- #1
DWill
- 70
- 0
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!
Integral of (x dx dy), limits of integration are from 0 <= y <= 6, 0 <= x <= y.
---------------
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!