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"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n\
questions to: drake@math.umn.edu or rogness@math.umn.edu"
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"\tBatman:\"There's an eclipse of the sun due.\"\n\tRobin:\"But that's only \
for half a minute!\"\n\tBatman:\"That's all we'll need, if my calculus is \
correct...\"\n\nBy 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 ",
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". What about surface integrals? That is, what about integrating a vector \
field over some parametrized ",
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" in three-dimensional space?\n\nFor 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 ",
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" the surface?\n\nLet'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 in your \
text.\n\nTo 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\".\n\nThe precise definition of a surface integral (or \
flux integral; the two terms are interchangeable) of a vector field is in the \
next section."
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"Just for your reference, we'll repeat the definition of a surface integral \
of a vector field.\n\nLet ",
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"be a continuous vector field, and let M be a smooth surface lying in U that \
is parametrized by ",
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"(Note that, depending on which way the normal vector is supposed to point, \
you might use ",
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", so the only difference in your final answer is a negative sign.)\n\nThis \
definition is very similar to the one for surface integrals of scalar \
functions, which you can find in your textbook. 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.\n\nThere are at least two ways to \
convince yourself that this definition makes sense. As we said above, ",
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" represents the scalar triple product of the partial derivatives and ",
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", which we know is equal to the volume of the parallelepiped spanned by \
those vectors (up to a possible minus sign).\n\nAnother way to see this is to \
rewrite the dot product: ",
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particles at all through the surface; in that case, ",
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" has as much effect as it can.\n\nIt'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)."
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": to type the character \[CapitalPhi] in ",
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", type the following keys in order: Esc, P, h, i, Esc."
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"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.\n\n\
Suppose our windshield is a rectangular piece of glass, represented by the \
plane ",
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" over the rectangle 0\[LessEqual]x\[LessEqual]0.5, -1.25\[LessEqual]y\
\[LessEqual]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. Because the parameters really are just ",
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" and ",
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", we won't bother using ",
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" and ",
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" here."
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"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 \
m/s. So the velocity of the raindrops is described by the vector field ",
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"(x,y,z)=(0,0,-10). ",
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" can plot this vector field for us, as long as we load the correct package \
first:"
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" can also show us a picture of the windshield in the rain."
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"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.)\n\nThe steps we need to \
take to compute this integral are very similar to the ones we used last week. \
We need to find ",
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"(\[CapitalPhi](x,y)), but since ",
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" is a constant vector field (it never changes), we know that ",
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"(",
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"(x,y)) = (0,0,-10) at all points. Let's run the following command anyway so \
that ",
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" knows about it, and so that you know how to use the command in the \
future:"
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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:\
\>", "Text"],
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"And now we're ready to integrate. The integrand is ",
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"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 ",
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" 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."
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Here's a question for you to think about: what are the correct units for our \
answer?\
\>", "Text",
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"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. Since we expressed the speed of the rain in meters per second, we \
should do the same with the car; you can check that 50km/h is about 13.89 \
m/s.\n\nIn our earlier example, the velocity of the rain was described by the \
vector field (0,0,-10). In this case, relative to the car, the velocity of \
the rain is described by the vector field, ",
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". Before going on, convince yourself that this is the correct formula! \
(Think about the negative signs in particular.)\n\nHere's a picture of the \
new situation. Again, you can choose whether you want an interactive picture \
or not:"
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How much rain is hitting the windshield this time? Let's compute the \
integral:\
\>", "Text"],
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Cell["\<\
So there is nearly four 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!\
\>", "Text"]
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"\n\nLet'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. (For a \
picture of the original campy batmobile, and more information than you'd ever \
need to know, go to http://www.javelinamx.com/Batmobile/ in your browser.)\n\n\
(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\". \n\n(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).\n\n(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.\n\n(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).\n\n \
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."
}], "Text",
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Cell[TextData[{
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"\n\nHere's another \"driving in the rain\" problem. Batman is chasing the \
Penguin during a rainstorm. Penguin's getaway car doesn't have any wipers, \
so it's vitally important to figure out how much rain is hitting his \
windshield. If it's too much, he won't be able to drive fast enough to \
escape. No super villian (or super hero) would ever be caught dead driving a \
normal automobile, so Penguin's car doesn't have a normal windshield. Like \
the batmobile, his car has a cockpit with a \"bubble\" windshield shaped like \
the top half of the ellipsoid ",
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".\n\n(i) What is a parametrization of the whole windshield? Assume that \
the center of the ellipsoid is at the origin. Remember, we're only \
interested in the top half.\n\n(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 \
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". This is the same as in the example above.\n\n(iii) Batman is within \
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Knowing that Batman is scrupulous to a fault, the Penguin has no qualms about \
stopping; he knows Batman will stop as well, instead of catching up to him. \
If the rain's velocity is described by the vector field R(x,y,z) = (0,0,-5) \
meters / second, how much rain is hitting the Penguin's windshield while he \
sits at the light? Remember he's sitting still, so the rain is hitting his \
entire windshield.\n\n(iv) Once the light turns green, the Penguin takes off. \
His car is rocket propelled, and he's quickly driving 40 meters per second. \
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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).\n\
\n 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."
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"\nCommissioner 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 ",
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" (it's the yellow oval in the middle of the t-shirt).\n\n 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 ",
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". 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.\n\nBecause 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 \n\n",
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"\n\n(this is a rather poor model, but it makes the integrals reasonable).\n\
\n(i) Describe the parametrization of the surface of the clouds we're \
interested in. The surface of the clouds is described by ",
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annulus which we're using instead of the actual Bat symbol. Use an \
upward-pointing normal vector."
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"\n\nBatman 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\n\nB(x,y,z) = (-y,x, -5z/6),\n\nwhere (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 = 2; we're using the \
terminology from page 471 (Exercise 4) in your text.\n\n(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. \n\n(Hint: first assume the ABDSD is \
centered at (0,0,0), then translate your answer to center it at (5,5,5).)\n\n\
(ii) Calculate the flux of the Batman-detecting radiation across the ABDSD. \
Use an inward-pointing normal vector.\n\n(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?\n\n(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."
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"Exercise 5 (This exercise is worth zero points.)\n\nWhich 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 ",
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")?\n\nProvide rigorous mathematical reasoning to defend your choice. :)\n\n\
And keep in mind that the Burton movie prominently featured music from the \
most famous resident of the Suburb Currently Known As Chanhassen..."
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"The quote at the beginning of this lab is from \
http://www.showcase.ca/action/series/batman_quotes.asp. It's near the bottom \
of the page. Much of the expository material in this lab is by Jonathan \
Rogness; the \"Driving in the Rain\" example is a standard one and no claim \
of originality is made here. (In fact, I spent some time searching on the \
web to determine who first wrote this example down without any luck; at least \
5-10 other math courses throughout the country have posted similar examples, \
though.) The rest of the lab (including the \"Batman-ization\") is by Dan \
Drake. Comments and questions are welcome to either of us.\n\nUpdate (2004): \
Exercise 2 added by Jonathan Rogness; Minor updates in 2008 for ",
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" 6.0.\n\nThis lab is copyright 2002, 2004, 2008by Jonathan Rogness \
(rogness@math.umn.edu) and Dan Drake (drake@math.umn.edu), and is protected \
by the Creative Commons Attribution-NonCommercial-ShareAlike License. You \
can find more information on this license at \
http://creativecommons.org/licenses/by-nc-sa/1.0/\n\nAlthough it's not \
specifically required by the license, I'd appreciate it if you let me know if \
you use parts of our labs, just so I can keep track of it. Please send me \
any questions or comments!"
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