Let $S$ be the surface defined as: $$\Phi(u, v)=(u+v, u-v, u v)$$ Find the value of $$\iint_{S}x\:dS$$ for all points on $S$ over $R:x^2+y^2\leq 1$
My try: The surface is $x=u+v, y=u-v, z=uv$ Eliminating $u,v$ we get: $$4z=x^2-y^2$$
Also we have:
$\frac{\partial z}{\partial x}=\frac{x}{2} \quad \frac{\partial z}{\partial y}=\frac{-y}{2}$
So we have $$d S=\left(\sqrt{1+\frac{x^{2}}{4}+\frac{y^{2}}{4}}\right) d x d y$$
So $$\iint_{S}x\:dS=\iint_{R}x\left(\sqrt{1+\frac{x^{2}}{4}+\frac{y^{2}}{4}}\right)dxdy$$
Now by polar coordinates we get $$\iint_{S}x\:dS=\int_{\theta=0}^{2 \pi} \int_{r=0}^{1}(r \cos \theta) \sqrt{1+\frac{r^{2}}{4}}\: r d r d \theta=0$$
Is this correct?