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