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- May 24th 2011, 02:48 AMtransgalacticvariable substitution integral mmn 15 2c
- May 24th 2011, 03:32 AMProve It
After sketching the region

http://i22.photobucket.com/albums/b3...etchgraph2.jpg

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

$\displaystyle \displaystyle \int_0^{\sqrt[3]{28}}{\frac{1}{2}x^2 - \frac{1}{4}x^2\,dx} + \int_{\sqrt[3]{28}}^{\sqrt[3]{112}}{\sqrt{x} - \frac{1}{4}x^2\,dx}$. - May 24th 2011, 03:40 AMtransgalactic
coold you hekp me solve it by my way

because on the test it not effective to solve it like this - May 24th 2011, 03:51 AMHallsofIvy
Nice graph! Transgalactic, the point of the u, v substitution is that the curves $\displaystyle x^2= 2y$, $\displaystyle x^2= 4y$ and $\displaystyle y^2= 7x$, 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: $\displaystyle dxdy= \frac{\partial x\partial y}{\partial u\partial v} dudv$. And that symbol refers to the determinant, $\displaystyle \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|$.

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, $\displaystyle \frac{\partial u\partial v}{\partial x\partial y}= \left|\begin{array}{cc}\frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}\end{array}\right|$ and take the reciprocal. - May 27th 2011, 05:43 AMtransgalactic
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