# Integrating $\iint_{R} \frac{y}{x^2+y^2}\,dA$ where $R$ is bounded by $y=x,y=2x,x=2$

So I'm supposed to set an integral up for both orders of integration and evaluate using the "nicer" of the two

$$\iint_{R} \frac{y}{x^2+y^2} \,dA$$ for $$R$$ bounded by $$y=x, y = 2x$$, and $$x=2$$.

So what I have tried to do so far was figuring out the region which I'm integrating over which is

So this is what my two integrals are:

$$\int_{0}^{2}\int_{y=x}^{y=2x} \frac{y}{x^2+y^2} \,dy\, dx$$

$$\int_{0}^{2}\int_{x=\frac{y}{2}}^{x=y} \frac{y}{x^2+y^2}\, dx\, dy$$

I believe the easier integral would be the second one when you integrate with respect to $$dy$$ first, because since $$y$$ would be a constant, you can factor out $$x$$ and integrate $$\frac{1}{x^2+y^2}$$. I was wondering if my thought process is correct, and if it does can you help me integrate this?

• The integral you present in the title and the first formula is not the same as the two integrals at the end (look at the numerator). Which did you intend? – rogerl Oct 23 '18 at 18:15
• If $y$ is in the numerator, then it is easier to integrate first by $dy$. Check the title and content of your question as @rogerl suggested. – the_candyman Oct 23 '18 at 18:16
• @rogerl I think I fixed it now! – DummKorf Oct 23 '18 at 18:18
• Are my bounds correct? – DummKorf Oct 23 '18 at 18:18
• The lower bound on the second integral should be $\frac{1}{2}y$, not $\frac{1}{2}$. I suspect that was a typo. As noted above, integrating first with respect to $y$ is easier (the numerator is close to the derivative of the denominator). – rogerl Oct 23 '18 at 18:19

Your bounds on the second one are not right; you cover a region which is smaller than intended, namely the one bounded by the lines $$y=x$$, $$y=2x$$ and $$y=2$$ (note $$y=2$$, not $$x=2$$).
You need to go all the way up to $$y=4$$, and take $$\int_{x=y/2}^{\min(1,y)}$$ in the inner integral instead, and then evaluate this by splitting the outer integral into cases: $$\int_{y=0}^2 + \int_{y=2}^4$$.
• I got $\ln{(\frac{5}{2})}$ as an answer using the first integral. Does that sound right? – DummKorf Oct 23 '18 at 19:09