How to compute the flux of the vector field I need help to solve this question from my textbook.
Compute the vector field $F=\langle 4x, 4y, 0\rangle$, through the surface $S$ which is the part of the surface $z=25-(x^2 +y^2)$ above the disk of radius $5$ entered at the origin, oriented upward.
I tried to use cylindrical coordinates:
$x=r\cos(\theta)$
$y=r\sin(\theta)$
$z=r $
for $0<\theta<2\pi$ and $0<r<5.$
I found the normal vector $\langle-r\cos(\theta), -r\sin(\theta), r\rangle$.
Then I solve for the integral of $F$ dot normal, but I keep getting the wrong answer.
Any leads?
 A: (Almost) full brute-force solution:
At a point $(x, y, z)$ on the paraboloid, one normal vector is $(2x, 2y, 1)$ (you can find this by rewriting the surface equation as $x^2+y^2+z-25=0$, and taking the gradient of the left-hand side). Then $$n(x, y, z)=\left(\frac{2x}{\sqrt{4x^2+4y^2+1}}, \frac{2y}{\sqrt{4x^2+4y^2+1}}, \frac{1}{\sqrt{4x^2+4y^2+1}}\right)$$is the normalized normal vector oriended upwards. We want to integrate the dot product of this with $F$ over the entire paraboloid. This dot product is
$$
F(x, y, z)\cdot n(x, y, z) = \frac{8x^2+8y^2}{\sqrt{4x^2+4y^2+1}}=\frac{8r^2}{\sqrt{4r^2+1}}
$$
Given a small region of the bounding disc in the $xy$-plane of "area" $dA$, the "area" of the piece of the paraboloid directly above it is $\sqrt{4r^2+1}\,dA$ (where that factor is the reciprocal of the $z$ component of the normalized normal vector). This will allow us to integrate over the disc and not the paraboloid (presumably, parametrizing the paraboloid by the points in the disc and substituting correctly would yield a corresponding result, but much slower).
So we get
$$
\iint_{\text{paraboloid}}F(x, y, z)\cdot n(x, y, z)\,dA =\iint_{\text{disc}}F(x, y, z)\cdot n(x, y, z)\sqrt{4r^2+1}\,dA\\
=\iint_{\text{disc}}8r^2\,dA=\int_0^{2\pi}\int_0^5 8r^2\cdot r\,drd\theta=2\pi\int_0^58r^2\cdot rdr=\boxed{2500\pi}
$$
Maybe a little smarter solution:
The vector field is entirely parallel to the $xy$ plane, so we can slice our paraboloid into thin horizontal slices of "height" $dz$, calculate the flux through each of them, and then integrate it all up afterwards. This lets us ignore that the sides of each slice is slanted, and instead pretend they are vertical, which sounds nice.
(A little more rigorously, any given slice has a small range of different circular cross sections, and we could pretend that the side of the slice was vertical with a cross section from that range. Some such choices will give a flux that's larger than the true flux, some will yield a flux that's smaller than the true flux, and some choice will give exactly the right flux. The correctness of Riemann / Darboux integration as below follows eventually from this line of reasoning, perhaps you recognize the outline of an upper sum versus lower sum argument here. Note that if there was any vertical component to $F$, this wouldn't work as easily.)
At height $z$, the parabola slice has radius $\sqrt{25-z}$. Since $F$ is normal to the sides of this slice, and the magnitude of $F$ is constantly equal to $4r=4\sqrt{25-z}$, that means the flux through that slice is $4\sqrt{25-z}\cdot 2\pi\sqrt{25-z}\,dz$. We integrate this over the whole height of the paraboloid, and we get
$$
\int_0^{25}4\sqrt{25-z}\cdot 2\pi\sqrt{25-z}\,dz=\boxed{2500\pi}
$$
A whole different approach:
The flux of $F$ out of a closed body is equal to the integral of the divergence of $F$ over the whole body. Your paraboloid together with the bottom disc is a closed body. The flux through the bottom disc is $0$. The divergence of $F$ is constantly equal to $8$. This tells us that the flux out of the paraboloid is equal to $8$ times the volume of the body.
All we need now is the volume of the part of the paraboloid that's above the $xy$ plane. There are several ways of doing this, some of them have fancy names like the washer method or the cylindrical shell method. Whichever one you use, you get that the volume is $\frac{625\pi}2$, and the total flux is therefore $\boxed{2500\pi}$.
