Change integral to polar coordinates and evaluate.

**Undefdisfigure** · January 26th 2011, 07:09 PM

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.

**dwsmith** · January 26th 2011, 07:14 PM

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?

**Undefdisfigure** · January 26th 2011, 07:59 PM

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.

**dwsmith** · January 26th 2011, 08:03 PM

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

$u=r\cos(\theta)$

**Prove It** · January 26th 2011, 08:10 PM

Originally Posted by dwsmith

U-Sub

$u=r\cos(\theta)$

Provided you are integrating with respect to $\displaystyle \theta$ first...

**Undefdisfigure** · January 26th 2011, 10:36 PM

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

**Prove It** · January 26th 2011, 11:39 PM

You'll have to switch the order...

**Undefdisfigure** · January 27th 2011, 09:30 PM

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?

**Prove It** · January 27th 2011, 09:32 PM

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

**Undefdisfigure** · January 28th 2011, 12:12 PM

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?

**TheEmptySet** · January 28th 2011, 12:23 PM

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

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

Going to polar gives

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

Then using the recommend substitution gives

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

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

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