# Double Integral Cartesian to Polar Coordinates

1. May 6, 2012

### jerzey101

1. The problem statement, all variables and given/known data
Use polar coordinates to evaluate:
sqrt(2)0sqrt(4-y2)y 1/(1+x2+y2) dxdy

2. Relevant equations

3. The attempt at a solution
I graphed it and I see r is the part of the elipse sqrt(4-y2) and goes from 0 to ∏/4. I'm not sure how to make the bounds for r or how to covert the function to polar. Does it just convert to 1/(1+r2), then add the conversion factor r?
∏/40?? 1/(1+r2) r drdθ

2. May 6, 2012

### sharks

Use LaTeX formatting to make your integrals clearer and easier to understand.

Your double integral looks like this:
$$\int^{y=\sqrt 2}_{y=0} \int^{x=\sqrt {4-y^2}}_{x=y} \frac{1}{1+x^2+y^2} \,.dxdy$$
You have to plot the limits to see the area required and hence deduce the correct polar limits.
First, convert: $x=\sqrt {4-y^2}$ into a simpler form. This gives you: $x^2+y^2=4$, which is a circle with center (0,0) and radius = 2. The other limits are obvious.

I have attached the graph. If you are having trouble deducing the region required from the limits, save the graph, edit/shade it to show the required area and then attach it to your reply.

#### Attached Files:

• ###### area.png
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Last edited: May 6, 2012
3. May 6, 2012

### jerzey101

I'm just not sure what to put for the r limits. I know it is 0 to the circle x2+y2=4. So is it just 0 to 2?

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4. May 6, 2012

### sharks

You're correct.

5. May 6, 2012

### jerzey101

Thanks!!
I think I figured it all out now. Thanks again.