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

Question

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

Answer 1

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.

Answer 2

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:

$$\begin{gathered} \int\limits_{{u^2} + {v^2} \leqslant 1} {f(u,v) \cdot du \wedge dv} = \int\limits_0^{2 \cdot \pi } {\int\limits_0^1 {{r^2} \cdot r \cdot dr \wedge d\varphi } } \hfill \\ \hfill \\ = \int\limits_0^{2 \cdot \pi } {\int\limits_0^1 {{r^3} \cdot dr \cdot d\varphi } } = \int\limits_0^{2\pi } {\frac{1}{4} \cdot d\varphi } \hfill \\ \hfill \\ = \frac{1}{4} \cdot \int\limits_0^{2 \cdot \pi } {d\varphi } = \frac{1}{4} \cdot 2\pi \hfill \\ \hfill \\ = \frac{\pi }{2} \hfill \\ \end{gathered}$$