Question

Evaluate the double integral over the given region R
y/x^2y^2+1 R: 0 12 years ago

Answer 1

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

Answer 2

yep

Answer 3

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)

Answer 4

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 \(1\to x^2+1\):
\[\int_0^1\frac{1}{2x^2}\int_1^{x^2+1}\frac{du}{u}\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

I haven't learned conversion to polar yet.

Answer 10

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

Answer 11

partial derivatives. partial integration

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

hmmm I'm guessing