$\int \int_ A \sin\left(\frac{y-x}{x+y}\right)dA$
where $A$ is the trapzeium with vertices $(1,1), (2,2), (2,0), (4,0)$.
I decided to use $u=y-x$ and $v=y+x$ as my variables, and from attempting to change variables I got,
$\frac{1}{2}\int_{-2}^{0} \int_{?}^{4} \sin\left(\frac{u}{v}\right)dvdu$
I'm confused on how to work out the lower? limit for $v$. To work out my previous limits I drew the area, worked out each equation for each line and subbed in $u$ and $v$, but one line was $x=0$, hence the confusion as $u$ and $v$ can't be subbed in here.
Thanks in advance for any tips or help!
Best Answer
You've chosen to put $u$ as the last variable to integrate. This is not how I would personally go about it, but let's continue.
Let's transform these points in terms of $u$ and $v$. Our points become $$(u, v) = (0, 2), (0, 4), (-4, 4), (-2, 2).$$ Plot these on the $u$-$v$ plane. Note that $u$ ranges not from $-2$ to $0$, but $-4$ to $0$. The upper bound for $v$ is indeed $4$, but the lower bound is made up of two line segments, which have a cusp at $u = -2$. So, we should probably split the domain up.
For $u$ between $-4$ and $-2$, the lower bound follows the line $v = -u$. For $u$ between $-2$ and $0$, the line is $v = 0$. So, our integral is split like so:
$$\iint_A = \int_{-4}^{-2} \int_{-u}^4 + \int_{-2}^{0} \int_0^4.$$
If you instead put $v$ on the outside, you get something a lot nicer:
$$\int_2^4 \int_{-v}^0$$
Don't forget the Jacobean too!