Double integral again

kessie

I've been playing around with this for a while. Anyone can help me out?

**Posted:** Wed Feb 09, 2005 1:20 am

blahblahblah

Over the region, we want to integrate:

$\int^5_0\int^{\sqrt {25-y^2}}_0 xe^ydxdy$

The innermost integral evaluates to:

$\frac {25-y^2}{2}e^y$

You can do the rest (it's straightforward integration by parts). It might be faster the other way, but I stopped as soon as I saw that this would work.

Oh, and the answer's $\frac{8e^5-23}{2}$

**Posted:** Wed Feb 09, 2005 1:40 am

liyi

hmm.
Polar transformation:
$x=r\sin\theta$ and $y=r\cos\theta$ , and $r\in[0,5], \theta\in[0,\pi/2]$ .

$\int_D y e^x dx dy = \int_0^5 \int_0^{\pi/2} r\cos\theta e^{r\sin\theta} rd\theta dr = \int_0^5 r \left.e^{r\sin\theta}\righ...$

**Posted:** Wed Feb 09, 2005 2:19 am