$ at $x = 0$. So to find flux through the spherical surface, we must subtract flux through the disk at $x=0$. Please note the outward unit normal vector to the disk $y^2+z^2 = 9$ at $x = 0$ is $(-1, 0, 0)$. But the $x$ component of the vector field is $0$ at $x = 0$. So there is no flux through the disk. Hence your final answer is as calculated above.