12.3 Surface area of a function graph

Surface Area Formula and Double Integral

The surface area formula lets you measure how much area a curved surface $z = f(x,y)$ occupies in three-dimensional space. This extends the idea of double integrals beyond volume and mass to geometry of surfaces themselves.

Calculating Surface Area Using the Surface Area Formula

For a surface $z = f(x,y)$ defined over a region $R$ in the $xy$-plane, the surface area is:

$S = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA$

The intuition here: you're summing up the areas of tiny surface patches across the entire region $R$. Each patch is a small piece of the actual curved surface, not just its flat projection onto the $xy$-plane.

The integrand $\sqrt{1 + f_x^2 + f_y^2}$ acts as a correction factor. A flat surface (where both partial derivatives are zero) gives $\sqrt{1} = 1$, so the surface area equals the area of $R$ itself. The steeper the surface, the larger the partial derivatives, and the more the integrand exceeds 1.

Step-by-step process for computing surface area:

1. Identify the surface $z = f(x,y)$ and the region $R$ over which you're integrating.
2. Compute the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.
3. Form the integrand: $\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}$.
4. Set up the double integral over $R$ with appropriate bounds.
5. Evaluate the integral (switching to polar coordinates if $R$ is circular often simplifies things).

Example: Find the surface area of $z = x^2 + y^2$ over the disk $x^2 + y^2 \leq 1$.

1. $f_x = 2x$, $f_y = 2y$
2. Integrand: $\sqrt{1 + 4x^2 + 4y^2}$
3. Switch to polar: $\sqrt{1 + 4r^2} \cdot r \, dr \, d\theta$, with $0 \leq r \leq 1$, $0 \leq \theta \leq 2\pi$
4. Evaluate: $\int_0^{2\pi}\int_0^1 r\sqrt{1+4r^2}\,dr\,d\theta = 2\pi \cdot \frac{1}{12}\left(5\sqrt{5}-1\right) = \frac{\pi}{6}\left(5\sqrt{5}-1\right)$

The Correction Factor in the Integrand

The factor $\sqrt{1 + f_x^2 + f_y^2}$ is sometimes loosely called a "Jacobian-like" term, but it's more precisely the area magnification factor. It accounts for how much a small flat rectangle $dA$ in the $xy$-plane gets stretched when you lift it onto the curved surface.

• It measures the ratio of actual surface area to projected area in the $xy$-plane at each point.
• Where the surface is nearly horizontal, this factor is close to 1. Where the surface is steep, it's much larger.
• Without this factor, you'd just be computing the area of $R$ itself, ignoring the curvature entirely.

This is analogous to the arc length factor $\sqrt{1 + (f')^2}$ from single-variable calculus. The surface area formula is its natural two-dimensional generalization.

Parametric Surfaces

Representing Surfaces Using Parametric Equations

Not every surface can be written conveniently as $z = f(x,y)$. A sphere, for instance, fails the vertical line test. Parametric surfaces handle this by expressing all three coordinates as functions of two parameters $u$ and $v$:

$x = x(u,v), \quad y = y(u,v), \quad z = z(u,v)$

or equivalently as a position vector $\vec{r}(u,v) = \langle x(u,v),\, y(u,v),\, z(u,v) \rangle$, where $(u,v)$ ranges over some domain $D$ in the $uv$-plane.

Common examples:

• Sphere of radius $a$: $\vec{r}(\theta, \phi) = \langle a\sin\phi\cos\theta,\, a\sin\phi\sin\theta,\, a\cos\phi \rangle$
• Cylinder of radius $a$: $\vec{r}(\theta, z) = \langle a\cos\theta,\, a\sin\theta,\, z \rangle$
• A function graph $z = f(x,y)$ is itself a parametric surface with $\vec{r}(x,y) = \langle x,\, y,\, f(x,y) \rangle$

Surface Area of Parametric Surfaces

For a parametric surface $\vec{r}(u,v)$, the surface area formula becomes:

$S = \iint_D \left\| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right\| \, dA$

The two partial derivatives $\frac{\partial \vec{r}}{\partial u}$ and $\frac{\partial \vec{r}}{\partial v}$ are tangent vectors to the surface. Their cross product produces a vector perpendicular to the surface, and the magnitude of that cross product gives the area of the infinitesimal parallelogram spanned by those tangent vectors.

Step-by-step process:

1. Compute $\vec{r}_u = \frac{\partial \vec{r}}{\partial u}$ and $\vec{r}_v = \frac{\partial \vec{r}}{\partial v}$ (these are vectors with three components each).
2. Take the cross product $\vec{r}_u \times \vec{r}_v$.
3. Find its magnitude $\left\| \vec{r}_u \times \vec{r}_v \right\|$.
4. Integrate this magnitude over the parameter domain $D$.

You can verify that when $\vec{r}(x,y) = \langle x, y, f(x,y) \rangle$, this cross product formula reduces to $\sqrt{1 + f_x^2 + f_y^2}$, recovering the formula from the first section.

Normal Vectors

Definition and Significance of Normal Vectors

A normal vector to a surface at a point is any vector perpendicular to the surface there. It's typically denoted $\vec{n}$. The direction of $\vec{n}$ tells you which way the surface is "facing" at that point.

Normal vectors show up throughout multivariable calculus and physics:

• They define the orientation needed for surface integrals (flux integrals in particular).
• In physics, they determine the direction of force on a surface in fluid pressure or electromagnetic problems.
• They're essential for finding tangent planes: the tangent plane at a point has $\vec{n}$ as its normal.

Calculating Normal Vectors for Parametric Surfaces

For a parametric surface $\vec{r}(u,v)$, the normal vector comes directly from the cross product of the tangent vectors:

$\vec{N} = \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$

This works because $\vec{r}_u$ and $\vec{r}_v$ are both tangent to the surface, so their cross product is perpendicular to both, hence perpendicular to the surface.

To get a unit normal vector, divide by the magnitude:

$\hat{n} = \frac{\vec{r}_u \times \vec{r}_v}{\left\| \vec{r}_u \times \vec{r}_v \right\|}$

A few things to keep in mind:

• The cross product $\vec{r}_u \times \vec{r}_v$ and $\vec{r}_v \times \vec{r}_u$ point in opposite directions. Your choice of ordering determines whether $\vec{n}$ points "outward" or "inward" for a closed surface. Be consistent with whatever convention the problem uses.
• For a function graph $z = f(x,y)$, using $\vec{r}(x,y) = \langle x, y, f(x,y) \rangle$ gives $\vec{N} = \langle -f_x, -f_y, 1 \rangle$. This always has a positive $z$-component, so it points upward.
• The magnitude $\left\| \vec{r}_u \times \vec{r}_v \right\|$ is exactly the integrand in the parametric surface area formula. So computing the normal vector and computing surface area are closely related tasks.