1 1-x
∫ ∫ (sqrt(x+y)(y-2x)^2)dydx
0 0
$$ \int_0^1 \int_0^{1-x} \sqrt{x+y} \, (y-2x)^2 \,dy \, dx $$
I've determined that $u = x+y$ and $v = y-2x$ and that the jacobian is $= 1/3$. and that $x = (u-v)/3$ and that $y = (2u+v)/3$.
but I am having problems finding the boundaries of the new integral.
what I have attempted so far after that is:
x = 1 --> u = v+3
x= 0 --> u = v
y = 1-x --> v = (3-7u)/2
y = 0 --> v = -2u
Are these boundaries correct? if not, can someone help me on how to find them? I'm having a lot of trouble here.
The boundary of the region in the $xy$-plane is given by the lines $x=0$, $y=0$, and $y=1-x$.
You're right to say that $x=0$ becomes the line $u=v$ and $y=0$ becomes $v=-2u$. However, notice that $y=1-x$ corresponds to the line $u=1$ in the $uv$-plane. Graph these three lines in the $uv$-plane to determine your new region of integration.
You will find the new bounds should be:
$$\begin{align}-2u<v<u\\0<u<1\end{align}$$