# Thread: variable substitution integral mmn 15 2c

1. ## variable substitution integral mmn 15 2c

2. After sketching the region

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

$\displaystyle \int_0^{\sqrt[3]{28}}{\frac{1}{2}x^2 - \frac{1}{4}x^2\,dx} + \int_{\sqrt[3]{28}}^{\sqrt[3]{112}}{\sqrt{x} - \frac{1}{4}x^2\,dx}$.

3. coold you hekp me solve it by my way
because on the test it not effective to solve it like this

4. 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.

5. 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