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Math Help - Double integral with transformations.

  1. #1
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    Double integral with transformations.

    I need help in determine the limits of the double integrals where I had to use transformations.

    The double integral is \int\int_R (4x + 8y) dA where R is a parallelogram with vertices (-1,3), (1,-3), (3,-1) and (1,5).

    The transformations were given as: x = \frac{1}{4}(u+v), y=\frac{1}{4} (v-3u).

    Using the transformation equations of u=x-y and v=3x+y (solving simultaneoulsy from above), the points in the transformed plane T^{-1} become (-4,-2), (-4,-6), (4,10) and (4,6).

    I calculated the Jacobian as \mid 4 \mid and the integrand becomes 3v-6u.

    So,
    \int\int_R (4x+8y) dA = \int\int_r (3v-6u)   \mid 4 \mid dv du where r is the region in the transformed plane.
    = 12 \int_{-4}^{4} \int_*^* (v-2u) dv du

    Could someone please help to me with what the * * limits of the inner integral would be and correct any errors I have made in getting this far?

    Thanks in advance,
    Tammy
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  2. #2
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    Re: Double integral with transformations.

    Sorry for putting anyone out. I think I found the problem. Simple math errors have brought me undone.

    First off I calculated the transformed points wrong. They should of been (-4,0), (-4,8), (4,0) and (4,8) which gives a nice square.

    Also the integrand should of been 3v - 5u.

    So,
    \int_{-4}^{4}\int_0^8 (3v-5u) \mid4\mid dv du = 3072

    Thanks to anyone you read this. Please help with my other post about the TRIPLE INTEGRAL.
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