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?

  • 2
    $\begingroup$ 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

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: $\int_{y=0}^2 + \int_{y=2}^4$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.