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Math Help - coordinate transformation double integral

  1. #1
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    coordinate transformation double integral

    Hey everyone, I am stuck on this question, I just can't get anymore done. Any help would be great.

    Evaluate \int_1^2 \int_{1/y}^y  \ \sqrt{\frac{y}{x}} \cdot e^{\sqrt{xy}} \  dx \ dy by making the coordinate transformation x=\frac{u}{v} and y=uv.

    My working so far:
    xy=u^2

    \frac{y}{x}=v^2

    Jacobian  \ = \frac{2u}{v}

    So:
    \int_1^2 \int_{1/uv}^{uv}  \ \sqrt{v^2} \cdot e^{\sqrt{u^2}} \cdot \frac{2u}{v} \  du \ dv

    But I don't get how to go from here, with the limits of one of the integrals having terminals with u and v in it.

    If anyone can help that would be fantastic!

    Thanks
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  2. #2
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    The limits of integration, in terms of x and y are: y= 1, y= 2, and x= 1/y, x= y.

    Since x= u/v and y= uv, those become uv= 1, uv= 2, and u/v= 1/uv, u/v= uv. The first two give v= 1/u and and v= 2/u while the last two reduce to u^2= 1 and v^2= 1.

    Drawing those lines on a uv- graph, you can see that u= 1 and v= 1 intersect at (1, 1) on the graph of v= 1/u and intersect the graph of y= 2/v at u= 1, v= 2 and u= 2, v= 1. Essentially, then, the region you want to integrate over, in the uv-plane, is bounded by u= 1, v= 1, and v= 2/u. If u ranges from 1 to 2 then, for each u, v ranges from 1 to 2/u.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    The limits of integration, in terms of x and y are: y= 1, y= 2, and x= 1/y, x= y.

    Since x= u/v and y= uv, those become uv= 1, uv= 2, and u/v= 1/uv, u/v= uv. The first two give v= 1/u and and v= 2/u while the last two reduce to u^2= 1 and v^2= 1.

    Drawing those lines on a uv- graph, you can see that u= 1 and v= 1 intersect at (1, 1) on the graph of v= 1/u and intersect the graph of y= 2/v at u= 1, v= 2 and u= 2, v= 1. Essentially, then, the region you want to integrate over, in the uv-plane, is bounded by u= 1, v= 1, and v= 2/u. If u ranges from 1 to 2 then, for each u, v ranges from 1 to 2/u.
    Oh yes, thanks HallsofIvy, so the double integral is: \int_1^2 \int_1^{2/u} 2u \cdot e^u \ dvdu
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  4. #4
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    Quote Originally Posted by Nguyen View Post
    Oh yes, thanks HallsofIvy, so the double integral is: \int_1^2 \int_1^{2/u} 2u \cdot e^u \ dvdu

    Yes...yet the original integral is pretty easy to evaluate directly.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    Yes...yet the original integral is pretty easy to evaluate directly.

    Tonio
    Thanks Tonio!
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