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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

enter image description here

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?

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    $\begingroup$ 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? $\endgroup$ – rogerl Oct 23 '18 at 18:15
  • $\begingroup$ 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. $\endgroup$ – the_candyman Oct 23 '18 at 18:16
  • $\begingroup$ @rogerl I think I fixed it now! $\endgroup$ – DummKorf Oct 23 '18 at 18:18
  • $\begingroup$ Are my bounds correct? $\endgroup$ – DummKorf Oct 23 '18 at 18:18
  • $\begingroup$ 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). $\endgroup$ – rogerl Oct 23 '18 at 18:19
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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: $\int_{y=0}^2 + \int_{y=2}^4$.

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  • $\begingroup$ So... the second integral is not the easier one to integrate over? $\endgroup$ – DummKorf Oct 23 '18 at 19:01
  • $\begingroup$ Correct, the first one seems more convenient. $\endgroup$ – Hans Lundmark Oct 23 '18 at 19:02
  • $\begingroup$ I got $\ln{(\frac{5}{2})}$ as an answer using the first integral. Does that sound right? $\endgroup$ – DummKorf Oct 23 '18 at 19:09
  • $\begingroup$ @DummKorf: Yes. $\endgroup$ – Hans Lundmark Oct 24 '18 at 8:50

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