(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 35540, 915] NotebookOptionsPosition[ 33307, 844] NotebookOutlinePosition[ 34067, 872] CellTagsIndexPosition[ 34024, 869] WindowFrame->Normal ContainsDynamic->False*) (* Beginning of Notebook Content *) Notebook[{ Cell[TextData[{ StyleBox["Lab 6B - Surface Integrals of Vector Fields", FontSize->24, FontWeight->"Bold", FontVariations->{"Underline"->True}], "\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n\ questions to: drake@math.umn.edu or rogness@math.umn.edu" }], "Text", CellFrame->True, TextAlignment->Center, FontColor->GrayLevel[1], Background->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Introduction", FontSize->14]], "Section", FontSize->12], Cell[TextData[{ "\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 ", StyleBox["circulation", FontSlant->"Italic"], ". What about surface integrals? That is, what about integrating a vector \ field over some parametrized ", StyleBox["surface", FontSlant->"Italic"], " 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 ", StyleBox["across", FontSlant->"Italic"], " 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." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Definition", FontSize->14]], "Section"], Cell[TextData[{ "Just for your reference, we'll repeat the definition of a surface integral \ of a vector field.\n\nLet ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox["F", FontWeight->"Bold"], ":", StyleBox[ RowBox[{"U", "\[Subset]", RowBox[{ SuperscriptBox["\[DoubleStruckCapitalR]", "3"], "\[LongRightArrow]", SuperscriptBox["\[DoubleStruckCapitalR]", "3"]}]}], FontSize->14]}], TraditionalForm]]], "be a continuous vector field, and let M be a smooth surface lying in U that \ is parametrized by ", StyleBox["\[CapitalPhi]", FontWeight->"Bold"], "(u,v), where ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{"u", ",", "v"}], ")"}], " ", "\[Subset]", " ", RowBox[{"R", " ", "\[Epsilon]", " ", SuperscriptBox["\[DoubleStruckCapitalR]", "2"]}]}], TraditionalForm]]], ". The ", StyleBox["surface integral", FontWeight->"Bold"], " (or ", StyleBox["flux integral", FontWeight->"Bold"], ")", StyleBox[" of F over M", FontWeight->"Bold"], " is:\n" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"\[Integral]", RowBox[{ SubscriptBox["\[Integral]", "M"], RowBox[{ StyleBox[ RowBox[{"F", " ", "\[CenterDot]", "n"}], FontWeight->"Bold"], RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}]}]}]}], "=", RowBox[{"\[Integral]", RowBox[{ SubscriptBox["\[Integral]", "R"], RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{"\[CapitalPhi]", RowBox[{"(", RowBox[{"u", ",", "v"}], ")"}]}], ")"}], "\[CenterDot]", RowBox[{"(", FormBox[ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", "\[CapitalPhi]"}], RowBox[{"\[PartialD]", "u"}]], "\[Times]", FractionBox[ RowBox[{"\[PartialD]", "\[CapitalPhi]"}], RowBox[{"\[PartialD]", "v"}]]}], TraditionalForm], ")"}]}], "du", " ", "dv"}]}]}]}]], "DisplayFormula",\ FontSize->14], Cell[TextData[{ "(Note that, depending on which way the normal vector is supposed to point, \ you might use ", Cell[BoxData[ RowBox[{"(", FormBox[ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", "\[CapitalPhi]"}], RowBox[{"\[PartialD]", "v"}]], "\[Times]", FractionBox[ RowBox[{"\[PartialD]", "\[CapitalPhi]"}], RowBox[{"\[PartialD]", "u"}]]}], TraditionalForm], ")"}]], FontSize->14], " instead. This is equal to ", Cell[BoxData[ RowBox[{"-", RowBox[{"(", StyleBox[ FormBox[ RowBox[{ FractionBox[ RowBox[{"\[PartialD]", "\[CapitalPhi]"}], RowBox[{"\[PartialD]", "u"}]], "\[Times]", FractionBox[ RowBox[{"\[PartialD]", "\[CapitalPhi]"}], RowBox[{"\[PartialD]", "v"}]]}], TraditionalForm], FontSize->14], ")"}]}]]], ", 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, ", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{"F", "\[CenterDot]", "n"}], FontWeight->"Bold"], TraditionalForm]]], " represents the scalar triple product of the partial derivatives and ", Cell[BoxData[ FormBox[ StyleBox["F", FontWeight->"Bold"], TraditionalForm]]], ", 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: ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ RowBox[{"F", "\[CenterDot]", "n"}], FontWeight->"Bold"], " ", "=", " ", RowBox[{"||", StyleBox["F", FontWeight->"Bold"], "||", " ", "||", StyleBox["n", FontWeight->"Bold"], "||", " ", RowBox[{"cos", " ", "\[Theta]"}]}]}], TraditionalForm]]], ". 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 ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"], " 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 through the surface; in that case, ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"cos", " ", "\[Theta]"}], " ", "=", " ", RowBox[{ RowBox[{"cos", " ", RowBox[{"(", RowBox[{"\[Pi]", "/", "2"}], ")"}]}], " ", "=", " ", "0"}]}], TraditionalForm]]], " (remember that the normal vector will be perpendicular to ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"], " in this case). Similarly, if ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"], " is moving perpendicular to the surface, ", Cell[BoxData[ FormBox[ RowBox[{"||", StyleBox["F", FontWeight->"Bold"], "||", "||", StyleBox["n", FontWeight->"Bold"], "||"}], TraditionalForm]]], " will be multiplied by ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"cos", " ", "0"}], " ", "=", " ", "1"}], TraditionalForm]]], ", so ", StyleBox["F", FontWeight->"Bold", FontSlant->"Italic"], " 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)." }], "Text"], Cell[TextData[{ StyleBox["Tip", FontWeight->"Bold"], ": to type the character \[CapitalPhi] in ", StyleBox["Mathematica", FontSlant->"Italic"], ", type the following keys in order: Esc, P, h, i, Esc." }], "Text", CellFrame->True, Background->GrayLevel[0.833326]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Driving in the Rain", FontSize->14]], "Section"], Cell[TextData[{ "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 ", Cell[BoxData[ FormBox[ RowBox[{"z", "=", RowBox[{"2", "-", RowBox[{"2", "x"}]}]}], TraditionalForm]]], " 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 ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], ", we won't bother using ", StyleBox["u", FontSlant->"Italic"], " and ", StyleBox["v", FontSlant->"Italic"], " here." }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x_", ",", "y_"}], "]"}], "=", RowBox[{"{", RowBox[{"x", ",", "y", ",", RowBox[{"2", "-", RowBox[{"2", "x"}]}]}], "}"}]}], ";"}], "\[IndentingNewLine]", RowBox[{"windshield", "=", RowBox[{"ParametricPlot3D", "[", RowBox[{ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}], ",", RowBox[{"{", RowBox[{"x", ",", "0", ",", "0.5"}], "}"}], ",", RowBox[{"{", RowBox[{"y", ",", RowBox[{"-", "1.25"}], ",", "1.25"}], "}"}]}], "]"}]}]}], "Input", CellChangeTimes->{{3.412355979089212*^9, 3.412355979512249*^9}}], Cell[TextData[{ "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 ", StyleBox["G", FontWeight->"Bold"], "(x,y,z)=(0,0,-10). ", StyleBox["Mathematica", FontSlant->"Italic"], " can plot this vector field for us, as long as we load the correct package \ first:" }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{"G", "[", RowBox[{"x_", ",", "y_", ",", "z_"}], "]"}], " ", "=", " ", RowBox[{"{", RowBox[{"0", ",", "0", ",", RowBox[{"-", "10"}]}], "}"}]}], "\[IndentingNewLine]", RowBox[{"rain", " ", "=", " ", RowBox[{"VectorFieldPlot3D", "[", RowBox[{ RowBox[{"G", "[", RowBox[{"x", ",", "y", ",", "z"}], "]"}], ",", " ", RowBox[{"{", RowBox[{"x", ",", RowBox[{"-", ".5"}], ",", " ", "1"}], "}"}], ",", " ", RowBox[{"{", RowBox[{"y", ",", RowBox[{"-", "2"}], ",", "2"}], "}"}], ",", RowBox[{"{", RowBox[{"z", ",", ".5", ",", "2.5"}], "}"}]}], "]"}]}]}], "Input", CellChangeTimes->{{3.412355991526728*^9, 3.412355994399536*^9}}], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can also show us a picture of the windshield in the rain." }], "Text", CellChangeTimes->{{3.412356094315576*^9, 3.412356098502478*^9}}], Cell[BoxData[ RowBox[{"Show", "[", RowBox[{"windshield", ",", "rain", ",", RowBox[{"PlotRange", "\[Rule]", RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"-", ".5"}], ",", "1"}], "}"}], ",", RowBox[{"{", RowBox[{ RowBox[{"-", "2"}], ",", "2"}], "}"}], ",", RowBox[{"{", RowBox[{"1", ",", "2.5"}], "}"}]}], "}"}]}]}], "]"}]], "Input", CellChangeTimes->{{3.412356036866927*^9, 3.412356077810627*^9}}], Cell[TextData[{ "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 ", StyleBox["G", FontWeight->"Bold"], "(\[CapitalPhi](x,y)), but since ", StyleBox["G", FontWeight->"Bold"], " is a constant vector field (it never changes), we know that ", StyleBox["G", FontWeight->"Bold"], "(", "\[CapitalPhi]", "(x,y)) = (0,0,-10) at all points. Let's run the following command anyway so \ that ", StyleBox["Mathematica", FontSlant->"Italic"], " knows about it, and so that you know how to use the command in the \ future:" }], "Text"], Cell[BoxData[ RowBox[{"Apply", "[", RowBox[{"G", ",", RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}]}], "]"}]], "Input"], Cell["\<\ 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"], Cell[BoxData[ RowBox[{"n", "=", RowBox[{"Cross", "[", RowBox[{ RowBox[{"D", "[", RowBox[{ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}], ",", "x"}], "]"}], ",", RowBox[{"D", "[", RowBox[{ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}], ",", "y"}], "]"}]}], "]"}]}]], "Input"], Cell[TextData[{ "And now we're ready to integrate. The integrand is ", StyleBox["G", FontWeight->"Bold"], "(", "\[CapitalPhi]", "(x,y))\[CenterDot]", Cell[BoxData[ FormBox[ OverscriptBox["n", "\[RightVector]"], TraditionalForm]]], "." }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{"integrand", "=", RowBox[{ RowBox[{"Apply", "[", RowBox[{"G", ",", RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}]}], "]"}], ".", "n"}]}], ";"}], "\[IndentingNewLine]", RowBox[{"Integrate", "[", RowBox[{"integrand", ",", RowBox[{"{", RowBox[{"x", ",", "0", ",", ".5"}], "}"}], ",", RowBox[{"{", RowBox[{"y", ",", RowBox[{"-", "1.25"}], ",", "1.25"}], "}"}]}], "]"}]}], "Input"], Cell[TextData[{ "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 ", Cell[BoxData[ FormBox[ OverscriptBox["n", "\[RightVector]"], TraditionalForm]]], " 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." }], "Text"], Cell["\<\ Here's a question for you to think about: what are the correct units for our \ answer?\ \>", "Text", CellFrame->True, Background->GrayLevel[0.849989]], Cell[TextData[{ "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, ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"H", "(", RowBox[{"x", ",", "y", ",", "z"}], ")"}], " ", "=", " ", RowBox[{"(", RowBox[{ RowBox[{"-", "13.89"}], ",", "0", ",", RowBox[{"-", "10"}]}], ")"}]}], TraditionalForm]]], ". 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:" }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{ RowBox[{"H", "[", RowBox[{"x_", ",", "y_", ",", "z_"}], "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"-", "13.89"}], ",", "0", ",", RowBox[{"-", "10"}]}], "}"}]}], ";"}], "\[IndentingNewLine]", RowBox[{"newrain", "=", RowBox[{"VectorFieldPlot3D", "[", RowBox[{ RowBox[{"H", "[", RowBox[{"x", ",", "y", ",", "z"}], "]"}], ",", RowBox[{"{", RowBox[{"x", ",", RowBox[{"-", "1"}], ",", "1"}], "}"}], ",", RowBox[{"{", RowBox[{"y", ",", RowBox[{"-", "2"}], ",", "2"}], "}"}], ",", RowBox[{"{", RowBox[{"z", ",", "0", ",", "3"}], "}"}]}], "]"}]}], "\[IndentingNewLine]", RowBox[{"Show", "[", RowBox[{"windshield", ",", "newrain", ",", RowBox[{"PlotRange", "\[Rule]", RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"-", ".5"}], ",", "1"}], "}"}], ",", RowBox[{"{", RowBox[{ RowBox[{"-", "2"}], ",", "2"}], "}"}], ",", RowBox[{"{", RowBox[{"1", ",", "2.5"}], "}"}]}], "}"}]}]}], "]"}]}], "Input", CellChangeTimes->{{3.41235614920285*^9, 3.412356176877342*^9}}], Cell["\<\ How much rain is hitting the windshield this time? Let's compute the \ integral:\ \>", "Text"], Cell[BoxData[ RowBox[{ RowBox[{"(*", " ", RowBox[{"New", " ", "vector", " ", "field", " ", "is", " ", "H"}], " ", "*)"}], "\[IndentingNewLine]", RowBox[{"(*", " ", RowBox[{"Parametrization", " ", RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}], " ", "already", " ", "defined"}], " ", "*)"}], "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"integrand", "=", RowBox[{ RowBox[{"Apply", "[", RowBox[{"H", ",", RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}]}], "]"}], ".", RowBox[{"Cross", "[", RowBox[{ RowBox[{"D", "[", RowBox[{ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}], ",", "x"}], "]"}], ",", " ", RowBox[{"D", "[", RowBox[{ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}], ",", "y"}], "]"}]}], "]"}]}]}], ";"}], "\[IndentingNewLine]", RowBox[{"Integrate", "[", RowBox[{"integrand", ",", RowBox[{"{", RowBox[{"x", ",", "0", ",", ".5"}], "}"}], ",", RowBox[{"{", RowBox[{"y", ",", RowBox[{"-", "1.25"}], ",", "1.25"}], "}"}]}], "]"}]}]}]], "Input"], 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"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Exercises", FontSize->14]], "Section"], Cell[TextData[{ StyleBox["Exercise 1", FontWeight->"Bold"], "\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", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]], Cell[TextData[{ StyleBox["Exercise 2", FontWeight->"Bold"], "\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 ", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{ RowBox[{ FractionBox[ SuperscriptBox["x", "2"], "9"], "+", FractionBox[ SuperscriptBox["y", "2"], "4"], "+", SuperscriptBox["z", "2"]}], "=", "1"}], FontSize->16], TraditionalForm]]], ".\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 \ parameters a bit from your parametrization in (i). (We'll define \"front \ half\" as being the portion of the windshields where ", Cell[BoxData[ FormBox[ RowBox[{"x", ">", "0"}], TraditionalForm]]], ". This is the same as in the example above.\n\n(iii) Batman is within \ sight of the Penguin when they both approach intersections with red lights. \ 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. \ 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", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]], Cell[TextData[{ StyleBox["Exercise 3\n", FontWeight->"Bold"], "\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 ", ButtonBox["http://www.math.umn.edu/~drake/batman.gif", BaseStyle->"Hyperlink", ButtonData:>{ URL["http://www.math.umn.edu/~drake/batman.gif"], None}], " (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 ", StyleBox["annulus", FontSlant->"Italic"], ". 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", Cell[BoxData[ RowBox[{ RowBox[{"L", RowBox[{"(", RowBox[{"x", ",", "y", ",", "z"}], ")"}]}], " ", "=", " ", RowBox[{"(", RowBox[{"0", ",", "0", ",", SuperscriptBox[ RowBox[{"(", RowBox[{"x", "+", "y"}], ")"}], "2"]}], ")"}]}]]], "\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 ", Cell[BoxData[ FormBox[ RowBox[{"sin", "(", RowBox[{"x", "+", "y"}], ")"}], TraditionalForm]]], "; the portion of the xy plane in which we're interested is the annulus from \ 2 to 5.\n\n(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." }], "Text", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]], Cell[TextData[{ StyleBox["Exercise 4", FontWeight->"Bold"], "\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." }], "Text", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]], Cell[TextData[{ "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 ", StyleBox["The Shining", FontSlant->"Italic"], ")?\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..." }], "Text", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]], Cell[CellGroupData[{ Cell["Credits", "Subsection"], Cell[TextData[{ "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 ", StyleBox["Mathematica", FontSlant->"Italic"], " 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!" }], "Text", CellChangeTimes->{{3.412356231265384*^9, 3.412356238904645*^9}}] }, Closed]] }, Closed]] }, ScreenStyleEnvironment->"Working", PrintingStyleEnvironment->"Printout", WindowSize->{775, 789}, WindowMargins->{{Automatic, 279}, {Automatic, 53}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"Magnification"->1, "PaperOrientation"->"Portrait", "PaperSize"->{612, 792}, "PostScriptOutputFile":>"", "PrintCellBrackets"->False, "PrintMultipleHorizontalPages"->False, "PrintRegistrationMarks"->True, "PrintingMargins"->{{54, 54}, {72, 72}}}, CellLabelAutoDelete->True, Magnification->1, FrontEndVersion->"6.0 for Linux x86 (32-bit) (April 20, 2007)", StyleDefinitions->"Default.nb" ] (* End of Notebook Content *) (* Internal cache information *) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[568, 21, 397, 11, 113, "Text"], Cell[CellGroupData[{ Cell[990, 36, 81, 2, 65, "Section"], Cell[1074, 40, 1844, 31, 443, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[2955, 76, 64, 1, 35, "Section"], Cell[3022, 79, 1059, 36, 101, "Text"], Cell[4084, 117, 988, 34, 43, "DisplayFormula"], Cell[5075, 153, 4050, 111, 438, "Text"], Cell[9128, 266, 272, 9, 45, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[9437, 280, 73, 1, 35, "Section"], Cell[9513, 283, 1213, 30, 173, "Text"], Cell[10729, 315, 666, 19, 52, "Input"], Cell[11398, 336, 451, 11, 65, "Text"], Cell[11852, 349, 733, 20, 52, "Input"], Cell[12588, 371, 204, 5, 29, "Text"], Cell[12795, 378, 483, 14, 31, "Input"], Cell[13281, 394, 981, 24, 155, "Text"], Cell[14265, 420, 147, 4, 31, "Input"], Cell[14415, 426, 275, 5, 47, "Text"], Cell[14693, 433, 361, 11, 31, "Input"], Cell[15057, 446, 261, 11, 31, "Text"], Cell[15321, 459, 486, 15, 52, "Input"], Cell[15810, 476, 677, 13, 103, "Text"], Cell[16490, 491, 162, 5, 45, "Text"], Cell[16655, 498, 1040, 21, 173, "Text"], Cell[17698, 521, 1169, 36, 72, "Input"], Cell[18870, 559, 105, 3, 29, "Text"], Cell[18978, 564, 1257, 35, 112, "Input"], Cell[20238, 601, 757, 13, 155, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[21032, 619, 63, 1, 35, "Section"], Cell[21098, 622, 2396, 35, 513, "Text"], Cell[23497, 659, 2869, 51, 540, "Text"], Cell[26369, 712, 2608, 52, 511, "Text"], Cell[28980, 766, 2048, 30, 531, "Text"], Cell[31031, 798, 776, 14, 225, "Text"], Cell[CellGroupData[{ Cell[31832, 816, 29, 0, 38, "Subsection"], Cell[31864, 818, 1415, 22, 281, "Text"] }, Closed]] }, Closed]] } ] *) (* End of internal cache information *)