Given the portion of the sphere $S$ which is in the first octant ' $S:$ $x^2+y^2+z^2=1$ ; $ x,y,z$ $\geq0$ , let $n$ be the unit normal vector (outwards) to $S$ . also given the vector field $F=(z,-y,x)$ . Find the flux of $F$ on $S$
We have the unit sphere $X^2+y^2+z^2=1$ and the field $F=(z,-y,x)$ so $div=-1$ but our sphere is not closed , what I did was try to close it with the plane $z=0$ and then we can use gauss theorem
$\iint_S F \cdot n ds+\iint_P(z,-y,x) \cdot (0,0,-1)ds=\iiint_V -1 dV$
the triple integram is the sphere volume so I got $\frac{-4\pi}{3}$ but when I try to calculate $\iint_P(z,-y,x) \cdot (0,0,-1)ds$ I keep getting $-$ because $\iint_P -xds$ and since $z=0$ we get $x^2+y^2=1$ so the integral will be with $0\leq \theta \leq \pi$ and $0 \leq r \leq 1$ and since I moved to polar coordiantes $J=r$ and $x=rcos\theta$ but this way the integral will be $0$ and the final answer in the book is $ \frac{4-\pi}{6}$
The way it was solved in the book was by solving a normal surface integral but I want to try in gauss theorem since it is the way I would think of.
I believe it should look like this
Thanks for any tips and help!
Best Answer
Close the surface with quarter disks in planes $x = 0, y = 0, z = 0$ and then apply Divergence theorem.
$\nabla \cdot \vec F = - 1$. The volume of the unit sphere in first octant is $\dfrac{\pi}{6}$.
So the net outward flux through the closed surface is $ - \dfrac{\pi}{6}$.
Now surface integral over quarter disk in $y = 0$ plane is zero as $\vec F = (z, 0, x)$ and $\hat n = (0, -1, 0)$. So $\vec F \cdot \hat n = 0$
Surface integral over quarter disk in $z = 0$ plane:
$\vec F = (0, -y, x)$ and $\hat n = (0, 0, -1)$.
Parametrize the disk as,
$(r \cos t, r \sin t, 0), \ \ 0 \leq r \leq 1, 0 \leq t \leq \frac{\pi}{2}$
So integral is,
$\displaystyle \int_0^{\pi/2} \int_0^1 (- r \cos t) r \ dr \ dt = - \dfrac{1}{3}$
You can see why it would be the same for the disk in plane $x = 0$.
So subtracting flux through disks, we get the flux through the spherical surface as,
$\dfrac{2}{3} - \dfrac{\pi}{6}$