Motivating Surface and Flux Integrals

Understanding 1D Integrals

Let's start with the first integral learned in Calculus.

Understanding 2D Integrals

Understanding 3D Integrals

5) Surface Integrals

We now have a fully motivated surface integral. Just as an arc-length integral has a speed factor in it to correct for the difference between accumulation of time and accumulation of distance, surface integrals have a "speed" factor to correct for the difference between the accumulation of our parameters and the accumulation of area along the surface.

We'll start by finishing the integral in red above:

$$ \int_0^1 \int_0^{1-t} f(r(s,t)) \ \\\| \vec{r}_s(s,t) \times \vec{r}_t(s,t) \\\| \ ds \ dt $$ $$ = \int_0^1 \int_0^{1-t} f(r(s,t)) \ \\\| \langle 1, 0, -1 \rangle \times \langle 0, 1, -1 \rangle \\\| \ ds \ dt$$ $$ = \int_0^1 \int_0^{1-t} f(r(s,t)) \ \\\| \langle 1, 1, 1 \rangle \\\| \ ds \ dt$$ $$ = \int_0^1 \int_0^{1-t} f(r(s,t)) \ \sqrt{3} \ ds \ dt$$ The $\sqrt{3}$ in the integral above tells us that the cross-section on the surface above is $\sqrt{3}$ times bigger than the $\Delta s$-$\Delta t$ rectangle you see in the $xy$-plane below it.
This factor was constant because of our surface and parameterization. But it is not constant for many surfaces: The factor is usually an expression of $s$ and $t$ telling us the stretch factor at that point.

Surface Integral Calculation

With the example we started with, let's compute the surface integral for the function $f(x,y,z) = 6xy$. $$ \iint_R 6xy \ dA = \int_0^1 \int_0^{1-t} 6st \ \sqrt{3} \ ds \ dt $$ $$ = 3 \sqrt{3} \int_0^1 t \left( \int_0^{1-t} 2s \ ds \right) \ dt $$ $$ = 3 \sqrt{3} \int_0^1 t \left( \ \ \left. s^2 \ \ \right|_0^{1-t} \ \right) \ dt $$ $$ = 3 \sqrt{3} \int_0^1 t (1-t)^2 \ dt $$ $$ = 3 \sqrt{3} \int_0^1 t^3 - 2t^2 + t \ dt $$ $$ = 3 \sqrt{3} \left( \left. \frac{1}{4}t^4 - \frac{2}{3}t^3 + \frac{1}{2}t^2 \ \ \right|_0^1 \right) $$ $$ = 3 \sqrt{3} \left( \frac{1}{4} - \frac{2}{3} + \frac{1}{2} \right) $$ $$ = 3 \sqrt{3} \left( \frac{1}{12} \right) = \frac{\sqrt{3}}{4}. $$

One More Surface Visualization

Below is a surface parameterized by $s$ and $t$, only $s = e^x$ and $t = e^y$, making changes in $s$ and $t$ grow as the $x$ and $y$ coordinates get bigger.
The area of the parallelogram showing by default approximates the accumulation of surface area from a given change in $x$ and $y$. If you change it to $s$ and $t$, you can see that the area change is very fast in some areas and very slow in others.

Example 6: Tangent Plane with Parameter Area

Surface Integral Problems

Problem 1: Find the surface integral of the function $f(x,y,z) = z$ over the surface $r$ with (implicit) equation $y = 1 - z^2$ and bounds $ 0 \leq x \leq 7$ and $ 0 \leq z \leq 7$.

Solution: We have an implicit equation that we need to parameterize. Since $y$ can be expressed as a function of $x$ and $z$ (in particular $z$), we can let $$ \vec{r}(s,t) = \langle s, 1 - t^2, t \rangle. $$ Then the integral becomes $$ \int_0^7 \int_0^7 f(r(s,t)) \ \\\| \vec{r}_s(s,t) \times \vec{r}_t(s,t) \\\| \ ds \ dt $$ $$ = \int_0^7 \int_0^7 t \ \\\| \langle 1, 0, 0 \rangle \times \langle 0, -2t, 1 \rangle \\\| \ ds \ dt $$ $$ = \int_0^7 \int_0^7 t \sqrt{(0)^2 + (-1)^2 + (-2t)^2} \ dt \ ds $$ $$ = \int_0^7 \int_0^7 t \sqrt{1+4t^2} \ ds \ dt $$ $$ = \int_0^7 \left( t \sqrt{1+4t^2} \right) \left( \int_0^7 1 \ ds \right) \ dt $$ $$ = \int_0^7 \left( t \sqrt{1+4t^2} \right) \left( 7 \right) \ dt $$ $$ = \frac{7}{8} \int_0^7 \frac{d}{dt} \left( \frac{2}{3} (1+4t^2)^{3/2} \right) \ dt $$ $$ = \frac{7}{8} \left( \left. \frac{2}{3} (1+4t^2)^{\frac{3}{2}} \ \ \right|_0^7 \right)$$ $$ = \frac{7}{12} \left( 197^{3/2} - 1 \right). $$

Problem 2: Find the surface integral of the function $f(x,y,z) = e^{-z}$ over the cylinder $x^2 + y^2 = 16$ with the bounds $0 \leq z \leq 1$.

Solution: This is a perfect problem for cylindrical coordinates. This is because $r$ is fixed:$r = \sqrt{x^2 + y^2} = 4$, so we can let $\theta$ and $z$ cover all of the points in question. $$ \vec{r}(\theta, z) = \langle 4 \cos(\theta), 4 \sin(\theta), z \rangle. $$

Now, we can solve this problem computationally or we can solve it with less computation using Geometry. We'll solve it both ways, starting with the Geometric way.

Geometric Way

Because changing the $\theta$-value of a point moves it parallel to the $xy$-plane and changin the $z$-value moves it parallel to the $z$-axis, these directions of motion are perpendicular. So we only need to multiply their magnitides and don't have to take a cross product. Furthermore, $z$ is a variable in space, so its change in length is in no way distorted by our parameterization. Finally, a small change in $\theta$, $\Delta \theta$ translates to a distance travelled of $r \Delta\theta = 4 \Delta\theta$ as we approach the limit. This Geometric explanation also works for the Jacobian used in polar coordinates: $dx dy = r \ dr d\theta$.

So our integral is just $$ \iint_R f \ dA = \int_0^1 \int_0^{2\pi} e^{-z} \ |4| \ d\theta \ dz. $$ Solving... $$ \int_0^1 4e^{-z} \left( \int_0^{2\pi} 1 \ d\theta \right) \ dz = \int_0^1 4e^{-z} \left( 2\pi \right) \ dz = 8\pi \int_0^1 e^{-z} \ dz $$ $$ = 8\pi \left( \left. -e^{-z} \ \ \right|_0^1 \ \right) = 8\pi \left( 1 - \frac{1}{e} \right). $$

Regular Way

$$\vec{r}_\theta = \frac{\partial}{\partial \theta} \left\langle 4 \cos(\theta), 4 \sin(\theta), z \right\rangle = \langle -4 \sin(\theta), 4 \cos(\theta), 0 \rangle$$ $$\vec{r}_z = \frac{\partial}{\partial z} \left\langle 4 \cos(\theta), 4 \sin(\theta), z \right\rangle = \langle 0, 0, 1 \rangle$$ $$\vec{r}_\theta \times \vec{r}_z = \langle 4\cos(\theta), 4\sin(\theta), 0 \rangle$$ $$ \\\| \vec{r}_\theta \times \vec{r}_z \\\| = 4.$$ $$ \iint_R f \ dA = \int_0^1 \int_0^{2\pi} e^{-z} \ 4 \ d\theta \ dz = \cdots = 8\pi \left( 1 - \frac{1}{e} \right). $$
Note that we had the exact same integral as above. We were just able to skip the computation of the cross product with the geometric argument.

Flux Integrals

Geometric Motivation

Flux integrals have a very close approach to surface integrals. To see how close they are, we'll begin with one more visualization of a surface integral. The intention of this example is to motivate a visual interpretation of the surface integral that is very, very close to the flux integral. In fact, the only difference will be an angle.

Example 7: Visualizing the Whole Surface Integral at Once

Here a sphere is the surface used and we want to integrate over the surface for a function $f(x,y,z) = |x| + |y| + |z|$. The numerical length, which is not uniform over the sphere, is represented physically with a pole sticking straight out of the surface at that point. The surface integral can then be estimated with Riemann sums that have changes in two physical directions and the height just mentioned.

Example 8: Breaking it Up into "Boxes"

The example above used Riemann Sums with dimensions in space. But our integrals have changes in terms of our parameters $s$ and $t$ (or $\theta$ and $\phi$ in this case).
The picture above ties together all of the previous visualizations to tell you exactly what is in the integrand. The area of the parallelogram tangent to the surface is $\\\| \vec{r}_s(s,t) \times \vec{r}_t(s,t) \\\|$ while the height is the function's value $f(x,y,z) = f(\vec{r}(s,t))$. This is exactly the volume formula.


The surface integral of a function $f(x,y,z)$ over the surface $\vec{r}(s,t)$ is given by $$ \iint_R (\text{height}) (\text{parallelogram area}) \ ds \ dt $$ $$ = \iint_R f(r(s,t)) \ \\\| \vec{r}_s(s,t) \times \vec{r}_t(s,t) \\\| \ ds \ dt $$

Flux Integral "Boxes"

When we switch from scalar fields $f(x,y,z)$ to vector fields $\vec{F}(x,y,z)$, the object we are integrating is no longer a scalar. So it wouldn't make any sense to use an integral like $ = \iint_R f(r(s,t)) \ \\\| \vec{r}_s(s,t) \times \vec{r}_t(s,t) \\\| \ ds \ dt $ anymore. But take a look at what happens when we replace the height of our boxes with the vector field's vector at a given point on our surface.

Example 9: Vector Field "Boxes"

Replacing the "height" line segments for a scalar field with vectors still allows us to compute the area of this box. Except this box is really a parallelepiped.

Flux Integral Computation

As you know from Linear Algebra, the way to compute the (signed) volume of a parallelepiped is with a determinant: $$ V = \left| \ \ \ \ \ \begin{array} \vec{F}(x,y,z) \\ \vec{r}_s \\ \vec{r}_t \end{array} \ \ \ \ \ \right|. $$ You might also have seen that the cross product $\vec{r}_s(s,t) \times \vec{r}_t(s,t)$ gave us the signed area of the parallelogram on the surface represented as an oriented normal vector to the surface. Taking the dot product with $\vec{F}(x,y,z)$ gives us exactly the same thing. In fact, the equality below is where we get the cross product formula from: $$ \vec{F}(\vec{r}(s,t)) \cdot \left( \vec{r}_s(s,t) \times \vec{r}_t(s,t) \right) = \left| \ \ \ \ \ \begin{array} \vec{F}(x,y,z) \\ \vec{r}_s \\ \vec{r}_t \end{array} \ \ \ \ \ \right|. $$ The quantity $\vec{F}(\vec{r}(s,t)) \cdot \left( \vec{r}_s(s,t) \times \vec{r}_t(s,t) \right)$ is often called the "Box Product".

What does this mean for computing flux integrals? It means we have almost exactly the same formula. In fact, the switch from surface integrals to flux integrals is remarkably close to the switch from path integrals to contour integrals.


The Flux of a vector field $\vec{F}(x,y,z)$ through the surface $\vec{r}(s,t)$ is given by $$ \iint_R (\overrightarrow{\text{height}}) \cdot (\overrightarrow{\text{parallelogram area vector}}) \ ds \ dt $$ $$ = \iint_R \vec{F}(\vec{r}(s,t)) \cdot \left( \vec{r}_s(s,t) \times \vec{r}_t(s,t) \right) \ ds \ dt. $$