How to find the surface integral of $f(x,y,z)=e^{-z}$

Question

How can I find the surface integral of $f(x,y,z)=e^{-z}$ given that $x^2+y^2=9$ and $0\leq z \leq 2$.

I started by changing the $x$ and $y$ coordinates to polar coordinates:

$x=3\cos(\theta),\ y=3\sin(\theta)$

I am confused as to what to do from this point.

Answer 1

One option is to parameterize the cylinder $S$ in cylindrical coordinates using the vector function

$$\vec s(\theta,z) = 3\cos(\theta)\,\vec\imath + 3\sin(\theta)\,\vec\jmath + z\,\vec k$$

with $0\le \theta\le2\pi$ and $0\le z\le2$. Then the surface integral is

$$\iint_S e^{-z} \, dA = \int_0^2 \int_0^{2\pi} e^{-z} \left\|\frac{\partial\vec s}{\partial \theta} \times \frac{\partial\vec s}{\partial z}\right\| \, d\theta \, dz$$

Answer 2

You can use the formula $$\iint_S f(\vec{r}(\theta,z))\biggr|\biggr|\frac{\partial\vec{r}}{\partial\theta}\times \frac{\partial\vec{r}}{\partial z}\biggr|\biggr|d\theta dz$$ for a given parametrization $\vec{r}:\mathbb{R}^2\rightarrow\mathbb{R}^3$ of a surface $S$ and $f:\mathbb{R}^3\rightarrow \mathbb{R}$. In this case you have $\vec{r}(\theta,z) = (3\cos\theta, 3\sin\theta,z)$ for $0\leq \theta < 2\pi, 0 \leq z \leq 2$. Calculating $\frac{d\vec{r}}{d\theta}$ and $\frac{d\vec{r}}{dz}$: $$\frac{\partial\vec{r}}{\partial\theta} = (-3\sin \theta, 3\cos \theta, 0)$$ $$\frac{\partial\vec{r}}{\partial z} = (0,0,1)$$ and $$\frac{\partial\vec{r}}{\partial\theta}\times \frac{\partial\vec{r}}{\partial z} = (3\cos\theta,3\sin\theta,0), \biggr|\biggr|\frac{\partial\vec{r}}{\partial\theta}\times \frac{\partial\vec{r}}{\partial z}\biggr|\biggr| = 3$$ With this and $f(\vec{r}(\theta,z))=e^{-z}$, you can calculate the integral over the whole surface $S$. $$\int_0^{2\pi}\int_0^2 3e^{-z}dzd\theta = 6\pi(1-e^{-2})$$