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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!

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  • $\begingroup$ You are probably expected to use a change of variable. A common one is $u=x-y$, $v=x+y$. $\endgroup$ – Umberto P. Feb 11 '18 at 18:36
  • $\begingroup$ Another possible change of variables is $u=\frac{x-y}{x+y}$, $v=x+y$. $\endgroup$ – StubbornAtom Feb 11 '18 at 18:39
  • $\begingroup$ What are the bounds in that case? Do I still keep the double integral? By book doesn't discuss change of variables under this chapter $\endgroup$ – novo Feb 11 '18 at 18:41
  • $\begingroup$ You'll still have a double integral, but you have to figure out the bounds. If you use the substitution $u=x-y, v=x+y,$ it maps lines to lines, so that the image of the trapezoid will be the quadrilateral with vertices $(0,2), (0,4), (4,4) \text{ and } (2,2),$ if my arithmetic is correct. This is the region you need to integrate over, since it's where $u,v$ take their values. It's just like figuring out the new limits after changing variables in a definite integral with one variable. $\endgroup$ – saulspatz Feb 11 '18 at 18:54
  • $\begingroup$ @novo Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png $\endgroup$ – gimusi Feb 13 '18 at 23:49
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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!

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  • $\begingroup$ Thank you! After some reading and a lot of head scratching I ended up with this. From here on out, it should be, as you say, elementary calculus. I was preparing for a lecture, but it appears the change in variables are not subject until next week, hence my confusion in the first place. Thank you for the assistance! $\endgroup$ – novo Feb 11 '18 at 20:30
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You need to divide the integral, for example as follow

$$\int_0^2 dy \int_{y-2}^{y-4}\sin\left(\frac{x-y}{x+y}\right)dy+\int_2^4 dy \int_{y}^{y-4}\sin\left(\frac{x-y}{x+y}\right)dy$$

enter image description here

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