"Evaluate the surface integral for the portion of a cylinder $x^2+y^2=1$ between the planes $z=1$ and $z=4$."
My Work
The major issue that I'm having is with the parameterization, as this cylinder involves 3 relevant parameters $r$,$z$, and $\theta$. The integral is:
$\int\int|r_u\times r_v|dudv$
I originally thought that I might be able to just say $r=1$ therefore $x=rcos\theta=cos\theta,y=rsin\theta=sin\theta,z=z$ and end with r$= ⟨cos\theta,sin\theta,z ⟩$. However, if I do this, I get:
r$_z= ⟨ 0,0,1⟩$
r$_\theta = ⟨ -sin\theta,cos\theta,0⟩$
Which, when crossed, results in:
$ ⟨ cos\theta,0,0⟩$
Therefore resulting in the integral
$\int_1^4\int_0^{2\pi} cos\theta d\theta dz$
Which is $0$. Clearly, this is wrong, and probably a result of my setup, but I'm not sure how to fix it. Would anyone be able to point out where I'm going wrong?
$r$ does equal 1.
I amd going to use $S$ for the surface instead of $\mathbf r$
$S(z,\theta) = (\cos\theta, \sin\theta, 0) + (0,0,z)\\ \frac {dS}{dz} \times \frac {dS}{d\theta} = (\cos\theta, \sin\theta, 0)\ dz\ d\theta$
but what we really need is:
$\|\frac {dS}{dz} \times \frac {dS}{d\theta}\| = 1 \ dz\ d\theta$