How do you find the vector field that correlates with the surface in divergence thereom?

Question

Express the surface area of the paraboloid $$z = 1 - x^2 - y^2\quad (z\geq0)$$ as a triple integral using the Divergence theorem and by choosing an appropriate vector field, use the Divergence Theorem to find the surface area of the paraboloid.

Answer 1

Hint: Let $z=f\left(x,y\right)$ be a part of a closed surface $\partial V$. Then by the divergence theorem

$$\iint_{\partial V}\boldsymbol{\Phi}\cdot\boldsymbol{n}{\rm d}S=\iiint_{V}\boldsymbol{\nabla}\cdot\boldsymbol{\Phi}{\rm d}V$$

where the normal to the $z=f\left(x,y\right)$ part of the surface satisfies $\boldsymbol{n}\propto\left(-f_{x},-f_{y},1\right)$. You want $\boldsymbol{\Phi}\cdot\boldsymbol{n}=1$ for the LHS to be the area. Can you take it from here?

Answer 2

Change to cylindrical coordinates to have your paraboloid (call it $P$) as $z = 1- r^2, z\geq 0, $. So $P$'s area is $\int\int_P r \cdot n dr d\theta$, with $n$ a unitary vector that is normal to $P$, for instance, $n=\frac{(-2r^2 \cos\theta, 2r^2 \sin \theta, r)}{4r^4 + r^2}$. Mind that $n$ is actually associating a vector to each point on $P$, so $n$ perfectly qualifies as a vector field, and so does $r\cdot n$. Now, use Gauss's divergence theorem with the vector field $n\cdot r$ and you're done!