Definition: Area
Let \(\phi: \mathcal{D} \subseteq \mathbb{R}^2 \to \mathbb{R}^3\) be a totally differentiable real parametric surface.
The area of \(\phi\) is the integral of the magnitude of its normal vector (if it exists):
\[\iint_{\mathcal{D}} ||\partial_u \phi(u, v) \times \partial_v \phi(u,v)||\,\mathrm{d}\mathcal{D}\]
Notation
\[A(\phi) \qquad S(\phi)\]
Example: Area of Graph
Let \(f: \mathcal{D} \subseteq \mathbb{R}^2 \to \mathbb{R}\) be a real scalar field and let \(\phi: \mathcal{D} \to \mathbb{R}^3\) be a parametric surface which parameterizes \(f\)'s graph:
\[\phi(u,v) = \begin{bmatrix}u \\ v \\ f(u, v)\end{bmatrix}\]
If \(f\) is totally differentiable, then \(\phi\) is also totally differentiable and for \(\phi\)'s partial derivatives, we have:
\[\partial_u \phi(u,v) = \begin{bmatrix} 1 \\ 0 \\ \partial_u f(u,v) \end{bmatrix} \qquad \partial_v \phi(u,v) = \begin{bmatrix} 0 \\ 1 \\ \partial_v f(u,v) \end{bmatrix}\]
For \(\phi\)'s normal vector, we have:
\[\partial_u \phi(u,v) \times \partial_v \phi(u,v) = \begin{bmatrix} 1 \\ 0 \\ \partial_u f(u,v) \end{bmatrix} \times \begin{bmatrix} 0 \\ 1 \\ \partial_v f(u,v) \end{bmatrix} = \begin{bmatrix} - \partial_u f(u, v) \\ - \partial_v f(u, v) \\ 1 \end{bmatrix}\]
\[||\partial_u \phi(u,v) \times \partial_v \phi(u,v)|| = \sqrt{1 + \partial_u f(x,y)^2 + \partial_v f(u,v)^2}\]
The area of \(\phi\) is thus given by the following integral (if it exists):
\[\iint_{\mathcal{D}} \sqrt{1 + \partial_u f(x,y)^2 + \partial_v f(u,v)^2} \,\mathrm{d}(u,v)\]
Example: Surface of a Sphere
The surface of a sphere of radius \(R\) centered at the origin in \(\mathbb{R}^3\) can be parameterized via the real parametric surface \(\phi: [0, \uppi] \times [0, 2\uppi] \to \mathbb{R}^3\) defined as follows:
\[\phi(u,v) = \begin{bmatrix}R \sin u \cos v \\ R \sin u \sin v \\ R \cos u\end{bmatrix}\]
We see that \(\phi\) is totally differentiable. For its partial derivatives, we have
\[\partial_u \phi(u, v) = \begin{bmatrix} R \cos u \cos v \\ R \cos u \sin v \\ -R \sin u \end{bmatrix} \qquad \partial_v \phi(u, v) = \begin{bmatrix} -R \sin u \sin v \\ R \sin u \cos v \\ 0 \end{bmatrix}\]
and so its normal vector is the following:
\[\begin{aligned}\partial_u \phi(u, v) \times \partial_v \phi(u, v) & = \begin{bmatrix} R \cos u \cos v \\ R \cos u \sin v \\ -R \sin u \end{bmatrix} \times \begin{bmatrix} -R \sin u \sin v \\ R \sin u \cos v \\ 0 \end{bmatrix} \\ & = R^2 \sin u \begin{bmatrix}\sin u \cos v \\ \sin u \sin v \\ \cos u\end{bmatrix}\end{aligned}\]
We have:
\[||\partial_u \phi(u, v) \times \partial_v \phi(u, v)|| = R^2 \sin u\]
The area of \(\phi\) is thus given by the following integral:
\[F(\phi) = \iint_{\mathcal{D}} ||\partial_u \phi(u, v) \times \partial_v \phi(u, v)|| \,\mathrm{d}\mathcal{D} = \iint_{[0, \uppi] \times [0, 2\uppi]} R^2 \sin u\,\mathrm{d}\mathcal{D}\]
Since \(R^2 \sin u\) is continuous, we can compute it via an iterated integral:
\[\begin{aligned}\iint_{[0, \uppi] \times [0, 2\uppi]} R^2 \sin u\,\mathrm{d}\mathcal{D} & = \int_0^{2\uppi} \int_{0}^{\uppi} R^2 \sin u \,\mathrm{d}u \,\mathrm{d}v \\ & = R^2\int_0^{2\uppi} \int_{0}^{\uppi} \sin u \,\mathrm{d}u \,\mathrm{d}v \\ & = R^2 \int_{0}^{2\uppi} \left(\left.-\cos u \right\vert_{0}^{\uppi}\right) \,\mathrm{d}v \\ & = R^2 \int_{0}^{2\uppi} 2 \,\mathrm{d}v \\ & = 4\uppi R^2\end{aligned}\]