1. ## Polar Double Integral

Find the double integral of y(e^x) over the region in the first quadrant enclosed by the circle x^2+y^2=25.

For this problem, I think I set up the integral properly, with the radius going from 0 to 5, and the angle going from 0 to pi/2. I set y=r*sin(theta) and x=r*cos(theta), but the first antiderivative from tabular integration by parts (w.r.t. r), was very messy, and the second antiderivative w.r.t. theta looks to be messy as well. I'm not sure whether to keep on brute forcing it, or whether there is a more elegant solution.

I would appreciate any help. Mathematica returns 4e^5 - 23/2 as the answer by the way.

2. Originally Posted by MathTooHard
Find the double integral of y(e^x) over the region in the first quadrant enclosed by the circle x^2+y^2=25.

For this problem, I think I set up the integral properly, with the radius going from 0 to 5, and the angle going from 0 to pi/2. I set y=r*sin(theta) and x=r*cos(theta), but the first antiderivative from tabular integration by parts (w.r.t. r), was very messy, and the second antiderivative w.r.t. theta looks to be messy as well. I'm not sure whether to keep on brute forcing it, or whether there is a more elegant solution.

I would appreciate any help. Mathematica returns 4e^5 - 23/2 as the answer by the way.
The integral will be, of course, $\displaystyle \int \int (r \sin (\theta)) e^{r cos(\theta)} \, (r dr \, d\theta)= \int \int r^2 e^{r \cos(\theta)} \sin(\theta) \, d\theta \, dr$. Integrate with respect to $\displaystyle \theta$ first, by letting $\displaystyle u= r \cos(\theta)$.

3. Are you saying that I just have to deal with extremely messy antiderivatives?

4. Originally Posted by MathTooHard
Are you saying that I just have to deal with extremely messy antiderivatives?
I don't see the difficulty with integrating first with respect to $\displaystyle \theta$. You make a straightforward u-substitution. Where is your trouble here? (Note that HoI did not include the integral terminals, but they should be obvious ....)

5. Oh thanks! The code didn't show up the first time so I didn't get it. I never thought of changing the orders of integration.