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Math Help - Double integral

  1. #1
    Senior Member Dinkydoe's Avatar
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    Double integral

    I'm asked to calculate the double integral

    \int{\int_{D}(x+y)^2}dxdy with D=\left\{(x,y)\in \mathbb{R}^2: (x-2y)^2+(3x+3y)^2\leq 9\right\}

    It seems the best to me, to calculate this with a substitution and adjusted domain S:

    Take u=x-2y, v=3x+3y so that our new domain S will be \left\{(u,v):u^2+v^2\leq 9\right\}

    and we have x=\frac{1}{9}(3u+2v), y=\frac{1}{9}(v-3u)

    And our new integral becomes \frac{1}{9}\int\int_{S}v^2dS. However I wasn't sure I calculated the Area element dS correctly.

    I found dS= 9dudv. Is this all correct?
    Last edited by Dinkydoe; April 11th 2010 at 11:22 AM.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    I'm asked to calculate the double integral

    \int{\int_{D}(x+y)^2}dxdy with D=\left\{(x,y)\in \mathbb{R}^2: (x-2y)^2+(3x+3y)^2\leq 9\right\}

    It seems the best to me, to calculate this with a substitution and adjusted domain S:

    Take u=x-2y, v=3x+3y so that our new domain S will be \left\{(u,v):u^2+v^2\leq 9\right\}

    and we have x=\frac{1}{9}(3u+2v), y=\frac{1}{9}(v-3u)

    And our new integral becomes \frac{1}{9}\int\int_{S}v^2dS. However I wasn't sure I calculated the Area element dS correctly.

    I found dS= 9dudv. Is this all correct?
    In 2D space the change of variables means that we now have

    \int{\int_{D}(g(u,v))}det|d(x,y)/d(u,v)|dudv

    So take the determinent. This means we want

    |x1,x2/y1,y2|dudv = 1/9

    So you are correct.
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