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Math Help - Integral using polar co-ordinates

  1. #1
    Super Member Showcase_22's Avatar
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    Integral using polar co-ordinates

    Solve \int \int_{x^2+y^2 \leq 1} (1+x^2+y^2)^{\frac{1}{3}}~dx~dy

    Substituting x=r \cos \theta and y=r \sin \theta:

    \int_0^{2 \pi} \int_0^1 r(1+r^2)^{\frac{1}{3}}~dr~d \theta

    \int_0^{2 \pi} \left[ \frac{3}{8}(1+r^2)^{\frac{4}{3}} \right]_0^1~d \theta

    \int_0^{2 \pi} \frac{3}{8} \left( 2^{\frac{4}{3}}-1 \right)~d \theta

    =\frac{3 \pi}{4} \left(2^{\frac{4}{3}}-1 \right)

    This doesn't really look like a really good answer. =S

    So, is this right?
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    it's correct, so relax.
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