I need to find the integral $\iint_D e^{x^2+y^2} \, dA$ over the disk $D = \{ (x,y) \; \vert \; x^2+y^2 \le 1\}$ by using polar coordinates.
First I inserted the $x= r \cdot \cos(\theta)$ and $y = r \cdot \sin(\theta)$ and simplified to find:
$\int_{0}^{2\pi}\int_{0}^{1} r\cdot e^{r^2}\,dr d\theta$
Then I used substituted $\frac{du}{2} = r \cdot dr$ to find:
$\int_{0}^{2\pi}\frac{1}{2}\int_{0}^{1} e^u \, du d\theta \rightarrow \int_{0}^{2\pi}\frac{1}{2}(e-1) \, d\theta$
But I am not sure whether to proceed with this now:
$\frac{1}{2}(e-1) \theta \bigg/_{\!\!\! 0}^{\,2\pi} = \frac{(e-1)2\pi}{2} = \pi(e-1)$
Should I proceed that way? Or am I calculating the integral of $\theta$ wrong?
It looks fine. I think that you are complicationg things a bit too much when you compute that last integral. Note that$$\int_0^{2\pi}\frac{e-1}2\,\mathrm d\theta$$is the integral of a constant function. So, the value of integral is just that constant times the length of the interval. That is, it is equal to $2\pi\times\frac{e-1}2=\pi(e-1)$.