Determine the flux of the field (x,y,z) through the surface z=$\sqrt(x^2+y^2-15)$ , 16 ≤$x^2+y^2$≤40 with outward pointing unitnormal.
I understand that you need to use Gauss's divergence theorem, but I can't seem to workout the integrals. I bound the region with three surfaces, two circles at the top and bottom and then the surface z=$\sqrt(x^2+y^2-15)$.
I am having troubles working out the flux through the top and bottom surfaces. I am using the formula \begin{equation} \iint \limits_{}^{} u N dS \end{equation}
For the bottom surface I get \begin{equation} \iint \limits_{}^{} (x,y,z)(0,0,-1) dxdy= \iint \ -z dxdy \end{equation}
I get stuck here. Any help would be appreciated!
Yes close the surface with disks at top (in plane $z = 5$) and bottom (in plane $z = 1$) and apply divergence theorem.
Divergence of the vector field $\nabla \cdot (x,y,z) = 3$.
Using cylindrical coordinates, the surface is
$z = \sqrt{r^2-15} \implies r^2 = z^2 + 15$.
We also know $16 \leq r^2 \leq 40$. That leads to $1 \leq z \leq 5$.
So total flux can be calculated as
$\displaystyle 3 \int_1^5 \pi(z^2+15) \ dz$.
Now we will find flux through the bottom disk.
$\vec F \cdot \hat{n} = (x,y,1) \cdot (0, 0, -1) = -1$
So the surface integral will be equal to the area of the disk with minus sign, which is $ - 16 \pi$.
Similarly for flux through the top surface,
$\vec F \cdot \hat{n} = (x,y,5) \cdot (0, 0, 1) = 5$
So the flux is $5$ times the area of the disk which is $200 \pi$.
Net outward flux through both top and bottom disks $ = 184 \pi$.
Now subtract that from the total flux through closed surface to get the answer.