Evaluate the double integral over the given region R
y/x^2y^2+1 R: 0

Mathematics
- UsukiDoll

Evaluate the double integral over the given region R
y/x^2y^2+1 R: 0

Mathematics
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- jamiebookeater

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- UsukiDoll

|dw:1378000632564:dw|

- anonymous

\[\int_0^1\int_0^1\frac{y}{x^2y^2+1}~dx~dy~~?\]

- UsukiDoll

yep

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## More answers

- 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

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

missing a \(dx\) there, but you get the idea.

- anonymous

and a \(y~dy\), now that I notice it...

- UsukiDoll

oh a natural log. 1/u is ln u + c

- 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

I haven't learned conversion to polar yet.

- 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

partial derivatives. partial integration

- 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

**converting probably wouldn't make it easier in this case anyway.

- UsukiDoll

I'm gonna do this in scratch paper and see what's up

- 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

|dw:1378001763252:dw|

- Loser66

\[\int_{0}^{1} \frac{y}{x^2y^2+1}dx\]
let u = xy --> du =ydx.
does it work? I don't know

- UsukiDoll

hmmm I'm guessing
|dw:1378002161827:dw|

- 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

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

|dw:1378002494214:dw|

- UsukiDoll

|dw:1378002546265:dw|

- UsukiDoll

hmmmm.... @dan815

- UsukiDoll

|dw:1378002725536:dw|

- UsukiDoll

|dw:1378002795137:dw|

- UsukiDoll

arghhhhhhhhhhhh looked legit though

- UsukiDoll

from the bottom

- 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

There should be a 1/2 in front of the limit.

- UsukiDoll

|dw:1378003703584:dw|

- UsukiDoll

|dw:1378003804811:dw|

- UsukiDoll

|dw:1378003936137:dw|

- 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

wait how did you get pi/4 is it from evaluating the first integral?

- anonymous

\[\int_0^1\frac{dx}{x^2+1}=\frac{\pi}{4}\]

- UsukiDoll

?????????????????????? oh tan ^-1 x

- UsukiDoll

OH YEAH now I remember F(1) - F(0)
tan^-1 1 - tan^-1 0
tan^-1 1 is pi/4

- anonymous

Yep

- UsukiDoll

awesome. yeah it's been a while since I've touched those. I'm gonna brush up on em

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