I'm having difficulty integrating the double integral $\int\int_R\frac{3x-y}{3x+y} dA$ where R is the region inside the triangle with vertices $(0,0), (1,3)$ and $(2,0)$.

I have tried converting it to $dx dy$ form, integrating $\int_0^3\int_{\frac13y}^{-\frac{y}{3}+2} \frac{3x-y}{3x+y} dx dy$ instead, but I keep getting the answer 0, instead of the correct answer $\frac32$. Does anyone have any idea what might have gone wrong and how to solve this problem?

Here's what I tried doing in greater detail: $\int_0^3\int_{\frac13y}^{-\frac{y}{3}+2} \frac{3x-y}{3x+y} dx dy = \int_0^3\int_{\frac13y}^{-\frac{y}{3}+2} 1 - \frac{2y}{3x+y} dx dy = \int_0^3 -\frac23y + 2 - \frac23ln6y + \frac23yln(2y) dy = -3 + 6 - 3ln6 + 3ln6 - 3 = 3-3 = 0$.

Any help is much appreciated!

  • $\begingroup$ You should show the work you've done, so that someone can pinpoint your error. $\endgroup$ – Tim Raczkowski Jan 15 '15 at 5:01
  • $\begingroup$ I've added in some working $\endgroup$ – inggumnator Jan 15 '15 at 5:31
  • $\begingroup$ After $\int_0^3\int_{y/3}^{2-y/3}\left(1-{2y\over 3x+y}\right)\,dxdy$, I get $\int_0^3\left(x-{2y\over 3}\ln(3x+y)\right)|_{x=y/3}^{x=2-y/3}\,dy$. $\endgroup$ – Tim Raczkowski Jan 15 '15 at 5:40
  • $\begingroup$ In the next line you should have $\ln6$ instead of $\ln6y$. Also remember that $\ln0$ is undefined. Finally you need to find the antiderivative of $\frac{2y}3\ln2y$. $\endgroup$ – Tim Raczkowski Jan 15 '15 at 5:54

I have given the solution in handwritten form and I hope it is legible.



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