Skip to main content

Course Notes for Multivariable Calculus

Remkes Kooistra

Section 11.2 Flux Integrals

Subsection 11.2.1 Fields and Surfaces

Given a vector field and a parametric curve, the line integral measured the interaction of the field and the curve. It told how much the field helped or hindered movement along the curve. This was measured by a dot product of the field with the tangent to the curve, since the dot product nicely measures the interaction of two vector. Now, instead of asking for the action of a field along a curve, I’m interesting in the action of a field through a surface.
The biggest difference here is that a surface doesn’t have the notion of movement. Instead, the surface is treated as a static object; I’m interesting in how the field moves through the surface. That said, parametrization still matters. A parametric surface has a normal calculated in the two parameters. Again, I look to the dot product. The dot product of the vector field with the normal to the surface shows the interaction.
If the field and the normal point roughly in the same direction, then the field is passing through the surfaces, transversely, at that point. This is a strong positive reaction. If the field and normal have nearly opposite direction, then there is a strong negative reaction. The parametrization and the normal gives an orientation to the surface: there is a differnce (in sign) between moving through the surface with the normal or against the normal. I can think of this of choosing sides of the surface: say, moving from below to above is positive and moving from above to below is negative. Unlike a locus, a parametric surface has this orientation, this choice of above and below, intrinsically built into the definition.
If the vector field and the normal to the surface are nearly perpendicualr, then the field mostly flows along the surface, tangentially. This is measured by a small dot prodcut.
The unit tangent was useful in the definition of the line integral. I’m going to need the equivalent definition here of the unit normal.

Definition 11.2.1.

Let \(\sigma(u,v)\) be a parametric surface in \(\RR^3\) defined over a simply-connected domain \(D\text{.}\) The unit normal \(N\) of the surface is the vector \(\frac{\sigma_u \times \sigma_v}{|\sigma_u \times \sigma_v|}\text{.}\) (Note that in the definition of a parmetric surface, the normal can never be zero. Division by zero is avoided by this stipulation.)
For parametric curves, the parametrization by arclength is a unique parametrization that can be assigned to any particular curve shape. It has the property that its tangent is aways the unit tangent. I can also define a special parametrization of a surface, \(\sigma(s,t)\text{,}\) where the \(\sigma_s(s,t) \times \sigma_t(s,t) = N\text{.}\) This is the unique parametrization where the normal is always the unit normal. Just like I used the parametrization by arclength to define line integrals, I use this special parametrization to defined integrals over parametric surfacees.

Definition 11.2.2.

Let \(\sigma(s,t)\) be a parametric surface in \(\RR^3\) defined over a simply-connected domain \(D\text{.}\) Assume the parametrization in parameters \((s,t)\) is the unique parametrization where \(\sigma_s \times \sigma_t = N\text{.}\) Let \(F\) be a vector field defined on a neighbourhood of the surface. (A neighbourhood is any open set which contains the entire surface.) The flux integral is defined to be the following integral.
\begin{equation*} \int_{\sigma} F \cdot N dA = \int_{\sigma} F \cdot dA = \int_D F(\sigma(s,t)) \cdot N(s,t) ds dt \end{equation*}
The first notation is more complete, but the second notation is conventional.
As with curves, I don’t want to calculate with the special parametrizations, since it may be very difficult to actually produce that parametrization. I’d rather calculate with an arbitrary parametrization. In an arbitrary parametrization, \(dA = |\sigma_u \times \sigma_v| du dv\text{.}\) I can perform a change of variables on the integral in the definition.
\begin{align*} \int_{\sigma} F \cdot N dA \amp = \int_D F(\sigma(u,v)) \cdot \frac{\sigma_u \times \sigma_v}{|\sigma_u \times \sigma_v|} |\sigma_u \times \sigma_v| du dv\\ \amp = \int_D F(\sigma(u,v)) \cdot (\sigma_u \times \sigma_v) du dv \end{align*}
The result of the change of variables is a flux integral calcuted in any arbitrary parametrization. The length of the norml has conveniently cancelled off (like the length of the tangent cancelled off in the definition of a line integral.) The result is a very reasonable way to calcualte flux integrals in any arbitrary parametrizations with the knowledge that the integral doesn’t depend on the parametrizations.

Subsection 11.2.2

\(F(x,y,z) = (0,0,k)\) is a constant vertical flow with the same flow rate. I can imagine this flow through a vertical pipe with radius \(a\) around the \(z\) axis by restricting the field to the domain where \(x^2 + y^2 \leq a\text{.}\) At some point in the pipe (say \(z=0\)), I can ask how much fluid is flowing through the pipe. The cross section of the pipe is a circle of radius \(a\) at height \(z=0\text{.}\) I can parametrize this circle as \(\sigma(r,\theta) = (r \cos \theta, r \sin \theta,0)\) for \((r,\theta) \in [0,a] \times [0,2\pi]\text{.}\) Then I can calculate the flux integral to determine how much water is flowing through the pipe at this point.
\begin{align*} \sigma_r \amp = (\cos \theta, \sin \theta, 0)\\ \sigma_\theta \amp = (-r \sin \theta, r \cos \theta, 0)\\ \sigma_r \times \sigma_\theta \amp = (0, 0, r)\\ \int_\sigma F \cdot dA \amp = \int_0^a \int_0^{2\pi} k r d\theta dr\\ \amp = \frac{2\pi a^2k}{2} = \pi a^2 k = (\pi a^2) k \end{align*}
This answer is unsurpring. \(\pi a^2\) is the cross-sectional area and \(k\) is the rate of flow. The rate of flow is constant everywhere, so it makes sense that the result of the integral is just the product of this cross sectional area \(\pi a^2\) and the flow rate \(k\text{.}\)
Let me introduce a small variation of the previous example. I’ll stick with the same pipe, but change the field so that the flow through the pipe is not uniform. Consider this field.
\begin{equation*} F(x,y,z) = \left( 0, 0, \frac{k (a^2 - x^2 - y^2)}{a^2} \right) \end{equation*}
In the very middle of the pipe, this flow has the same flow rate \(k\) as the previous example. However, the flow gets slower closer to the edge of the pipe, becoming no flow at all at the edge itself. I’ll calculate the flux integral to measure the total flow of this new field through a cross-section of the pipe.
\begin{align*} \int_\sigma F \cdot dA \amp = \int_0^a \int_0^{2\pi} \frac{k(a^2-r^2)}{a^2} r d\theta dr\\ \amp = \frac{2\pi k}{a^2} \int_0^a a^2r - r^3 dr\\ \amp = \frac{2\pi k}{a^2} \left. \left( \frac{a^2 r^2}{2} - \frac{r^2}{4} \right) \right|_0^a\\ \amp = \frac{2\pi k}{a^2} \left(\frac{a^4}{2} - \frac{a^4}{4} \right) = \frac{\pi k a^2}{2} \end{align*}
This is half the original flux. The lower rate of flow makes sense; near the centre, the flow was similar to the previous example, but near the edge, the flow is much slower. The flux integral should produce a smaller result.
Still sticking with this pipe and measuring the flux through a cross-sectional circle, I’ll complicate the field yet more. The following is a turbulent flow, where other directions are introduced as the flow may includes vortices.
\begin{align*} F(x,y,z) \amp = \left( \sin \left( \left( \frac{a-\sqrt{x^2 + y^2}}{a} \right) \pi \right), \sin \left( \left( \frac{a-\sqrt{x^2 + y^2}}{a} \right) \pi \right), \right. \\ \amp \left. \cos \left( \left( \frac{a-\sqrt{x^2 + y^2}}{a} \right) \pi \right) \right) \end{align*}
Again, I’ll calculate the flux integral to see what happens with the net flow rate through a cross-section for this turbulent flow.
\begin{align*} F \cdot \sigma_r \times \sigma_z \amp = k \cos \left( \left( \frac{a-r}{a} \right) \pi \right) r\\ \int_\sigma F \cdot dA \amp = \int_0^a \int_0^{2\pi} k \cos \left( \left( \frac{a-r}{a} \right) \pi \right) r d\theta dr\\ \amp = 2\pi k \int_0^a \cos \left( \pi - \frac{\pi r}{a} \right) r dr\\ \amp = 2\pi k\int_0^a -r \cos \left( \frac{\pi r}{a} \right) dr\\ \amp = -2\pi k \frac{a}{\pi} \int_0^a \frac{\pi r}{a} \cos \left( \frac{\pi r}{a} \right) dr\\ u \amp = \frac{\pi r}{a}\\ \amp = -2\pi k \frac{a}{\pi} \int_0^{\pi} u \cos u \frac{a du }{\pi}\\ \amp = \frac{-2\pi a^2}{\pi} \left( \cos u + u \sin u \right) \bigg|_0^{\pi} = \frac{4k a^2}{\pi} \end{align*}
Consider a paddle with a roughly rectangular cross section 15cm wide and 30cm tall. Say it moves through the water in the \(z\) direction with speed \(v\text{.}\) I can reinterpret the situation by letting the water move with field \(F = (0,0,v)\) in which the paddle is stationary. I can think of the force the paddle causes in terms of the flux of this field through the paddle.
Now, I can perform a paddle stroke directly perpendicular to the direction of movement or at an angle \(\theta\text{.}\) My question is: how does our forward force (hence flux) vary due to \(\theta\text{?}\)
If the paddle is angled at angle \(\theta\text{,}\) then it is represented as a surface with normal \((\sin \theta, 0, \cos \theta)\text{.}\) Therefore \(F \cdot N = v \cos \theta\text{.}\) (I can make this calculation without even have to do all the formal details of parametrization, since only the normal is important. The paddle is a part of a plane, so it has a constant normal.) Then I calculate the flux integral.
\begin{equation*} \int_\sigma F \cdot N dA = \int_D V \cos \theta dA = v \cos \theta \int_D 1 dA = 450 v \cos \theta \end{equation*}
Compared to the perpendicular force or \(450v\) the force due to a paddle stroke at an angle is this original force multiplied by \(\cos \theta\text{.}\) Cosine is the approprate result here: when there is no adjustment to the angle (\(\theta = 0\)), cosine is \(1\) and the full force is applied; when the paddle turned a full quarter turn, cosine is \(0\) and there is no force at all.
<Prev^TopNext>