I've been asked to solve the following integral using variable substitution:
$$\int_0^{3/5}\int_y^{3-4y}(x+4y)\cdot e^{y-x}\;dx\;dy$$
Letting $u=x+4y$ and $v=y-x$, the integrand (without the Jacobian) becomes the following:
$$u\cdot e^v$$
Isolating $x$ and $y$, we get the following:
$$x=\frac{u-4v}{5}$$
$$y=\frac{u+v}{5}$$
Now to find the Jacobian:
$$J(u,v)=\begin{vmatrix}\frac{1}{5} & \frac{-4}{5} \\ \frac{1}{5} & \frac{1}{5} \end{vmatrix}=\frac{1}{5}$$
Lastly,to find the new bounds. This is where I have been struggling. When I try to get my book to explain the new bounds of integration, it only does three of them and a $0$ appears as the last bound and I can't for the life of me figure out why.
Substituting $x=y$, I get the following bound:
$$5v=0\implies v=0$$
Substituting $x=3-4y$, I get the following bound:
$$5u=15\implies u=3$$
Substituting $y=0$, I get the next bound:
$$0=\frac{u+v}{5}\implies u=-v$$
But the book doesn't show how I get my other limit of integration. When I set $y=\frac{3}{5}$, I get the following:
$$u+v=3$$
But then the book immediately gives the following as my new iterated integral:
$$\frac{1}{5}\int_0^3\int_{-u}^0u\cdot e^v\;dv\;du$$
I just want to know where the second $0$ comes from as I can't for the life of me figure it out. My only idea is that they've plugged $u=-v$ into the last equation to get $u=0$.
Best Answer
The red region is the region of integration in the $uv$ coordinate system. The additional bound $u+v=3$ (the green line) does not further bound the region.
You can see that $v$ is bounded below by $v=-u$ and above $v=0$ and that $u$ is restricted to the interval $[0,3]$.