UsukiDoll
  • UsukiDoll
Evaluate the double integral over the given region R y/x^2y^2+1 R: 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UsukiDoll
  • UsukiDoll
|dw:1378000632564:dw|
anonymous
  • anonymous
\[\int_0^1\int_0^1\frac{y}{x^2y^2+1}~dx~dy~~?\]
UsukiDoll
  • UsukiDoll
yep

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UsukiDoll
  • UsukiDoll
why do I get the feeling that I want to use tan^-1 x because I see x^2y^2 so if that's the u it would be xy and the a would be sqroot(1)
anonymous
  • anonymous
You could do that, or integrate with respect to y first. It would involve a substitution. \[\int_0^1\int_0^1\frac{y}{x^2y^2+1}~dy~dx\] Let \(u=x^2y^2+1\), then \(du=2x^2y~dy~~\Rightarrow~~\dfrac{1}{2x^2}du=dy\), and limits change from \(0\to1\) to \(1\to x^2+1\): \[\int_0^1\frac{1}{2x^2}\int_1^{x^2+1}\frac{du}{u}\]
anonymous
  • anonymous
missing a \(dx\) there, but you get the idea.
anonymous
  • anonymous
and a \(y~dy\), now that I notice it...
UsukiDoll
  • UsukiDoll
oh a natural log. 1/u is ln u + c
anonymous
  • anonymous
I'm not sure this really makes it easier compared to your idea of a trig sub, but maybe integrating by parts afterward may work. If not, then you'll probably have to convert to polar to evaluate.
UsukiDoll
  • UsukiDoll
I haven't learned conversion to polar yet.
UsukiDoll
  • UsukiDoll
it's a double integral meaning I have to integrate in terms of y or x first. then plug in the values and then to another integration
UsukiDoll
  • UsukiDoll
partial derivatives. partial integration
anonymous
  • anonymous
Right, I'm aware of that. Converting to polar is just a way of rewriting an integrand and limits so that it's easier to compute. The final answer, according to Wolfram, involves \(\pi\) and a logarithm, so I think we're headed in the right direction.
anonymous
  • anonymous
**converting probably wouldn't make it easier in this case anyway.
UsukiDoll
  • UsukiDoll
I'm gonna do this in scratch paper and see what's up
Loser66
  • Loser66
to me , I will let u = xy , and change the integral respect to x first, I mean dx inside. so du = ydx , the inside integral turns to du/u^2 +1
UsukiDoll
  • UsukiDoll
|dw:1378001763252:dw|
Loser66
  • Loser66
\[\int_{0}^{1} \frac{y}{x^2y^2+1}dx\] let u = xy --> du =ydx. does it work? I don't know
UsukiDoll
  • UsukiDoll
hmmm I'm guessing |dw:1378002161827:dw|
Loser66
  • Loser66
when taking integral respect to x, y turns constant, right? so, mine = tan^(-1) u = tan^(-1) xy from 0 to 1 . but didn't think about the second integral yet. hihihi..
UsukiDoll
  • UsukiDoll
that's what I was thinking..using tan^-1x subsitutuion since a = sqroot(x^2y^20 would be xy and the u would be one oops typos
UsukiDoll
  • UsukiDoll
|dw:1378002494214:dw|
UsukiDoll
  • UsukiDoll
|dw:1378002546265:dw|
UsukiDoll
  • UsukiDoll
hmmmm.... @dan815
UsukiDoll
  • UsukiDoll
|dw:1378002725536:dw|
UsukiDoll
  • UsukiDoll
|dw:1378002795137:dw|
UsukiDoll
  • UsukiDoll
arghhhhhhhhhhhh looked legit though
UsukiDoll
  • UsukiDoll
from the bottom
anonymous
  • anonymous
Here's what I've got: \[\int_0^1\int_0^1\frac{y}{x^2y^2+1}~dy~dx\] Let \(t=x^2y^2+1\) so that \(dt=2x^2y~dy~~\iff~~\dfrac{1}{2x^2}dt=dy\): \[\int_0^1\frac{1}{2x^2}\left(\int_1^{x^2+1}\frac{du}{u}\right)~dx\\ \int_0^1\frac{1}{2x^2}\left[\ln|u|\right]_1^{x^2+1}~dx\\ \frac{1}{2}\int_0^1\frac{\ln(x^2+1)}{x^2}~dx\] Integrating by parts, let \[\begin{matrix}u=\ln(x^2+1)&&&dv=\frac{1}{x^2}~dx\\ du=\frac{2x}{x^2+1}&&&v=-\frac{1}{x}\end{matrix}\] Also, note that integrand is undefined for \(x=0\), so treat as an improper integral: \[\lim_{r\to0}\frac{1}{2}\left(\left[-\frac{\ln(x^2+1)}{x}\right]_r^1-\int_r^1\left(-\frac{1}{x}\right) \frac{2x}{x^2+1}~dx\right)\\ \lim_{r\to0}\left(\frac{\ln(r^2+1)}{r}-\ln2\right)+\int_0^1\frac{dx}{x^2+1}\] The limit works out nicely with a single application of L'Hopital's rule, and the remaining integral is quite simple to figure out as well.
anonymous
  • anonymous
There should be a 1/2 in front of the limit.
UsukiDoll
  • UsukiDoll
|dw:1378003703584:dw|
UsukiDoll
  • UsukiDoll
|dw:1378003804811:dw|
UsukiDoll
  • UsukiDoll
|dw:1378003936137:dw|
anonymous
  • anonymous
Looks like we get \(\dfrac{\pi}{4}-\dfrac{1}{2}\ln2\), and Wolfram says it's \(\dfrac{\pi}{4}-\dfrac{\ln4}{4}\), which is the same result!
UsukiDoll
  • UsukiDoll
wait how did you get pi/4 is it from evaluating the first integral?
anonymous
  • anonymous
\[\int_0^1\frac{dx}{x^2+1}=\frac{\pi}{4}\]
UsukiDoll
  • UsukiDoll
?????????????????????? oh tan ^-1 x
UsukiDoll
  • UsukiDoll
OH YEAH now I remember F(1) - F(0) tan^-1 1 - tan^-1 0 tan^-1 1 is pi/4
anonymous
  • anonymous
Yep
UsukiDoll
  • UsukiDoll
awesome. yeah it's been a while since I've touched those. I'm gonna brush up on em

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