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Math Help - Double Integration - Polar Co-ordinates

  1. #1
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    Double Integration - Polar Co-ordinates

    \iint(lower case B) e^-2((x^2)+(y^2)) dydx

    where B is the quarter of the circle of radius 1 in the upper right-hand quadrant

    ---------------------------------------------------------------------------------

    ok so e^-2((x^2)+(y^2)) becomes e^-2(r^2)

    the inner limits are between 0<r<1 since r=1

    and the integral goes from dydx to dr d(theta)

    the outer limit is 0<theta<(pi/2) since were are dealing with the top right hand quadrant?

    can the integral e^-2(r^2) dr be done? i cant see how, unless we change the order? could be way off there though.
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  2. #2
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    \displaystyle dy\, dx doesn't become \displaystyle dr\,d\theta, it becomes \displaystyle r\,dr\,d\theta, and \displaystyle r\,e^{-2r^2} CAN be integrated using a substitution.
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    with substitution i get -0.25e^(-2(r^2)) and evaluting between one and zero i get:

    (-0.25e^(-2))+0.25 d theta

    theta*((-0.25e^(-2))+0.25) then sub in pi/2

    seem correct?
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    Yes, so now evaluate \displaystyle \left \left(1-e^{-2}\right)\frac{\theta}{4}\right|_0^{\frac{\pi}{2}}
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