Region Of This Surface Area?

Question

A surface area is parameterized as $x=2uv$, $y=u^2-v^2$, $z=u^2 + v^2$

---> Where the domain is all points of the unit circle and its interior in the $uv$ plane, $u^2+v^2 ≤ 1$

Compute the surface area.

I'm actually okay for setting up the integral in all ways but one: the region of $u$ and $v$. As it uses the unit circle, both $u$ and $v$ will range from -1 to 1, but I'm not sure if I should take that as the region or be trying to put one in terms of the other. What would be the correct way to set up this region? Thanks for any assistance, coming back into calculus after a long absence.

Answer 1

The surface integral given $\bar{r}(u,v)$ is $\int_S\left |r_u\times r_v\right |\text{d}A$ So $r_u=\left \langle 2v, 2u, 2u\right \rangle$ and $r_v=\langle 2u, -2v, 2v\rangle$ so $r_u\times r_v=\langle 8uv, 4u^2-4v^2, -4v^2-4u^2\rangle$ now the magnitude of this takes a bit of working out but we get $4\sqrt2(u^2+v^2)$ so our integral is now $4\sqrt2\iint_Du^2+v^2\text{d}A$ because we are in the unit circle converting to polar offers a very nice solution $4\sqrt2\int_{0}^{2\pi}\int_{0}^{1}r*r^2\text{d}r\text{d}\theta$ and the rest is easy