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Thread: evaluate the double integral

  1. #1
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    evaluate the double integral

    Using polar coordinates, \int\int_{R} \sin(x^2+y^2)dA evaluate the integral where R is the region  1 \leq x^2+y^2 \leq 64.

    Heres what I got sofar, I need help integrating the inner integral. I'm lost from here.
    1 \leq x^2+y^2 \leq 64
    1 \leq r^2 \leq 8
    1 \leq r \leq 8

    \int_{0}^{2pi}\int_{1}^{8}\sin(r^2) rdr d\theta
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  2. #2
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    Try a u-substitution u = r^2.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Because ...

    \frac {d}{dr} \cos (r^{2}) = - 2\cdot r\cdot \sin (r^{2})

    ... is...

    \int r\cdot \sin (r^{2}) \cdot dr = - \frac{1}{2}\cdot \cos (r^{2}) + c

    ... so that...

    \int_{0}^{2\cdot \pi} \int_{1}^{8} \sin (r^{2})\cdot r\cdot dr\cdot d\theta = \pi \cdot (\cos 1 - \cos 64) \simeq .46635958\dots

    Kind regards

    \chi \sigma
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