Simple Surface Integral problem

Question

Hi here's the question I am working on:

Calculate the surface integral of $F \cdot dS$: $F(x,y,z) = (2x, 2y, 2z)$

$S$ is the side of the cylinder $x^2 + y^2 = 9$ and $0 < z < 5$ , without top and bottom, oriented outwards.

I parametrized $S$ using $r(\theta,z) = (3\cdot\cos(\theta), 3\cdot\sin(\theta),z)$ and found $F \cdot (r_\theta\times r_z)$ is $18$.

Thus I have $18(5)(2\pi) = 180\pi$

but I have a different answer from the teacher.

Any insights as to what I did wrong?

Answer 1

\begin{align*}\int_S \vec{F} \cdot d\vec{S} &= \iint_D \vec{F}(r(\theta,z)) \cdot \vec{n}(\theta,z) \ dA \\ \\ & = \int_{z=0}^{z=5}\Bigg( \int_{\theta = 0}^{\theta = 2\pi} \vec{F}(r(\theta,z)) \cdot r_\theta \times r_z \ d\theta \Bigg)\ dz\end{align*}

Answer 2

I think you forgot that the radius of the cylinder was $3$, not $1$.

You don't really need a parametrization, but if you use one, you should get that $$ \mathbf{n}(x,y,z) = \frac{1}{3}\left<x,y,0\right> $$ Therefore $$ \mathbf{F} \cdot\mathbf{n} = \left<2x,2y,2z\right> \cdot \frac{1}{3}\left<x,y,0\right> = \frac{2}{3}(x^2 + y^2) = 6 $$ (not $18$, as you have). So $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = 6 \cdot\operatorname{Area}(S) = 6 \cdot 5\cdot 6\pi = 180\pi $$ (again, the circumference is $6\pi$, not $2\pi$).