For a question, I am asked to find the flux of $F=\langle 3x,0,2\rangle$ across the surface of $x^2+y^2+z^2=4, x>0, \ y<0,\ z<0$. I tried solving this with cylindrical and polar coordinate systems, but both give different answers.
Spherical coordinate system: The parameterization of the surface is given as $r(u,v)=\langle2\sin(u)\cos(v),-2\sin(u)\sin(v),-2\cos(u)\rangle$ with $0 \leq u \leq \frac{\pi}2$, and $0 \leq u \leq \frac{\pi}2$. With this, I got $-6\pi$.
Cylindrical coordinate system: The parameterization of the surface is given as $r(u,v)=\langle u\cos(v),-u\sin(v),-\sqrt{4-u^2} \rangle$, with $0 \leq u \leq 2$ and $0 \leq v \leq \frac{\pi}2$. With this, I got $-2\pi$.
Where did I mess up?
I figured out the problem. Here's what it should be:
Spherical coordinate system: The parameterization of the surface is given as $r(u,v)=\langle2\sin(u)\cos(v),2\sin(u)\sin(v),2\cos(u)\rangle$ with $-\frac{\pi}2 \leq u \leq 0$, and $-\frac{\pi}2 \leq v \leq 0$. With this, I get $2\pi$.
Cylindrical coordinate system: The parameterization of the surface is given as $r(u,v)=\langle u\cos(v),u\sin(v),-\sqrt{4-u^2} \rangle$, with $0 \leq u \leq 2$ and $ -\frac{\pi}2 \leq v \leq 0$. With this, I get $2\pi$.