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Math Help - Double Integration with polar co-ordinates

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    Double Integration with polar co-ordinates

    Let D be the disk of radius 6 centered at the origin. Find the
    double integral of the function f (x, y) = e^(x^2 +y^2) over D

    When I do this i get a function that says e^r^2, and then I'm uncertain as to how to integrate this, I don't think I'm using polar co-ordinates correctly. Any suggestions/ideas?
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    MHF Contributor Calculus26's Avatar
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    recall dA = rdrdtheta

    so you have re^r^2) which can be integrated with a simple substituion
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    what's the substitution? sorry my integration is a bit rusty...
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    MHF Contributor Calculus26's Avatar
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    you can let u = e^(r^2) or simply u = r^2
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    Ted
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    Quote Originally Posted by llamagoogle View Post
    Let D be the disk of radius 6 centered at the origin. Find the
    double integral of the function f (x, y) = e^(x^2 +y^2) over D

    When I do this i get a function that says e^r^2, and then I'm uncertain as to how to integrate this, I don't think I'm using polar co-ordinates correctly. Any suggestions/ideas?
    Calculus26 told you dA=rdrd\theta, probably you forgot the additional r.
    Converting to polar:
    \int_0^{2\pi} \int_0^6 e^{r^2} r dr d\theta.
    Use the substitution u=r^2 to solve the inner integral.
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    Quote Originally Posted by Ted View Post
    Calculus26 told you dA=rdrd\theta, probably you forgot the additional r.
    Converting to polar:
    \int_0^{2\pi} \int_0^6 e^{r^2} r dr d\theta.
    Use the substitution u=r^2 to solve the inner integral.
    Thanks for your reply, I think I was quite unclear in my question though, my problem lay with integrating a function that was e^u,(integrating the substitution), and when I tried to carry out the second integral, it did not have any variables with theta hence, I was unsure how to proceed...
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  7. #7
    o_O
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    Recall that: \int a dx = ax + C

    i.e. after doing the inner integral, treat the result as a constant in terms of \theta.
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  8. #8
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    Quote Originally Posted by o_O View Post
    Recall that: \int a dx = ax + C

    i.e. after doing the inner integral, treat the result as a constant in terms of \theta.
    oh yes sorry, seems like a dumb question now, thanks!
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