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Math Help - Double Integrals w/ Polar Coordinates

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    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 \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 y=\sqrt{1-x^2} to the line y=x, and x goes from 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 \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.
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