Surface integral problem: ##\iint_S (x^2+y^2)dS##

[A330NEO]

Homework Statement

$\iint_S (x^2+y^2)dS$, $S$is the surface with vector equation $r(u, v)$ = $(2uv, u^2-v^2, u^2+v^2)$, $u^2+v^2 \leq 1$

Homework Equations

Surface Integral. $\iint_S f(x, y, z)dS = \iint f(r(u, v))\left | r_u \times r_v \right |dA$,

The Attempt at a Solution

First, I tried to shift the form of $\iint_S (x^2+y^2)dS$. $x^2+y^2$.
$x^2+y^2$ = $u^4v^4+2u^2v^2+v^4$, and $\left | r_u \times r_v \right |$ = $\sqrt{32v^4+64u^2v^2+32u^4}$
Thus, the initial integral becomes $\iint_S 2^2sqrt{2}(u^2+v^2)^3 dudv$
I used polar coordinates, as u = rsin$\theta$ and v = $rsin\theta$. $0\leq r\leq 1$, $0 \leq \theta \leq 2\pi$. As result, the answer came to be $2^3\sqrt{2}/5*\pi$ but the answer sheet says its zero. Am I missing something?[/B]

 

[RUber]

I checked your algebra, and all looks good up to the switch to polar coords.
Did you try without the substitution?

 

[A330NEO]

I tried, but it was way too complexed. Anyway, if my algebra and switghing to polar coords is not bad, is it presumable that the 'answer sheet' is wrong?

 

[LCKurtz]

You know immediately just looking at the problem that the answer must be positive. The integrand is nonnegative. So, yes, the answer sheet is wrong.