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Math Help - double integration with polar coordinates

  1. #1
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    double integration with polar coordinates

    By changing to polar coordinates, evaluate the integral (where a>0):

    int[a,0]int[sqrt(a^2 - x^2), 0](x^2 + y^2)dydx
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  2. #2
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    May aswell post here!

    Draw a picture of the region of the limits.

    The lower limits are x = 0 and y = 0, which tells you that both x > 0 and y > 0, which means you're dealing with the 1st quadrant of the cartesian axis system.

    Sketching the upper y limit  y = \sqrt{a^2-x^2}

    gives:  y^2 + x^2 = a^2

    Circle with centre (0,0), radius a.

    So from that sketch you should see that, in polar, the  \theta limits range from  0 \to \frac{\pi}{2} .

    And the radius ranges from  0 \to a . Which converts to:

    I =  \int_0^{\frac{\pi}{2}} \int_0^{a} ((rcos(\theta))^2+(rsin(\theta))^2).|\frac{d(x,y)}  {d(r,\theta)}| dr d\theta

    |\frac{d(x,y)}{d(r,\theta)}| = r

     I = \int_0^{\frac{\pi}{2}} \int_0^{a} r^3 dr d\theta
    Last edited by Mush; January 1st 2009 at 07:12 PM.
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