Multivariable calculus and vector analysis

Surface integrals

We have stressed that parametrized surfaces are similar to parametrized paths. In the same way, surface integrals are similar to path integrals. Recall that there are two types of integrals over paths: path integrals of scalar-valued functions and line integrals of vector fields. For this course, we tend to stress the line integral of vector fields since it appears in our fundamental theorems.

As you probably guessed, there are two types of surface integrals: surface integrals of scalar-valued functions and surface integrals of vector fields. And, yes, in this course, we do tend to stress the surface integral of vector fields since it appears in fundamental theorems.

Surface integrals of scalar-valued functions

Let $S$ be a surface parametrized by $\mathbf{\Phi }\left(u,v\right)$ for $\left(u,v\right)$ in some region $D$.

Imagine you had a scalar-valued function $f\left(\mathbf{x}\right)$ so that $f\left(\mathbf{x}\right)$ was the density of the surface at point $\mathbf{x}$. Just like we did for paths, we want to find the total mass of the surface by integrating the density $f\left(\mathbf{x}\right)$ over the surface.

To obtain mass from density, we need to multiply density times surface area. From our surface area calculation we know that

$$\begin{array}{lll}\hfill ∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥\Delta v\Delta v& & \hfill \end{array}$$

is the surface area of a small section of the surface. If we multiply this by the density $f\left(\mathbf{\Phi }\left(u,v\right)\right)$, we will obtain the mass of that small section. We could approximate the mass of the whole surface by a Riemann sum of such terms. Then, by letting $\Delta s$ and $\Delta t$ go to zero, we would discover that the mass of the whole surface is the integral

$$\begin{array}{lll}\hfill {\iint }_{S}gdS={\iint }_{D}f\left(\mathbf{\Phi }\left(u,v\right)\right)∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥dudv.& & \hfill \text{ (1)}\end{array}$$

This integral is a two-dimensional analog of the path integral of a scalar-valued function, except that $\left|\right|{\mathbf{\Phi }}^{\text{'}}\left(t\right)\left|\right|$ is replaced by

$$\begin{array}{lll}\hfill ∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥.& & \hfill \end{array}$$

Surface integrals of vector fields

The line integral of a vector field $\mathbf{F}$ could be interpreted as the work done by the force field $\mathbf{F}$ on a particle moving along the path. The surface integral of a vector field $\mathbf{F}$ actually has a simpler explanation. If the vector field $\mathbf{F}$ represents the flow of a fluid, then the surface integral of $\mathbf{F}$ will represent the amount of fluid flowing through the surface (per unit time).

The amount of the fluid flowing through the surface per unit time is also called the flux of fluid through the surface. For this reason, we often call the surface integral of a vector field a flux integral.

If water is flowing perpendicular to the surface, a lot of water will flow through the surface and the flux will be large. On the other hand, if water is flowing parallel to the surface, water will not flow through the surface, and the flux will be zero. So, to calculate the total amount of water flowing through the surface, we want to add up the component of the vector $\mathbf{F}$ that is perpendicular to the surface.

Let $\mathbf{n}$ be a normal vector to the surface. The flux of fluid through the surface is determined by the component of $\mathbf{F}$ that is in the direction of $\mathbf{n}$, i.e. by $\mathbf{F}\cdot \mathbf{n}$. (Note that $\mathbf{F}\cdot \mathbf{n}$ will be zero if $\mathbf{F}$ and $\mathbf{n}$ are perpendicular, positive if $\mathbf{F}$ and $\mathbf{n}$ are pointing the same direction, and negative if $\mathbf{F}$ and $\mathbf{n}$ are pointing in opposite directions).

To illustrate, let’s look at our helicoid again. Here’s the helicoid with a normal vector, shown in blue. In this case, we are using the upward pointing normal vector. (We could have used the downward point normal instead. If we did, our fluid flux calculation we have the opposite sign.)

Given some fluid flow $\mathbf{F}$, if we integrate $\mathbf{F}\cdot \mathbf{n}$, we will determine the total flux of fluid through the helicoid, counting flux in the direction of $\mathbf{n}$ as positive and flux in the opposite direction as negative.

Imagine that the fluid flow is in the direction given by the magenta arrows

It appears that the fluid is flowing generally in the same direction as $\mathbf{n}$ (for the most part $\mathbf{F}$ and $\mathbf{n}$ are closer to pointing in the same direction than pointing in the opposite direction). However, notice, for example, that when $s=0$ and $t=2\pi$ (or when $s=0$ and $t=0$), the fluid is flowing in the opposite direction of $\mathbf{n}$ (at least the flow is closer to the opposite direction than the same direction). At these points, the fluid is crossing the surface in the opposite direction than it is at most points on the surface.

The below figure demonstrates this more clearly. Here, you can see the fluid vector $\mathbf{F}$ (in magenta) at the same point as the normal vector (in blue). The blue dot on the slider shows the dot product $\mathbf{F}\cdot \mathbf{n}$. Note that $\mathbf{F}\cdot \mathbf{n}$ is usually positive, but is negative at a few points, such as those mentioned above. When $\mathbf{F}\cdot \mathbf{n}=0$, what is the relationship between the fluid vector $\mathbf{F}$ and the surface?

Using equation (1) with $g=\mathbf{F}\cdot \mathbf{n}$, the total fluid flow is

$$\begin{array}{lll}\hfill {\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_S\mathbf{F}\cdot \mathbf{n}dS={\iint }_D\left(\mathbf{F}\cdot \mathbf{n}\right)∥\frac{\partial \mathbf{\Phi }}{\partial u}\times \frac{\partial \mathbf{\Phi }}{\partial v}∥dudv.& & \hfill \text{ (2)}\end{array}$$

The formula for a unit normal vector of the surface is

$$\begin{array}{lll}\hfill \mathbf{n}=\frac{\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)}{∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥}.& & \hfill \end{array}$$

If we plug this expression for $\mathbf{n}$ into equation (2), the $∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥$ factors cancel, and we obtain

$$\begin{array}{lll}\hfill {\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_D\mathbf{F}\left(\mathbf{\Phi }\left(u,v\right)\right)\cdot \left(\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)\right)dudv.& & \hfill \text{ (3)}\end{array}$$

Note that equation (3) is similar to the line integral of a vector field

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_a^b\mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)dt.& & \hfill \end{array}$$

For line integrals, we integrate the component of the vector field in the tangent direction given by ${\mathbf{c}}^{\text{'}}\left(t\right)$. For surface integrals, we integrate the component of the vector field in the normal direction given by $\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)$.

You can read some examples here.

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

4d. Path integrals and line integrals

4e. Surface integrals

• Surface integrals
• Surface integral examples

5. The fundamental theorems of vector calculus

6. Review material