# Changing cartesian integral to polar integral

1. Jun 25, 2008

### DWill

Change the Cartesian integral to the equivalent polar integral and evaluate:

Integral of (x dx dy), limits of integration are from 0 <= y <= 6, 0 <= x <= y.

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I don't need help as much in evaluating the integral as just setting it up right. To change this to a polar integral do I change the integrand to [r^2 * cos(theta) dr d(theta)]? I'm not sure how exactly to change the limits of integration.

For the x limits, I see that the top limit is x = y, and if I substitute I get r cos(theta) = r sin(theta), which simplifies to theta = pi/4. So I thought this integral should have limits of 0 <= theta <= pi/4.

For y I'm more not sure, I tried the same approach as above. Since the top limit is y = 6, I thought of substituting to get r sin(theta) = 6, and simplifying to r = 6/sin(theta). So I thought the limits on r would be 0 <= r <= 6/sin(theta)

Well I solved the integral with what I tried figuring out above and got -18, while the correct answer is 36. Which parts did I do wrong? thanks!

2. Jun 25, 2008

### nicksauce

Ok you have one theta limit, but what about the other one?

When you have x = 0, and y = anything, you have theta = pi/2, not theta = 0.

It might help if you carefully draw out your region of integration.