6.3 Surface Area and Parametric Surfaces

Parameterization and Surface Area

Parameterization lets you describe a 3D surface using just two parameters, essentially mapping a flat region in the $uv$-plane onto a curved surface in space. This is the foundation for computing surface area, because once you have a parameterization, you can use partial derivatives and cross products to measure how much the surface stretches and bends. These ideas connect directly to line integrals and set the stage for surface integrals and flux later in the course.

Parameterization of 3D Surfaces

A parametric surface is defined by a vector-valued function that takes two inputs and produces a point in 3D:

$\mathbf{r}(u,v) = \langle x(u,v),\; y(u,v),\; z(u,v) \rangle$

The parameters $u$ and $v$ range over some domain $D$ in the $uv$-plane. As $(u,v)$ varies across $D$, the function $\mathbf{r}$ traces out the surface in space.

Common parameterizations you should know:

• Planes: $\mathbf{r}(u,v) = \mathbf{p} + u\,\mathbf{a} + v\,\mathbf{b}$, where $\mathbf{p}$ is a point on the plane and $\mathbf{a}$, $\mathbf{b}$ are two non-parallel vectors lying in the plane. This is the surface analog of a parametric line.
• Spheres (radius $a$): $\mathbf{r}(\theta,\phi) = \langle a\sin\phi\cos\theta,\; a\sin\phi\sin\theta,\; a\cos\phi \rangle$, where $\phi \in [0,\pi]$ is the polar angle (measured from the positive $z$-axis) and $\theta \in [0,2\pi)$ is the azimuthal angle. Be careful: some textbooks swap the roles of $\theta$ and $\phi$, so always check your convention.
• Cylinders (radius $a$): $\mathbf{r}(\theta,z) = \langle a\cos\theta,\; a\sin\theta,\; z \rangle$. Here $\theta$ wraps around the cylinder and $z$ runs along its height.
• Cones (half-angle $\alpha$): $\mathbf{r}(\theta,z) = \langle z\tan\alpha\cos\theta,\; z\tan\alpha\sin\theta,\; z \rangle$. The radius of each circular cross-section grows linearly with $z$.

When choosing a parameterization, look for the surface's natural symmetry. Spherical symmetry suggests using angles $\theta$ and $\phi$; cylindrical symmetry suggests $\theta$ and $z$. For a surface given explicitly as $z = f(x,y)$, the simplest choice is to use $x$ and $y$ themselves as parameters.

Surface Area of Parametric Surfaces

The key idea: at each point on the surface, the partial derivatives $\mathbf{r}_u$ and $\mathbf{r}_v$ are tangent vectors. Their cross product $\mathbf{r}_u \times \mathbf{r}_v$ is normal to the surface, and its magnitude tells you how much area a tiny rectangle $du \times dv$ in the parameter domain gets stretched into on the surface.

The surface area formula is:

$SA = \iint_D |\mathbf{r}_u \times \mathbf{r}_v| \, du \, dv$

Step-by-step process for computing surface area:

1. Write down the parameterization $\mathbf{r}(u,v)$ and identify the parameter domain $D$.
2. Compute the partial derivatives $\mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u}$ and $\mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v}$, differentiating each component separately.
3. Take the cross product $\mathbf{r}_u \times \mathbf{r}_v$. This gives a vector normal to the surface at each point.
4. Find the magnitude $|\mathbf{r}_u \times \mathbf{r}_v|$. Simplify before integrating if possible; algebraic cleanup here saves a lot of pain in the integral.
5. Evaluate the double integral over $D$. Choose an integration order that matches the domain's shape, and use substitution or symmetry to simplify.

Surface Area Formula Derivation

The formula comes from approximating the surface with tiny parallelograms.

1. Fix a point $(u_0, v_0)$. A small change $\Delta u$ in the first parameter moves you along the surface by approximately $\mathbf{r}_u \,\Delta u$. Similarly, $\Delta v$ moves you by $\mathbf{r}_v \,\Delta v$.
2. These two vectors span a small parallelogram on the surface. The area of a parallelogram with sides $\mathbf{A}$ and $\mathbf{B}$ is $|\mathbf{A} \times \mathbf{B}|$, so the patch has area approximately $|\mathbf{r}_u \times \mathbf{r}_v|\,\Delta u\,\Delta v$.
3. Summing over all patches and taking the limit as $\Delta u, \Delta v \to 0$ gives the integral $\iint_D |\mathbf{r}_u \times \mathbf{r}_v|\,du\,dv$.

There's a more general way to express this using the Jacobian matrix:

$J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \$4pt] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \$4pt] \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \end{bmatrix}$

This is a $3 \times 2$ matrix (not square), so it doesn't have a traditional determinant. Instead, the area element is:

$dA = \sqrt{\det(J^T J)}\, du\, dv$

You can verify that $\det(J^T J) = |\mathbf{r}_u \times \mathbf{r}_v|^2$, so this is equivalent to the cross product formula. The $J^T J$ form is the more general version (it works in any dimension), but for surfaces in $\mathbb{R}^3$, the cross product approach is usually more practical.

Surface Area in Different Coordinate Systems

Choosing coordinates that match the surface's symmetry can dramatically simplify the integral.

Cartesian coordinates work best for surfaces given as $z = f(x,y)$. Use the parameterization $\mathbf{r}(x,y) = \langle x,\; y,\; f(x,y) \rangle$, which gives:

$SA = \iint_D \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}\; dx\, dy$

This follows directly from the general formula. Compute $\mathbf{r}_x = \langle 1, 0, f_x \rangle$ and $\mathbf{r}_y = \langle 0, 1, f_y \rangle$, take their cross product, and you get $\langle -f_x, -f_y, 1 \rangle$ with magnitude $\sqrt{1 + f_x^2 + f_y^2}$.

Spherical coordinates are natural for surfaces with spherical symmetry (spheres, caps, portions of spheres). On a sphere of radius $\rho$:

• Conversion: $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$
• Surface area element: $dS = \rho^2 \sin\phi\; d\theta\, d\phi$

For example, the full sphere has $SA = \int_0^{2\pi}\int_0^{\pi} \rho^2 \sin\phi\; d\phi\, d\theta = 4\pi\rho^2$, which is a good sanity check.

Cylindrical coordinates simplify calculations for cylinders and surfaces of revolution. On a cylinder of radius $r$:

• Conversion: $x = r\cos\theta$, $y = r\sin\theta$, $z = z$
• Surface area element: $dS = r\; d\theta\, dz$

The lateral surface area of a cylinder of radius $r$ and height $h$ is then $SA = \int_0^{2\pi}\int_0^h r\; dz\, d\theta = 2\pi r h$, matching the familiar formula.

When deciding which coordinate system to use, ask yourself: does the surface have a natural axis of symmetry? If it's round in cross-section, cylindrical or spherical coordinates will likely make the cross product and the resulting integral much cleaner than Cartesian.