Problem :
Let $S$ be surface which parameterized $$S : r(u, v) = (2uv, u^2 - v^2, u^2 + v^2)\quad (u^2 +v^2 \leq 1)$$
Evaluate surface integral : $$\iint_S (x^2 + y^2) dS$$
I tried to find original surface which doesn't depend on $u, v$.
Since $$(2uv)^2 + (u^2 - v^2)^2 = (u^2 + v^2)^2$$ I find representation of $S$ without using $u, v$ like below : $$S : z^2 = x^2 + y^2$$
and, since $0\leq z = u^2 + v^2 \leq 1$, we can choose $z=\sqrt{x^2 + y^2}$ for surface $S$.
Then I tried to evaluate original surface integral, using $dS = \sqrt{2}dxdy$.
$$\begin{align} \iint_S (x^2 + y^2)dS &= \sqrt{2}\iint_{x^2 + y^2 \leq 1} (x^2 + y^2) dxdy \\ &= \sqrt{2} \times \frac{\pi}{2}\end{align}$$
But the answer of this problem is $\sqrt{2}\pi$, not $\frac{\sqrt{2}}{2}\pi$. I think there is no such error, but my result is different.
Is there any mistake or error what I didn't catch?
Thank you.
Presumably you are more interested in the integral $$ \int_S\,dS $$ which is the surface area rather than rather than $\iint_S(x^2+y^2)\,dS$ but this does not matter for the resolution of the riddle.
That surface is a cone which we normally parametrize with $$ s(x,y)=\begin{pmatrix}x\\y\\[2mm]\sqrt{x^2+y^2}\end{pmatrix}\,,\quad x^2+y^2\le 1\,. $$ With the parametrization $$ r(u,v)=\begin{pmatrix}2uv\\u^2-v^2\\u^2+v^2\end{pmatrix}\,,\quad u^2+v^2\le 1\,, $$ we trace the same cone but this happens twice since $$ r(u,v)=r(-u,-v)\,. $$ To get the surface element we calculate as usual $$ \partial_u\boldsymbol{r}\times\partial_v\boldsymbol{r}=4\begin{pmatrix}v\\u\\u\end{pmatrix}\times\begin{pmatrix}u\\-v\\v\end{pmatrix}=4\begin{pmatrix}2uv\\u^2-v^2\\-u^2-v^2\end{pmatrix}\,. $$ The length of this vector is $4\sqrt{2}\,(u^2+v^2)\,.$ Therefore $$ dS=4\sqrt{2}\,(u^2+v^2)\,du\,dv\,. $$ If we integrate naively over $u^2+v^2\le 1$ instead over half of that we think that the surface area is $$ A=4\sqrt{2}\iint\limits_{u^2+v^2\le 1}(u^2+v^2)\,du\,dv=4\,\sqrt{2}\,\int_0^{2\pi}\int_0^1r^3\,dr\,d\varphi=2\,\sqrt{2}\,\pi $$ which is two times the surface area of the lateral surface of the cone as expected.
You can check easily that the same procedure using the unambiguous classical parametrization $s(x,y)$ leads to $\sqrt{2}\pi$ as it must.