$$D=\{x^2+y^2+z^2\le 25,y\le -4\}. \qquad F=\{z^2,y^2,x^2\}$$
In order to find the total flux going outward D I need to evaluate two fluxes(or maybe not ? ):
the first one is the flux of the circular region surface $x^2+z^2\le 9$ (which is the intersection between the sphere and $y\le -4$)
and the second one is the surface created by cutting the sphere
I calculated the first flux by first parametrizing the surface and then by evaluating the double integral over if :
$$\iint\limits_{S_1} F\,ds$$
So I have $r(\theta)=(3\cos\theta,-4,3\sin\theta)$ with $0\le \theta \le 2\pi$
and the normal vector is $n=(0,-1,0)$. and eventually the flux will be = $-144\pi$
The problems start when I try to calculate the flux of the second surface :
So i have $r(\theta,\phi)=(p\sin\phi\sin\theta,p\cos\phi,p\sin\phi\cos\theta)$ with $0\le \theta \le 2\pi$,$\cos^{-1}{(-4/5)\le \phi \le \pi}$
The normal shoud be : $n = (-p^2\sin^2\phi\sin\theta,-p^2\sin\phi\cos\phi,-p^2\sin^2\phi\sin^2\theta)$
And when I multiply $F(r(\theta,\phi))n = -p^4\cos^2\theta(2\sin^4\phi\sin\theta+\sin\phi\cos\phi)$
Is that really What I should put in the integral? (in order to find tge flux of the second region)
What I'd like to get is the total flux! What would you do to?
By divergence theorem we have
$$\iint_S \vec F \cdot \vec n dS=\iiint_D \operatorname{div}\vec F dV=\iiint_D \operatorname 2y\, dV$$
and by cylindrical coordinates we have
$$\iiint_D \operatorname 2y\, dV=\int_0^{2\pi}\,d\theta \int_{-5}^{-4} 2y\, dy\int_0^{\sqrt{25-y^2}}r\,dr$$
(solution $-\frac{81}2 \pi$)