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Math Help - Double Integrals

  1. #1
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    Double Integrals

    Can't find an example of this, this was in the previous final exams, and with finals only a week away, I need an explaination of a few problems from this last section of the book.

    Using geometry, evaluate the double integral over D of [(32-x^2-y^2)^.5]dA over the circular disk D: x^2+y^2<=36

    I know the answer is 144pi, but how do I come about this answer?

    Thanks for the help, I will definitely have more questions over the next week!
    Last edited by Bracketology; November 30th 2011 at 04:09 PM.
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  2. #2
    MHF Contributor

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    Re: Double Integrals

    Are you sure you have written that correctly? For (x,y) close to the boundary of that disk, x^2+ y^2 will be close to 36 which will make 32- x^2- y^2 negative.

    I am going to assume the function you want to integrate is (36- x^2- y^2)^{1/2}. With that, since the problem says "using geometry", we can interpret this integral as the volume of a the upper half of a sphere of radius 6: z= (36- x^2- y^2)^{1/2} gives z^2= 36- x^2- y^2 so x^2+ y^2+ z^2= 36. Of course, on the xy-plane, z= 0 so we have x^2+ y^2= 36, the given circle. The volume of a sphere is \frac{4}{3}\pi r^3 so the volume of the hemisphere is \frac{2}{3}\pi r^3, which is, indeed, \frac{2}{3}\pi 6^3= 144\pi.

    It's very easy to do that in polar coordinates: \int_{\theta= 0}^{2\pi}\int_{r= 0}^6 \sqrt{36- r^2} rdrd\theta. Let u= 36- r^2 and it becomes \pi \int_{u= 0}^{36} u^{1/2}du
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