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Math Help - Another double integral

  1. #1
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    Another double integral

    Evaluate

    \int_0^1\int_0^{\pi/2}\frac{dx\,dy}{\sqrt{1-\sqrt y\cdot\sin^2x}}
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  2. #2
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    Reverse order

    If we attempt to do the x integral first, we get an elliptic integral, so we reverse the order of integration:
    \int_0^{\pi/2}\int_0^1\frac{dy\,dx}{\sqrt{1-\sqrt y\cdot\sin^2x}}
    Now, integrating with respect to y, we make the change of variables u=\sqrt{y}\to{dy}=2u\,du
    \int_0^{\pi/2}\int_0^1\frac{2u\,du\,dx}{\sqrt{1-u\cdot\sin^2x}}
    -\frac{4}{9}\int_0^{\pi/2}\left[\frac{(2+u\sin^2{x})\sqrt{1-u\sin^2{x}}}{\sin^4{x}}\right]_{u=0}^{1}\,dx (via a table of integrals)
    -\frac{4}{9}\int_0^{\pi/2}\frac{(2+\sin^2{x})\cos{x}-2}{\sin^4{x}}\,dx
    Using the substitution v=\sin{x}\to{dx}=\frac{dv}{\sqrt{1-v^2}}
    -\frac{4}{9}\int_0^{1}\frac{1}{v^2}+\frac{2}{v^4}-\frac{2}{v^4\sqrt{1-v^2}}\,dv
    This is an improper integral, with the integrand undefined at both limits. The limit as v\to0 does exist, but the limit as v\to1 is infinite. A little work, however, could be used to do this integral (if it exists)

    --Kevin C.
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  3. #3
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    Addendum

    The integral is finite: the antiderivative of the integrand is defined at v=1, and the limit as v\to0 also exists and is finite.

    --Kevin C.
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  4. #4
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    Quote Originally Posted by liyi View Post
    Evaluate

    \int_0^1\int_0^{\pi/2}\frac{dx\,dy}{\sqrt{1-\sqrt y\cdot\sin^2x}}
    Make the substitutions x = \theta & y = r^4 and apply Fubini

    \int_0^{\pi /2} {\int_0^1 {\frac{{4r^3 }}<br />
{{\sqrt {1 - r^2 \sin \theta } }}\,dr\,d\theta } } .

    This is a quarter of circle with radius 1. Cartesian transform

    4\int_0^1 {\int_0^{\sqrt {1 - y^2 } } {\frac{{x^2  + y^2 }}<br />
{{\sqrt {1 - y^2 } }}\,dx} \,dy}, and you have a routine integral.
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