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

*The problem statement, all variables and given/known data


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


*Relevant 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 $\iint_S 2^2\sqrt{2}(u^2+v^2)^3 dudv$
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?
 A: Given parametrization
$$\phi (u,v) = (2uv,{u^2} - {v^2},{u^2} + {v^2})$$
Given function:
$$f(u,v) = {u^2} + {v^2}$$
Using polar coordinates:
$$\begin{gathered}
  u = r \cdot cos(\varphi ) \hfill \\
  v = r \cdot \sin (\varphi ) \hfill \\ 
\end{gathered} $$
Transform function:
$$f(r \cdot cos(\varphi ),r \cdot \sin (\varphi )) = {r^2} \cdot co{s^2}(\varphi ) + {r^2} \cdot {\sin ^2}(\varphi ) = {r^2}$$
Transform surface volume element:
$$\begin{gathered}
  du = cos(\varphi ) \cdot dr - r\sin (\varphi ) \cdot d\varphi  \hfill \\
  dv = \sin (\varphi ) \cdot dr + r\cos (\varphi ) \cdot d\varphi  \hfill \\
   \hfill \\
  du \wedge dv = r \cdot dr \wedge d\varphi  \hfill \\ 
\end{gathered} $$
Integrate:
$$\begin{gathered}
  \int\limits_{{u^2} + {v^2} \leqslant 1} {f(u,v) \cdot du \wedge dv}  = \int\limits_0^{2 \cdot \pi } {\int\limits_0^1 {{r^2} \cdot r \cdot dr \wedge d\varphi } }  \hfill \\
   \hfill \\
   = \int\limits_0^{2 \cdot \pi } {\int\limits_0^1 {{r^3} \cdot dr \cdot d\varphi } }  = \int\limits_0^{2\pi } {\frac{1}{4} \cdot d\varphi }  \hfill \\
   \hfill \\
   = \frac{1}{4} \cdot \int\limits_0^{2 \cdot \pi } {d\varphi }  = \frac{1}{4} \cdot 2\pi  \hfill \\
   \hfill \\
   = \frac{\pi }{2} \hfill \\ 
\end{gathered}$$
A: It can't be $0$, because $x^2+y^2>0$ when $(x,y) \neq (0,0)$. I didn't check your answer but it seems that in any case, the answer sheet is wrong. 
