I'm having some trouble calculating vector fields through surfaces. After attempting a few and being dissapointed with a wrong answer multiple times I figured I must be doing something wrong in the process. I think its the parametrization I'm struggling with.
Oftentimes there's a surface I can easily parametrize in Cartesian coordinates (sometimes partially) which can also be written using cylindrical or spherical coordinates. It's this part that gives me trouble, I don't really how to combine a change of coordinates and a parametrization.
I looked around at a few examples online and tried applying what I learnt there without much positive result, here's one of the attempts:
Integrate the function $\vec F(x,y,z) = (2x,-3y,z)$ through the surface $\Sigma$ defined as the part of the cylinder $x^2+y^2=1$ between $z=0$ and $z=x+2$.
Parametrization $\vec\phi$:
$x = \cos\theta$
$y = \sin\theta$
$z = z$
$\frac{\partial\phi}{\partial z}\times\frac{\partial\phi}{\partial\theta}=(-\cos\theta,-\sin\theta,0)$Apply the formula: $\iint_\Sigma\vec F(x,y,z) .d\vec\sigma = \iint_K\vec F(\vec\phi(z,\theta)).(\frac{\partial\vec\phi}{\partial z}\times\frac{\partial\vec\phi}{\partial\theta}(z,\theta))$
Which gives $\int_0^{2\pi}d\theta\int_0^{\cos\theta+2}(2\cos\theta,-3\sin\theta,z).(-\cos\theta,-\sin\theta,0)dz$
Solving this integral doesn't give me the correct result, so I must've made an error along the way but I just can't find it. Again, this is just an example, it's this type of integrating that's giving me a hard time.
Here's the further solution:
$\int_0^{2\pi}d\theta\int_0^{\cos\theta+2}(-2\cos^2\theta+3\sin^2\theta)dz$
$=\int_0^{2\pi}(-2\cos^3\theta+3\sin^2\theta\cos\theta-4\cos^2\theta+6\sin^2\theta)d\theta$
$= 0 + 0 + (-4\pi) + 6\pi = 2\pi$
The correct answer according to my textbook is $-2\pi$