# Double Integrals w/ Polar Coordinates

• Mar 9th 2009, 08:08 PM
eigenvector11
Double Integrals w/ Polar Coordinates
I am given the following three double integrals for which to find the sum of. The question requires converting to polar coordinates (which is where I am having the most trouble):

First Integral: Integral(from 1 to (1/sqroot(2)))Integral(from x to sqroot(1-(x^2))) of x*y dydx.

Second Integral: Integral(from sqroot(2) to 1)Integral(from x to 0) of x*y dydx.

Third Integral: Integral(from 2 to sqroot(2))Integral(from sqroot(4-(x^2)) to 0) of x*y dydx.

Thanks for any ideas.
• Mar 10th 2009, 01:30 AM
Opalg
Quote:

Originally Posted by eigenvector11
I am given the following three double integrals for which to find the sum of. The question requires converting to polar coordinates (which is where I am having the most trouble):

First Integral: Integral(from 1 to (1/sqroot(2)))Integral(from x to sqroot(1-(x^2))) of x*y dydx.

Second Integral: Integral(from sqroot(2) to 1)Integral(from x to 0) of x*y dydx.

Third Integral: Integral(from 2 to sqroot(2))Integral(from sqroot(4-(x^2)) to 0) of x*y dydx.

Thanks for any ideas.

First comment is about the statement of the question. I don't know if this is just a cultural difference between Canada and Britain, but when I say "integral from a to b", I mean that a is the lower limit and b is the upper limit. These question all seem to be phrased the other way round, so that Integral(from 1 to (1/sqroot(2))) must presumably mean that 1/sqroot(2) is the lower limit and 1 is the upper limit.

So it looks as though the first integral should be $\displaystyle \int_{1/\sqrt2}^1\int_{\sqrt{1-x^2}}^xxy\,dy\,dx$.

The hardest part of the question is probably to figure out the region over which you are supposed to be integrating. In this case, y goes from the circle $\displaystyle y=\sqrt{1-x^2}$ to the line $\displaystyle y=x$, and x goes from $\displaystyle x=1/\sqrt2$ (which is where the line crosses the circle) to x=1. Draw a picture to see that the region of integration is the bounded by the unit circle, the line y=x and the line x=1. You will then be able to see from the picture that this region is described in terms of polar coordinates as follows: θ goes from 0 to π/4, and for each fixed value of θ, r goes from 1 to sec(θ). So the integral becomes $\displaystyle \int_0^{\pi/4}\int_1^{\sec\theta}(r\cos\theta)(r\sin\theta)r\, dr\,d\theta$. Once you have got that far, the actual evaluation of the integral should be easy.

Do the other questions the same way. The key step is to draw a picture of the region of integration.