# Thread: Change integral to polar coordinates and evaluate.

1. ## Change integral to polar coordinates and evaluate.

Hi, I am having difficulty picturing what to do with this double integral up to a certain point. I have scanned my work and ask that someone take a look at it and see where I got up to and tell me what I should do next. Integrate by parts? Substitution? I'm not sure. I have attached my scans to this thread. Thanks.

2. Originally Posted by Undefdisfigure
Hi, I am having difficulty picturing what to do with this double integral up to a certain point. I have scanned my work and ask that someone take a look at it and see where I got up to and tell me what I should do next. Integrate by parts? Substitution? I'm not sure. I have attached my scans to this thread. Thanks.
Shouldn't your circle be centered at the origin in image 1?

3. Yes it should, I'll fix that. But what should I do with this integral? Even with the image
fixed, the integral would be set up the same way.

4. Originally Posted by Undefdisfigure
Yes it should, I'll fix that. But what should I do with this integral? Even with the image
fixed, the integral would be set up the same way.
U-Sub

$\displaystyle u=r\cos(\theta)$

5. Originally Posted by dwsmith
U-Sub

$\displaystyle u=r\cos(\theta)$
Provided you are integrating with respect to $\displaystyle \displaystyle \theta$ first...

6. So do I have to switch the order of integration or is there a way to integrate with respect to r first?

7. You'll have to switch the order...

8. But du would equal -r sin(theta) and it's r^2 in front of the sin(theta). Do I remove the r (1/r) and put it in front of the integral?

9. When integrating with respect to $\displaystyle \displaystyle \theta$ you treat $\displaystyle \displaystyle r$ as a constant.

10. I sent a private message and I don't know if it was received but this is what it was:

In the question I asked, when you substitute u = r cos(theta), you get
du = - r sin(theta). When treating r as a constant, does that mean it doesn't matter if it's r^2 sin (theta) and you just integrate e^u du with respect to theta?

11. Originally Posted by Undefdisfigure
I sent a private message and I don't know if it was received but this is what it was:

In the question I asked, when you substitute u = r cos(theta), you get
du = - r sin(theta). When treating r as a constant, does that mean it doesn't matter if it's r^2 sin (theta) and you just integrate e^u du with respect to theta?
You have this integral

$\displaystyle \iint_{R}ye^{x}dA$

Going to polar gives

$\displaystyle \displaystyle \int_{0}^{5}\int_{0}^{\frac{\pi}{2}} r\sin(\theta)e^{r\cos(\theta)}r d\theta dr$

Then using the recommend substitution gives

$\displaystyle u=r\cos(\theta) \implies du=-r\sin(\theta)d\theta$
$\displaystyle u=r\cos(0) \implies u=r$ and $\displaystyle \displaystyle u=r\cos\left(\frac{\pi}{2}\right) \implies u=0$

$\displaystyle \displaystyle \int_{0}^{5} \left(-\int_{r}^{0}re^{u}du \right) dr =\int_{0}^{5}\int_{0}^{r}re^{u}dudr$