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Math Help - Double Integral

  1. #1
    Junior Member
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    Double Integral

    Hey there, i am having trouble doing some questions on double integrals, ive done most of them.
    Though im having trouble on question 2) ii) and question 3
    i was wondering if someone could point me in the right direction on how to start evaluating these integrals.

    thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    For the second question, you can see this thread.

    For the third one, the appropriate coordinates are the polar ones.
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  3. #3
    Super Member wingless's Avatar
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    \int e^{-x^2}~dx is called the Gaussian Integral. Even though the indefinite integral can't be evaluated, we can find the definite integral (from 0 to ∞) or (from -∞ to ∞).

    Let I = \int_0^{\infty}e^{-x^2}~dx.

    Then multiply it by itself: I.I = \int_0^{\infty}e^{-x^2}~dx \int_0^{\infty}e^{-x^2}~dx

    Change x in the second integral to y:

    Then multiply it by itself: I.I = \int_0^{\infty}e^{-x^2}~dx \int_0^{\infty}e^{-y^2}~dy

    This is,

    I^2 = \int_0^{\infty}\int_0^{\infty}e^{-x^2-y^2}~dx~dy

    Change x-y to polar coordinates.

    x= r\cos\theta
    y= r\sin\theta
    dx~dy =r~dr~d\theta

    I^2 = \int\int e^{-r^2}r~dr~d\theta

    I^2 = \int d\theta \int e^{-r^2}r~dr

    The integration area is now (0\leq\theta\leq \frac{\pi}{2}) and (0\leq r \leq \infty)

    I^2 = \int_0^{\frac{\pi}{2}} d\theta \int_0^{\infty} e^{-r^2}r~dr

    Calculate the integral and take its square root to find I.
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