How can I find the area of the surface $(u,v) \mapsto (u,v,uv)$ in the domain $A = \{(u, v) \in \mathbb R^2\colon\ u^2+v^2<3\}$?

Question

Consider the set $A = \{(u, v) \in \mathbb R^2\colon\ u^2+v^2<3\}$. What is the area of the surface $\sigma: A \to \mathbb R^3$ defined by $\sigma(u,v)=(u,v,uv)$?

$(A)14/ 3\quad\ (B) 14π/3 \quad\ (C) 9π\quad\ (D) 9$

Answer 1

I write the domain as $D$. It is known from multivariable calculus that the area $A$ which you want to calculate is given by

$$A = \int_D \| \sigma_u \times \sigma_v \| \ \mathrm d u \mathrm d v.$$

We have $\sigma_u = (1, 0, v)$ and $\sigma_v = (0, 1, u)$, hence $\| \sigma_u \times \sigma_v \| = \sqrt{u^2 + v^2 + 1}$. Now we introduce polar coordinates, then it follows that

$$A = \int_D \sqrt{u^2 + v^2 + 1} \ \mathrm d u \mathrm d v = \int_0^{2\pi} \int_0^{\sqrt{3}} \sqrt{r^2 + 1} \ r \mathrm d r \mathrm d \varphi = \frac{14\pi}{3}.$$