I want to find $$\int \int_ R \sin\left(\frac{x-y}{x+y}\right)dA$$ where $R$ is the trapezoid with vertices $(1,1)$, $(2,2)$, $(4,0)$ and $(2,0$). I've seen some examples with similar integrals in my calculus book using triangles instead, but I'm confused as to what happens when we decide to use some other geometric figure. The way I understood it from these examples was that you integrate $dx$ from the lowest x values to the highest (where y=0) ($2$ and $4$ here) , and for the other integral the bounds $x=1$ from $(1,1)$ and $-x+y$ (the line through $(1,1)$ and $(2,2)$
Thus I get $$\int_2^4 dx \int_1^{-x+y}\sin\left(\frac{x-y}{x+y}\right)dy$$. I'm not sure if my understanding of these bounds are correct though as I didn't succeed in trying to evaluate this with wolfram.
Appreciate any help!
Best Answer
Gimusi's answer gets you the right result, however you could use the change of variables $u = x+y , v = x-y$. Pay attention to what the domain $R$ looks like under this (linear!) transformation. I'll give you a hint as what the integral should look like when you're done $$\frac12\int_2^4 dv \int_0^v du \sin(\frac uv)$$ where $\frac12$ is the determinant resulting from the change of variables. From here it's elementary calculus.
Also, in your computation you can see at a glance that there's something wrong: you have the variable $y$ appearing in the upper extreme of the second integral, but you're integrating with respect to that same variable!