Changing a double integral to polar coordinates

[Ral]

Homework Statement

Rewrite by converting to polar coordinates, carefully drawing R.

\int^{2}_{0}\int^{\sqrt{2x-x^2}}_{0}\sqrt{x^2+y^2}dydx

Homework Equations

The Attempt at a Solution

I believe I have the inside part of it right. What I did was replace the x^2 and y^2 in \sqrt{x^2+y^2} with r^2cos^2\vartheta and r^2sin^2\vartheta, factored out the r^2, which left me with cos^2\vartheta+sin^2\vartheta, which equals 1. So then I would be left with \sqrt{r^2}rdrd\vartheta

What's confusing me is what my limits on the integrals should be. What I have right now is

\int^{\pi/2}_{0}\int^{2}}_{0}\sqrt{r^2}rdrd\vartheta
but I'm still unsure if that's correct or not.

 

[Mark44]

Have you drawn the region R? From your Cartesian integral, the limits for y range from 0 to sqrt(x^2 - 4). You show this as sqrt(x^2 - 2^2). Is there some reason you wrote 2^2 instead of 4?

Can you show us your reasoning in changing the limits of integration as you did?

One last thing. In your polar integral you should simplify sqrt(r^2) to r.

 

[Ral]

That limit should have been \sqrt{2x-x^2. I messed up typing it up, I edited it in the right limit.

 

[Mark44]

OK, I thought there might be a typo there. Your region R over which integration takes place is bounded by y = 0 and y = sqrt(2x - x^2), and by x = 0 and x = 2.

Can you draw this region? This problem also asks you to do that. The regions is a simple geometric shape, and one that will make the polar integral very easy.

Many times in these types of problems, the hardest part is understanding exactly what the region of integration looks like. If you can't do this, changing the limits will pretty much be insurmountable.

 

[Ral]

http://img96.imageshack.us/img96/5444/66431931.jpg

Here's what I had. The 1st quadrant would be shaded. The one that's confusing me the most is the integral for dr, while I think the integral for d\vartheta is correct.

 

[Mark44]

Yes, that's what R looks like.

The one that's confusing me the most is the integral for dr, while I think the integral for d\vartheta
is correct.

Do you mean the limits of integration for r and \vartheta?
Otherwise, I don't understand what you mean.

 

[Ral]

Yes, I mean the limits of integration for r and theta. So if that R is correct, then that means I had the correct answer, right?

 

[Mark44]

Yes, but I would write the polar integral as \int^{\pi/2}_{0}\int^{2}}_{0}r^2drd\vartheta

In the polar interpretation of R, r extends from 0 to 2. All points on the circle are 2 units from its center. \theta extends from 0 to \pi/2. That gives you the quarter circle in the first quadrant.

 

[Ral]

Yeah, I was going to do that simplification. Thanks for helping me clarify this.