# Thread: need help with double integral

1. ## need help with double integral

hey guy, I'm having a problem with double integral integration. Can you guys tell me where I went wrong?

The equation is $\displaystyle \int_0^1\int_0^y\frac{y^3}{x^2+y^2}dxdy$

let $\displaystyle U = x+y$
$\displaystyle \frac{dU}{dx} = 1$
$\displaystyle \int_0^1\int_0^y\frac{y^3}{U^2}dUdy$
$\displaystyle \int_0^1[\frac{-y^3}{x+y}]_0^y dy$
$\displaystyle \int_0^1 \frac{y^2}{2} dy$
$\displaystyle [\frac{y^3}{6}]_0^1 = \frac{1}{6}$

However, I seem like this answer is wrong, and the correct answer should be $\displaystyle \frac{pi}{12}$
Can anyone point out what I did wrong here?

Best Regards
Junks

2. ## Re: need help with double integral

$(x+y)^2 = x^2 + y^2$

Really? Try it again.

3. ## Re: need help with double integral

ahh, I see, I'll let U=x^2 + y^2
.................................................. ..........................................

So I let $\displaystyle U=x^2+y^2$
$\displaystyle \int_0^1\int_0^y\frac{y^3}{2*x*U}$
$\displaystyle \int_0^1[\frac{y^3}{2*x}*log(x^2+y^2)]_0^y$
$\displaystyle \int_0^1\frac{y^2}{2}*log(2y^2)$
let $\displaystyle U=log(2*y^2)$ and $\displaystyle \frac{dV}{dy}=\frac{y^2}{2}$
$\displaystyle \frac{dU}{dy}=\frac{2}{y}$ and $\displaystyle V=\frac{y^3}{6}$
using this rule $\displaystyle UV-\int \frac{dU}{dy}*\frac{dV}{dy}dy$
$\displaystyle [\frac{y^3}{6}*log(2*y^2)-\int y dy]_0^1$
$\displaystyle \frac{1}{6}*log(2)-\frac{1}{2}$ the answer turn out to be -0.38, but the correct answer seem to be pi/12 or 0.26. What did I do wrong in here?

4. ## Re: need help with double integral

$\displaystyle \int_0^1\int_0^y\frac{y^3}{x^2+y^2}dxdy=$

$\displaystyle \int_0^1 y^3 \int_0^y \dfrac{1}{x^2+y^2}~dx~dy=$

$\displaystyle \int_0^1 y^3 \left(\dfrac{\arctan(\frac{x}{y})}{y}|_0^y\right)~ dy=$

$\displaystyle \int_0^1 y^2 \left(\arctan(1)-\arctan(0)\right)~dy=$

$\displaystyle \int_0^1 y^2 \dfrac{\pi}{4}~dy=$

$\displaystyle \dfrac{\pi}{4}\left( \dfrac{y^3}{3}~|_0^1\right)=$

$\displaystyle \dfrac{\pi}{12}$