# Evaluating a double integral with change of coordinates. $\int\!\!\int_{R} (x+y)\,dA$

It looks simple, but I'm having a bit of difficulty with trying to obtain the proper change of coordinates.

Evaluate the following integral
$$\displaystyle\int\!\!\!\int_{R} (x+y)\,dA$$

with region R in the first quadrant bounded by $-1\le y-x\le 1$ and $1\le x^2+y^2 \le 4$

I've tried $u=x^2+y^2$ and $v=y-x$, but I can't seem to express $x+y$ in terms of $u$ and $v$. Is it possible to express $x+y$ in terms of $u$ and $v$ with the above substitutions? Or do we need another change of variables?

What other change of coordinates is possible to approach this problem? Based on what region $R$ looks like, I don't think polar coordinates would work for this even though there is an $x^2+y^2$ term (though I could be mistaken).

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Try to draw a picture – leo Jun 24 '12 at 4:47

Below is the region (both the blue and red) over which you are integrating. $$I = \int_{\Omega} (x+y) dA = \int_{\text{Red region}} (x+y) dA + \int_{\text{Blue region}} (x+y) dA$$ Denoting $\arctan((4-\sqrt{7})/3)$ as $\theta_1$ and $\arctan((4-\sqrt{7})/3)$ as $\theta_2$, we get that
In the red region, you have the $$\int_{\text{Red region}} = \int_{\theta = \theta_1}^{\theta = \theta_2} \int_{r=1}^{r=2} (\cdot) dr d \theta$$ In the blue region, you have the $$\int_{\text{Blue region}} = \underbrace{\int_{\theta = 0}^{\theta = \theta_1} \int_{r=1}^{r=f_1(\theta)} (\cdot) dr d \theta}_{\text{Blue region below}} + \underbrace{\int_{\theta = \theta_2}^{\theta = \pi/2} \int_{r=1}^{r=f_2(\theta)} (\cdot) dr d \theta}_{\text{Blue region above}}$$ Now all you need to find are the functions $f_1,f_2$ and perform the integral. The functions $f_1,f_2$ are easy to find since you know the equation of the line below is $x-y = 1$ and the equation of the line above is $y-x=1$. Hence, $$f_1(\theta) = \dfrac1{\cos(\theta) - \sin(\theta)} \text{ and } f_2(\theta) = \dfrac1{\sin(\theta) - \cos(\theta)}$$
Thank you for your solution. Eventually I found out that the substitution $u=x^2+y^2$ and $v=y-x$ works since $x+y$ cancels out with $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|$ and the integral becomes simple to compute, which was $3$. I didn't try compute your solution because of tedious computation, but did you get $3$ also? – tcmtan Jun 26 '12 at 8:18
I got 6 by direct integrating since $R={(x,y)|1\le x \le2, -1+x \le y \le 1+x}\implies \int_{1}^{2}\int_{-1+x}^{1+x}x+ydxdy=\int_1^2{4x}dx=6$ – Ale Jan 27 '14 at 10:49