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?
A partial answer, at least:
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: $\int_{y=0}^2 + \int_{y=2}^4$.