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Thread: variable substitution integral mmn 15 2c

  1. #1
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    variable substitution integral mmn 15 2c

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    After sketching the region



    it should be clear that you can evaluate the required area as

    \displaystyle \int_0^{\sqrt[3]{28}}{\frac{1}{2}x^2 - \frac{1}{4}x^2\,dx} + \int_{\sqrt[3]{28}}^{\sqrt[3]{112}}{\sqrt{x} - \frac{1}{4}x^2\,dx}.
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    coold you hekp me solve it by my way
    because on the test it not effective to solve it like this
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    Nice graph! Transgalactic, the point of the u, v substitution is that the curves x^2= 2y, x^2= 4y and y^2= 7x, in the xy-plane, become, in the uv-plane, u= 2, u= 4, and u= 7. Of course, those don't form a bounded figure. That's because, as x and y go to 0, u and v go to infinity. So your integral will be for u= 2 to 4, v= 7 to infinity.

    Now, when you substitute for x and y, you also have to substitute for u and v: dxdy= \frac{\partial x\partial y}{\partial u\partial v} dudv. And that symbol refers to the determinant, \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|.

    In order to find that directly, you would have to solve for x and y in terms of u and v. The hint says you don't have to do that. Find, instead, \frac{\partial u\partial v}{\partial x\partial y}= \left|\begin{array}{cc}\frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}\end{array}\right| and take the reciprocal.
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  5. #5
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    i put v to be the vertical axes
    and u to be the horizontal axes
    so i dont have 7 to infinity
    but 0 to 7
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