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Math Help - polar form of double integral

  1. #1
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    polar form of double integral

    I have to turn this in to polar

    the double integral (x^2+y^2 )^(3/2) from o to 3 and Sqrt(9-x^2)

    I THINK THE ANSWER IS (243PI /10)
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  2. #2
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    Quote Originally Posted by cchyfly View Post
    I have to turn this in to polar

    the double integral (x^2+y^2 )^(3/2) from o to 3 and Sqrt(9-x^2)

    I THINK THE ANSWER IS (243PI /10)
     <br />
I=\int_{x=0}^3 \int_{y=0}^{\sqrt{9-x^2}} (x^2+y^2)^{3/2} ~dy~dx<br />

    is the integral of r^3 over the first quadrant of the interior of the circle (x^2+y^2)=r^2=3, so:

     <br />
I=\int_{r=0}^3 \int_{\theta=0}^{\pi/2} (r^3)~r d \theta dr <br />

    RonL
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  3. #3
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    so actually in the last step it is r^4 and then evaluate the integral
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  4. #4
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    Quote Originally Posted by cchyfly View Post
    so actually in the last step it is r^4 and then evaluate the integral
    Yes, I left it in the form I did so you could more easily see where things came from.

    RonL
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