Lab 6B - Surface Integrals of Vector Fields
Math 2374 - University of Minnesota
http://www.math.umn.edu/math2374
questions to: drake@math.umn.edu or rogness@math.umn.edu

Introduction

    Batman:"There's an eclipse of the sun due."
    Robin:"But that's only for half a minute!"
    Batman:"That's all we'll need, if my calculus is correct..."

By now you're familiar with line integrals of vector fields, which in a certain sense measure how much the vector field is "pushing particles" along the path -- over a closed path, we called that circulation. What about surface integrals? That is, what about integrating a vector field over some parametrized surface in three-dimensional space?

For surfaces, it no longer makes much sense to ask how much the vector field is pushing particles along the surface. Instead, the question we'll want to answer is: how much is the vector field pushing particles across the surface?

Let's make that a little bit more precise. Say the vector field represents velocity. Consider a very small bit of the surface (a small parallelogram, for example), so small that we can consider the vector field to be constant over that parallelogram. In that case, the sides of the parallelogram and the vector representing the vector field form a parallelepiped. There's a picture and some more descriptions of this on pages 344-345 of your text.

To find the volume of that parallelepiped, we use the scalar triple product: we take the cross product of the sides of the parallelogram, and then find the dot product of that with the vector from the vector field. We interpret the result as "amount of fluid (or particles) moving across the surface per unit time".

The precise definition of a surface integral (or flux integral; the two terms are interchangeable) of a vector field is in the next section.

Definition

Just for your reference, we'll repeat the definition of a surface integral of a scalar function.  

Let F : U⊂^3⟶be a continuous vector field, and let M be a smooth surface lying in U that is parametrized by f(s,t), where (s, t) ϵR⊂^2.  The surface integral (or flux integral) of F over M is:

∫∫_MF  nσ = ∫∫_RF (f (s, t))  (∂f/s  ∂f/∂t) ds dt

Note that this definition is very similar to the one for surface integrals of scalar functions.  For scalar functions we multiplied the scalar function by the length of the normal vector.  In this case we'll take the dot product of our vector field with the normal vector.  Remember, the dot product of two vectors is a number, so our final integrand will actually still be a scalar function.

There are at least two ways to convince yourself that this definition makes sense. As we said above, F  n represents the scalar triple product of the partial derivatives and F, which we know is equal to the volume of the parallelepiped spanned by those vectors (up to a possible minus sign).

Another way to see this is to rewrite the dot product: F  n = || F || || n || cos θ. In this case, we see that the length of the normal vector is the area of the spanned parallelogram, which is multiplied by the length of F at that point. The cosine takes care of the ``tilt'': if the vector field is moving in the same direction as the surface, it isn't pushing any particles at all across the surface; in that case, cos θ = cos (π/2) = 0 (remember that the normal vector will be perpendicular to F in this case). Similarly, if F is moving perpendicular to the surface, || F || || n || will be multiplied by cos 0 = 1, so F has as much effect as it can.

It's important to get the orientation of the surface right. The integral represents the flux from the "inside" of the surface to the "outside" -- the direction in which the normal vector points is considered "outside". Every time you do a surface integral of a vector field, you should make sure you're using a normal vector that points in the correct direction (which should be specified in the problem).

Driving in the Rain

Suppose there's a nice spring shower outside and you're out driving a car.  You're sitting at a stoplight, and the wipers are at just the right speed to keep the windshield clean.  The light turns green, and you start driving at 30mph down the road.  All of a sudden you can't see a thing -- there's too much water on the windshield, and you have to turn your wipers to a higher setting.  Why is that?  We can actually explain it using flux integrals.

Suppose our windshield is a rectangular piece of glass, represented by the plane z = 2 - 2x over the rectangle 0≤x≤0.5, -1.25≤y≤1.25.  Here's a parametrization and picture of it; as you can guess by the slanting of the plane, we're assuming that the car is driving along the positive x-axis.

In[594]:=

f[x_, y_] = {x, y, 2 - 2x} ; RowBox[{windshield, =, RowBox[{ParametricPlot3D, [, RowBox[{f[x,  ... 25}], ,, 1.25}], }}], ,, AxesLabel {"x", "y", "z"}}], ]}]}]

Rain is really just a bunch of water falling from the sky.  If we're sitting at a stoplight, then the rain is falling straight down, say, at 10 km/h.  So the velocity of the raindrops is described by the vector field G(x,y,z)=(0,0,-10).  Mathematica can plot this vector field for us, as long as we load the correct package first:

In[620]:=

G[x_, y_, z_] = {0, 0, -10} RowBox[{rain,  , =,  , RowBox[{PlotVectorField3D, [, RowBox[{G[x,  ...  z], ,,  , {x, -.5, 1}, ,,  , {y, -2, 2}, ,, RowBox[{{, RowBox[{z, ,, -.5, ,, 2.5}], }}]}], ]}]}]

Out[620]=

{0, 0, -10}

[Graphics:HTMLFiles/index_14.gif]

Out[621]=

⁃Graphics3D⁃

Mathematica can also show us a picture of the windshield in the rain.

In[622]:=

Show[windshield, rain]

[Graphics:HTMLFiles/index_17.gif]

Out[622]=

⁃Graphics3D⁃

Now a flux integral measures the amount of fluid crossing a surface.  If we compute the flux integral of this vector field over the windshield, this will actually tell us how much rain is hitting the windshield.  (Because it's made of glass, it can't actually flow through, so instead it just splatters all over, and forces the driver to turn on the wipers.)

The steps we need to take to compute this integral are very similar to the ones we used last week.  We need to find G(f(x,y)), but since G is a constant vector field (it never changes), we know that G(f(x,y)) = (0,0,-10) at all points. Let's run the following command anyway so that Mathematica knows about it, and so that you know how to use the command in the future:

In[623]:=

Apply[G, f[x, y]]

Out[623]=

{0, 0, -10}

We need to compute the normal vector, which for a plane is easy (for this plane, it's (2,0,1)), but for more complicated situations, you'll have to use something like the following command to compute the cross product of the partial derivatives:

In[624]:=

n = Cross[D[f[x, y], x], D[f[x, y], y]]

Out[624]=

{2, 0, 1}

And now we're ready to integrate.  The integrand is G(f(x,y))·Overscript[n, ⇀].

In[625]:=

integrand = Apply[G, f[x, y]] . n ; RowBox[{Integrate, [, RowBox[{integrand, ,, {x, 0, .5}, ,, RowBox[{{, RowBox[{y, ,, RowBox[{-, 1.25}], ,, 1.25}], }}]}], ]}]

Out[626]=

RowBox[{-, 12.5}]

Why is the answer negative? The answer has to do with the normal vector.  Our normal vector is pointing in the direction that we're travelling -- look at Overscript[n, ⇀] and look at the picture of the windshield until you believe this.  In some sense the normal vector defines which direction is "outward."  Flux integrals measure the net flow from inside the surface to outside the surface, so in this case a positive answer would mean water was coming from the inside of the car; water hitting the windshield from the outside represents negative flow.

Here's a question for you to think about: what are the correct units for our answer?

Remember, this first part of the example was for when the car is sitting still.  Now suppose we're driving down the street at 50 km/h, or just over 31mph.  In this case, relative to the car, the velocity of the rain is described by the vector field, H(x, y, z) = (-50, 0, -10).  Before going on, convince yourself that this is the correct formula!  (Think about the negative signs in particular.)

Here's a picture of the new situation:

In[627]:=

H[x_, y_, z_] = {-50, 0, -10} ; newrain = PlotVectorField3D[H[x, y, z], {x, -1, 1}, {y, -2, 2}, {z, 0, 3}] Show[windshield, newrain]

[Graphics:HTMLFiles/index_29.gif]

Out[628]=

⁃Graphics3D⁃

[Graphics:HTMLFiles/index_31.gif]

Out[629]=

⁃Graphics3D⁃

How much rain is hitting the windshield this time?  Let's compute the integral:

In[630]:=

RowBox[{(* New vector field is H *), , (* Parametrization f[x, y] already defined *),  ... integrand, ,, {x, 0, .5}, ,, RowBox[{{, RowBox[{y, ,, RowBox[{-, 1.25}], ,, 1.25}], }}]}], ]}]}]}]

Out[631]=

RowBox[{-, 137.5}]

So there is more than 10 times as much water hitting the windshield when we're driving at 30 mph!  Now do you understand why you have to turn your wipers up?

The same principle applies, to a lesser extent, to other modes of transport.  If you've ever biked during a light shower, you know that you get a lot more wet than if you were walking.  On a slightly different note, when people are walking and it starts to rain, often they start running instead of walking.  If you run, you'll be out in the rain for a shorter time, but because you're going faster, you'll be hit by more water during the time that you're still outside.  It would be an interesting project to determine if it's really worth running or not!

Exercises

Exercise 1

Let's re-do the above example, but instead of driving a boring, everyday car, let's imagine you are driving the state-of-the-art, gadget-laden vehicle of the Dark Knight of Gotham City, the Batmobile. We'll assume that this Batmobile is somewhat similar to the version used in the campy 60's television series and has a hemispherical windshield.

(i) What is a parametrization of this hemisphere? Assume the radius of the hemisphere is 1 and that the center is at the origin. You can think of this shape as a sphere, but we're only interested in the part from the "equator" to the "north pole".

(ii) While driving, rain will only be hitting the front half of the windshield. Describe a parametrization of this portion of the windshield. You should be able to just restrict the domain of your parameters a bit from your parametrization in (i).

(iii) Let's say Batman is inside a building, foiling a nefarious plot by the Joker. It's raining, and the Batmobile is sitting outside the building ready for Batman to take off and return to the Batcave. If the rain's velocity is described by the vector field R(x,y,z) = (0,0,-5) meters / second, what is the flux integral of this vector field over the Batmobile's windshield? Remember that the vehicle is sitting still and the rain is hitting the entire hemisphere, not just the front half.

(iv) Batman leaves the building, hops into the Batmobile, and takes off. If he's driving at 50 meters / second, what is the flux integral of the rain over the windshield? We'll assume that now the rain is only hitting the front half of the windshield (this isn't entirely accurate, but oh well), so use your parametrization from part (ii).
     You will have to be careful about how you change the vector field that describes the rain; it will depend on the choice you make for the axis the Batmobile is travelling along and the parametrization you chose for the front half of the windshield. More succinctly: make sure your rain doesn't hit the side or back of the windshield.

Exercise 2

Commissioner Gordon is in dire need of Batman's help, so he activates the Bat symbol and shines it on the clouds above Gotham City. If you aren't familiar with what the Bat symbol looks like...well, you should be. There's a picture at http://www.math.umn.edu/~drake/batman.gif (it's the yellow oval in the middle of the t-shirt).

Let's model the surface of the clouds with the function z = sin(x + y). Let's say the spotlight for the Bat symbol is shining from the top of City Hall in gritty downtown Gotham City upwards to the swirling clouds above the troubled metropolis. In more mathematical terms, the spolight shines from the negative z-axis upwards onto a disk in the xy plane centered at the origin, of radius 5. Since it's hard to parametrize the lit-up portion of the Bat symbol (the yellow part in the picture), let's simplify the problem and assume the lit-up portion of the Bat symbol is actually just a disk with a smaller disk cut out of it; this is called an annulus. You can parametrize this 2-dimensional object the exact same way you parametrize a disk, but start the radius at something greater than 0. For this problem, let's use the annulus centered at the origin whose radius ranges from 2 to 5.

Because of fog (it always seems to be foggy in Gotham City), the vector field that describes the intensity of the photons hitting the clouds is not constant, but is described by

L (x, y, z) = (0, 0, (x + y)^2)

(this is a rather poor model, but it makes the integrals reasonable).

(i) Describe the parametrization of the surface of the clouds we're interested in. The surface of the clouds is described by sin(x + y); the portion of the xy plane in which we're interested is the annulus from 2 to 5.

(ii) Find the surface integral of the above vector field over the annulus which we're using instead of the actual Bat symbol. Use an upward-pointing normal vector.

Exercise 3

Batman is attempting to sneak into the Penguin's secret underground lair. The Penguin, suspecting this, has installed a Batman Detection System (BDS) which emits a special form of Batman-detecting radiation that can be described by the vector field

B(x,y,z) = (-y,x, -5z/6),

where (0,0,0) is the center of the Penguin's lair. Our hero is familiar with the Penguin's criminal ways, and has brought with him a special Anti-Batman Detection System Device (ABDSD) that will deactivate the Batman Detection System for a short while. The device is shaped like a torus with r = 1, and R = 2; we're using the terminology from problem 30 on page 338 in your text.

(i) Batman has placed the ABDSD so that it is centered at (5,5,5) in the Penguin's lair and is horizontal (i.e., the vertical line described by x=5, y=5 does not intersect any part of the torus; it "goes through the middle of the doughnut hole"). Parametrize the surface of the ABDSD. You've essentially done this problem in your homework.

(Hint: first assume the ABDSD is centered at (0,0,0), then translate your answer to center it at (5,5,5).)

(ii) Calculate the flux of the Batman-detecting radiation across the ABDSD. Use an inward-pointing normal vector.

(iii)The answer from (ii) represents units of radiation crossing the ABDSD per second. Let's say the ABDSD somehow absorbs all this radiation (thus preventing the Penguin from detecting Batman's presence), and that it can store a total of 1000 units of radiation before it stops working. How much time does Batman have before the ABDSD fails?

(iv) (OPTIONAL) You may want to come back to this after we've learned about the Divergence Theorem. Redo part (ii), but this time say the ABDSD is centered at the origin. Compare the flux you get in both cases. What is unusual or unexpected about these two answers? With the Divergence Theorem, we'll be able to explain this quite easily.

Pseudo-Exercise 4

Which popular culture Batman series is better, the campy 60's television series (which featured animated "POW!" balloons, and the excellent quote with which we began this lab), or the 1990 Tim Burton movie, which featured Michael Keaton (at the time better known as Mr. Mom or Beetlejuice) and Jack Nicholson (who had come a loooong way from The Shining)?

Provide rigorous mathematical reasoning to defend your choice. :)

And keep in mind that the Burton movie prominently featured music from the most famous resident of the Suburb Currently Known As Chanhassen...

Credits


Created by Mathematica  (November 6, 2004)