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Math Help - Substitutions in multiple integrals

  1. #1
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    Substitutions in multiple integrals

    R is a region in the first quadrant of the xy-plane bounded by xy=1, xy=9, y=x, and y=3x. Using the transformation v=\sqrt{\frac{y}{x}}, u=\sqrt{xy} with u>0, v>0 to rewrite the integral below over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G.

    \displaystyle \int\int_{R}(\sqrt{\frac{y}{x}}+\sqrt{xy})dxdy

    I'm totally lost on this. This is an exam review question but for all of our examples in the textbook the transformation is written in terms of x=... and y=...

    I have no idea what to do here.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by downthesun01 View Post
    R is a region in the first quadrant of the xy-plane bounded by xy=1, xy=9, y=x, and y=3x. Using the transformation v=\sqrt{\frac{y}{x}}, u=\sqrt{xy} with u>0, v>0 to rewrite the integral below over an appropriate region G in the uv-plane. Then evaluate the uv-integral over G.

    \displaystyle \int\int_{R}(\sqrt{\frac{y}{x}}+\sqrt{xy})dxdy

    I'm totally lost on this. This is an exam review question but for all of our examples in the textbook the transformation is written in terms of x=... and y=...

    I have no idea what to do here.
    Observe that xy=1\implies \sqrt{xy}=1 and xy=9\implies \sqrt{xy}=3 (we can't allow for -3 given the restriction on u). Thus, we see that 1\leq u\leq 3.

    Similarly, we see that y=x\implies \sqrt{\dfrac{y}{x}}=1 and y=3x\implies \sqrt{\dfrac{y}{x}}=\sqrt{3} (again, we can't allow for -\sqrt{3} due to the restriction on v). Thus we see that 1\leq v\leq \sqrt{3}. We now have the limits for the transformed integral.

    Next, we need to get x and y in terms of u and v and then find its Jacobian.

    Observe that

    (1) xv^2=y from the definition of v, and
    (2) u^2=xy from the definition of u.

    Substitute (1) into (2) to see that u^2=x^2v^2\implies x=\dfrac{u}{v}.

    Multiply the corresponding sides of (1) and (2) together to get xu^2v^2=xy^2\implies y=uv

    Now find the Jacobian. I leave it for you to verify that J(u,v)=\dfrac{2u}{v}.

    Therefore, \displaystyle\iint\limits_R\left(\sqrt{\dfrac{y}{x  }}+\sqrt{xy}\right)\,dx\,dy=\int_1^{\sqrt{3}}\int_  1^3\left(2u+\frac{2u^2}{v}\right)\,du\,dv

    Can you proceed? Does this all make sense?
    Last edited by Chris L T521; November 22nd 2010 at 11:09 PM.
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    You helped tons. Thanks a bunch.
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