# When can a double integral be interpreted as a surface area?

Let $$D$$ be a closed, bounded domain in $$\mathbb R^2$$ and let $$\vec r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$$ for $$(u, v) \in D$$ be a parametrization of a smooth surface $$S \subseteq \mathbb R^3$$. Then the area of $$S$$ is

$$\iint_D \| \vec r_u \times \vec r_v \| \, dA$$

I am interested in what might be considered a kind of converse of this: Is it always possible to interpret an integral $$\iint_D f \, dA$$ as the area of some surface?

More precisely, given a (non-negative) function $$f:\mathbb R^2 \to \mathbb R$$, defined on some domain $$D$$, under what conditions can we find a surface $$S \subseteq \mathbb R^3$$ and a parametrization $$\vec r(u,v): D \to S$$ such that

$$f(u,v) = \| \vec r_u \times \vec r_v \|$$ so that consequently $$\iint_D f \, dA = \iint_D \| \vec r_u \times \vec r_v \| \, dA \, ?$$

• Have you solved the (simpler) question for arclength of a parametrized curve? Commented Feb 8, 2023 at 2:36
• @TedShifrin Actually hadn't thought of that, but I'll give it a thought later today after I'm done teaching. Commented Feb 8, 2023 at 15:39

## 1 Answer

I assume $$f \geq 0$$ in $$D$$ and $$f > 0$$ almost everywhere in $$D$$. Otherwise it's not a reasonable representation of a surface area.

Suppose $$S$$ is actually flat: it can be represented by $$z = 0$$ on the domain $$\{(x(u,v),y(u,v)) \mid (u,v) \in D\}$$. Then

$$\vec r = \langle x, y, 0 \rangle$$ $$\vec{r}_u = \langle x_u, y_u, 0 \rangle$$ $$\vec{r}_v = \langle x_v, y_v, 0 \rangle$$ $$\vec{r}_u \times \vec{r}_v = \langle 0, 0, x_u y_v - x_v y_u \rangle$$ $$\|\vec{r}_u \times \vec{r}_v\| = |x_u y_v - x_v y_u|$$

(Not surprisingly, the factor in this transformation essentially between subsets of $$\mathbb{R}^2$$ is a Jacobian determinant.)

And one simple way to get $$f(u,v) = \|\vec{r}_u \times \vec{r}_v\| = |x_u y_v - x_v y_u|$$ is

\begin{align*} \tilde{f}(u,v) &= \begin{cases} f(u,v) & (u,v) \in D \\ 0 & (u,v) \notin D \end{cases} \\ x(u,v) &= \int_0^u \tilde{f}(t,v)\, dt \\ y(u,v) &= v \end{align*}

The only real requirement is that $$f$$ is integrable with respect to $$u$$ at each fixed $$v$$. (Similar solutions would exist if $$f$$ is instead integrable along some other set of level curves.)