(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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Cell["\<\ In last week's lab, we recalled the six different kinds of \ integrals we've studied this semester -- here they are again, for \ reference.\ \>", "Text"], Cell[BoxData[GridBox[{ { StyleBox["Type", FontVariations->{"Underline"->True}], StyleBox["Symbol", FontVariations->{"Underline"->True}], StyleBox["Region", FontVariations->{"Underline"->True}], StyleBox["Integrand", FontVariations->{"Underline"->True}], StyleBox[\(How\ to\ Do\ It\), FontVariations->{"Underline"->True}]}, {"double", \(\[Integral]\(\(\[Integral]\_R\%\ \)\ u\ \[DifferentialD]A\)\), \(R, \ a\ region\), \(scalar\ function\ u\), \(\[Integral]\((\[Integral]u \ \((x, y)\) \[DifferentialD]y)\) \[DifferentialD]x\)}, {"triple", \(\[Integral]\(\[Integral]\(\(\[Integral]\_S\%\ \)u\ \ \[DifferentialD]V\)\)\), \(S, \ a\ solid\), \(scalar\ function\ u\), \(\[Integral]\((\[Integral]\((\ \[Integral]u \((x, y, z)\) \[DifferentialD]z)\) \[DifferentialD]y)\) \ \[DifferentialD]x\)}, {\(line\ \((path)\)\), \(\(\[Integral]\_C\%\ \)u\ \ \[DifferentialD]s\ \ \ or\ \ \ \(\[ContourIntegral]\_C u\ \[DifferentialD]s\)\), RowBox[{\(C, \ a\ curve\ with\), "\[IndentingNewLine]", "parametrization", "\[IndentingNewLine]", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], StyleBox[\((t)\), FontWeight->"Plain"]}], StyleBox[",", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], RowBox[{ StyleBox["a", FontWeight->"Plain"], "\[LessEqual]", "t", "\[LessEqual]", "b"}]}]}], \(scalar\ function\ u\), RowBox[{ RowBox[{\(\[Integral]\_a\%b\), RowBox[{"u", RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], \((t)\)}], ")"}]}]}], "\[DoubleVerticalBar]", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}], \((t)\)}], "\[DoubleVerticalBar]", " ", \(\[DifferentialD]t\)}]}, {\(line\ \((path)\)\), RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}], StyleBox[" ", FontWeight->"Bold"], "or", " ", RowBox[{\(\[ContourIntegral]\_C\), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["x", FontWeight->"Bold"]}]}]}]}]}], RowBox[{\(C, \ a\ curve\ with\), "\[IndentingNewLine]", "parametrization", "\[IndentingNewLine]", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], StyleBox[\((t)\), FontWeight->"Plain"]}], StyleBox[",", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], RowBox[{ StyleBox["a", FontWeight->"Plain"], "\[LessEqual]", "t", "\[LessEqual]", "b"}]}]}], RowBox[{"vector", " ", "field", " ", StyleBox["F", FontWeight->"Bold"]}], RowBox[{\(\[Integral]\_a\%b\), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], \((t)\)}], ")"}], "\[CenterDot]", RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}]}], \((t)\), \(\[DifferentialD]t\)}]}]}, {"surface", \(\[Integral]\(\(\[Integral]\_M\%\ \)u \ \[DifferentialD]S\ \ \ or\ \ \ \(\[DoubleContourIntegral]\_M u \[DifferentialD]S\)\)\), RowBox[{\(M, a\ surface\ with\), "\[IndentingNewLine]", RowBox[{"parametrization", " ", StyleBox["f", FontWeight->"Bold"], StyleBox[\((s, t)\), FontWeight->"Plain"]}], "\[IndentingNewLine]", RowBox[{ StyleBox["and", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["parameter", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["range", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["R", FontWeight->"Plain"], "\[IndentingNewLine]"}]}], \(scalar\ function\ u\), RowBox[{ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_R\%\ \), RowBox[{"u", RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], RowBox[{ StyleBox["(", FontWeight->"Plain"], RowBox[{ StyleBox["s", FontWeight->"Plain"], ",", "t"}], ")"}]}], ")"}]}]}]}], "\[DoubleVerticalBar]", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontWeight->"Bold"]}], \(\[PartialD]s\)], "\[Times]", FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontWeight->"Bold"]}], \(\[PartialD]t\)]}], "\[DoubleVerticalBar]", \(\[DifferentialD]A\)}]}, {\(surface\ \((flux)\)\), RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_M\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}], StyleBox[ RowBox[{ StyleBox[" ", FontWeight->"Bold"], " "}]], "or", " ", RowBox[{\(\[DoubleContourIntegral]\_M\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}]}]}]}]}]}], RowBox[{\(M, a\ surface\ with\), "\[IndentingNewLine]", RowBox[{"parametrization", " ", StyleBox["f", FontWeight->"Bold"], StyleBox[\((s, t)\), FontWeight->"Plain"]}], "\[IndentingNewLine]", RowBox[{ StyleBox["and", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["parameter", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["range", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Plain"], StyleBox["R", FontWeight->"Plain"], "\[IndentingNewLine]"}]}], RowBox[{"vector", " ", "field", " ", StyleBox["F", FontWeight->"Bold"]}], RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_R\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], \((s, t)\)}], ")"}], "\[CenterDot]", RowBox[{"(", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontWeight->"Bold"]}], \(\[PartialD]s\)], "\[Times]", FractionBox[ RowBox[{"\[PartialD]", StyleBox["f", FontWeight->"Bold"]}], \(\[PartialD]t\)]}], ")"}]}], \(\[DifferentialD]A\)}]}]}]} }]], "Text", CellFrame->True, TextAlignment->Left, FontSize->13, Background->GrayLevel[0.849989]], Cell["\<\ We saw last week that Stokes's Theorem links line integrals of \ vector fields to surface integrals of vector fields via the curl. This week, \ we will work with the Divergence Theorem, which links surface integrals of \ vector fields to triple integrals via the divergence.\ \>", "Text"], Cell[TextData[{ StyleBox["Divergence Theorem", FontWeight->"Bold"], ": Suppose S is a solid with smooth boundary \[PartialD]S, and let ", StyleBox["F", FontWeight->"Bold"], " be a vector field which is smooth on and \"near\" S. Then ", Cell[BoxData[ RowBox[{ RowBox[{\(\[DoubleContourIntegral]\_\(\[PartialD]S\)\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], " ", \(\[DifferentialD]\[Sigma]\)}]}], "=", RowBox[{"\[Integral]", RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_S\), RowBox[{"div", " ", StyleBox["F", FontWeight->"Bold"], \(\[DifferentialD]V\)}]}]}]}]}]], "Text"], "." }], "Text", CellFrame->True, FontWeight->"Plain", Background->GrayLevel[0.849989]] }, Closed]], Cell[CellGroupData[{ Cell["Using the Divergence Theorem: Example 1", "Section", Background->None], Cell["\<\ Triple integrals of scalar functions are almost always easier to \ compute than surface integrals of vector fields. Therefore, we usually apply \ the Divergence Theorem to replace a surface integral with a triple integral. \ Of course, this only works if our surface is the boundary of some \ solid!\ \>", "Text"], Cell[CellGroupData[{ Cell["Example 1", "Subsection"], Cell[TextData[{ "Let ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], "(", \(x, y, z\), ")"}], "=", \((sinx\ \(cos\^2\) y, \ \(sin\^3\) y\ \(cos\^4\) z, \ \(sin\^5\) z\ \(cos\^6\) x)\)}], TraditionalForm]]], ", and let M be the surface of the cube in the first octant bounded by the \ coordinate planes and the planes ", Cell[BoxData[ RowBox[{ StyleBox["x", FontSlant->"Italic"], "=", \(\[Pi]\/2\)}]], "Text"], ", ", Cell[BoxData[ RowBox[{ StyleBox["y", FontSlant->"Italic"], "=", \(\[Pi]\/2\)}]]], ", and ", Cell[BoxData[ RowBox[{ StyleBox["z", FontSlant->"Italic"], "=", \(\[Pi]\/2\)}]]], ". Using the Divergence Theorem, calculate the outward flux of ", StyleBox["F", FontWeight->"Bold"], " across M." }], "Text"], Cell[TextData[{ "We need to calculate div ", StyleBox["F", FontWeight->"Bold"], "; we can do this by hand, or use ", StyleBox["Mathematica", FontSlant->"Italic"], ". Let's use ", StyleBox["Mathematica", FontSlant->"Italic"], ":" }], "Text"], Cell[BoxData[ \(F[x_, y_, z_] = {\((Sin[ x])\)*\((Cos[y]^2)\), \((Sin[y]^3)\)*\((Cos[z]^4)\), \((Sin[ z]^5)\)*\((Cos[x]^6)\)}\)], "Input"], Cell[TextData[{ "Remember: ", StyleBox["Mathematica", FontSlant->"Italic"], " will not understand our shorthand notation ", StyleBox["Sin^5[x]", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["!", FontFamily->"Courier"] }], "Text", CellFrame->True, Background->GrayLevel[0.849989]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can compute the divergence of a vector field using the command ", StyleBox["Div", FontWeight->"Bold"], ":" }], "Text"], Cell[BoxData[ \(d[x_, y_, z_] = Div[F[x, y, z]]\)], "Input"], Cell["\<\ Now, by the Divergence Theorem, we can just take the triple \ integral of our result.\ \>", "Text"], Cell[BoxData[ \(Integrate[ d[x, y, z], {x, 0, Pi/2}, {y, 0, Pi/2}, {z, 0, Pi/2}]\)], "Input"], Cell["\<\ If we didn't know the Divergence Theorem, we'd have to parametrize \ all six faces individually, check the normal vectors to be sure they pointed \ outward, evaluate the six integrals, and sum the values. If you'd like, you \ can do this to check our result.\ \>", "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Using the Divergence Theorem: Example 2", "Section", Background->None], Cell["\<\ Sometimes, we can use the Divergence Theorem to simplify a surface \ integral, even when the surface isn't the boundary of any solid. The idea is \ to invent another, simpler surface which completes the boundary of a solid. \ Then the Divergence Theorem will give the surface integral over the two \ surfaces together; we can calculate the simpler integral and subtract it from \ the total to get our solution. This is tricky, but it's an important idea. Re-read that last paragraph \ until it makes sense, or ask your TA to help you figure it out.\ \>", "Text"], Cell[CellGroupData[{ Cell["Example 2", "Subsection"], Cell[TextData[{ "Let ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], "(", \(x, y, z\), ")"}], "=", \((y\^2 - 2 y\ z, \ z\^2 - 2 x\ z, \ x\^2 - 2 x\ y)\)}], TraditionalForm]]], " and let M be the surface parametrized by ", StyleBox["f", FontWeight->"Bold"], "(r,\[Theta])=(r cos(\[Theta]), r sin(\[Theta]), cos(5\[Pi](1-r))-1, 0\ \[LessEqual]r\[LessEqual]1, 0\[LessEqual]\[Theta]\[LessEqual]2\[Pi]. Find \ the flux of ", StyleBox["F", FontWeight->"Bold"], " through M, using the upward normal vector." }], "Text"], Cell["This surface is terrible! Here's a graph:", "Text"], Cell[BoxData[{ \(\(f[r_, theta_] = {r*Cos[theta], r*Sin[theta], Cos[Sin[5 Pi*\((1 - r)\)]] - 1};\)\), "\[IndentingNewLine]", \(M = ParametricPlot3D[f[r, theta], {r, 0, 1}, {theta, 0, 2 Pi}, \ PlotPoints \[Rule] 40]\), "\[IndentingNewLine]", \(ShowLive[M]\)}], "Input"], Cell[TextData[{ "Of course, M is not the boundary of a solid. However, the boundary of M \ is the unit circle in the ", StyleBox["xy-", FontSlant->"Italic"], "plane (check this!). Therefore, M and the unit disk (call it \ \[CapitalDelta]) ", StyleBox["do", FontSlant->"Italic"], " form the boundary of a solid (call it S): try looking at the following \ graph from a few different points of view." }], "Text"], Cell[BoxData[{ \(\(g[r_, theta_] = {r*Cos[theta], r*Sin[theta], 0};\)\), "\[IndentingNewLine]", \(\[CapitalDelta] = ParametricPlot3D[ g[r, theta], {r, 0, 1}, {theta, 0, 2 Pi}]\), "\[IndentingNewLine]", \(S = ShowLive[M, \[CapitalDelta]]\)}], "Input"], Cell["\<\ Roughly speaking, in English here's what we're going to do in order \ to calculate the surface integral of our original surface:\ \>", "Text"], Cell[TextData[{ StyleBox["1. (Surface Integral over our original surface and the disk) = \ (triple integral of div ", FontFamily->"Helvetica"], StyleBox["F", FontFamily->"Helvetica", FontWeight->"Bold"], StyleBox[" over the solid region enclosed by the two surfaces)", FontFamily->"Helvetica"], "\t", StyleBox["[By the Div Them]", FontSlant->"Italic"], "\n\n2. ", StyleBox["(Surface Integral over original surface) + (Surface integral over \ disk) = (triple integral over the solid region)", FontFamily->"Helvetica"], "\t", StyleBox["[Computing the two surface integrals separately]", FontSlant->"Italic"], "\n\n3. ", StyleBox["(Surface Integral over original surface) = (triple integral over \ solid region) - (Surface Integral over disk)", FontFamily->"Helvetica"], "\t\t", StyleBox["[Solving for our original integral]", FontSlant->"Italic"] }], "Text", Background->RGBColor[1, 1, 0.733333]], Cell[TextData[{ "Remember, in the Divergence Theorem we use the ", StyleBox["outward", FontSlant->"Italic"], " pointing normal vector. In the problem, we're asked to find the integral \ over M with the upward pointing normal vector, but when we combine it with \ the unit disk, the outward pointing normal is the ", StyleBox["downward", FontSlant->"Italic"], " pointing normal vector. That means we have to replace M with -M in our \ calculations. (As you should recall from lecture, the only real difference \ is that it will multiply our answer by -1.)\n\nLet's rewrite the three \ equations above mathematically." }], "Text"], Cell[TextData[{ StyleBox["1. ", FontFamily->"Helvetica"], Cell[BoxData[ RowBox[{ StyleBox[ RowBox[{\(\[DoubleContourIntegral]\_\(\[CapitalDelta] + \ \((\(-M\))\)\)\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}], FontSize->14], StyleBox["=", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox[\(\[Integral]\_S\), FontSize->14], RowBox[{ StyleBox["div", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["F", FontSize->14, FontWeight->"Bold"], StyleBox[" ", FontSize->14], StyleBox[\(\[DifferentialD]V\), FontSize->14]}]}]}]}]}]], "Text"], "\t\t\t", StyleBox["[By the Div Them; -M because we've switched the normal]", FontSlant->"Italic"], "\n\n2. ", StyleBox[" ", FontFamily->"Helvetica"], Cell[BoxData[ RowBox[{ StyleBox[ RowBox[{ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\[CapitalDelta]\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight-> "Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}], "+", RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\(-M\)\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight-> "Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}]}], FontSize->14], StyleBox["=", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox[\(\[Integral]\_S\), FontSize->14], RowBox[{ StyleBox["div", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["F", FontSize->14, FontWeight->"Bold"], StyleBox[" ", FontSize->14], StyleBox[\(\[DifferentialD]V\), FontSize->14]}]}]}]}]}]], "Text"], "\t", StyleBox["[Computing the two surface integrals separately]", FontSlant->"Italic"], "\n\n3. ", Cell[BoxData[ RowBox[{ StyleBox[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\(-M\)\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}], FontSize->14], StyleBox["=", FontSize->14], RowBox[{ RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox[\(\[Integral]\_S\), FontSize->14], RowBox[{ StyleBox["div", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["F", FontSize->14, FontWeight->"Bold"], StyleBox[" ", FontSize->14], StyleBox[\(\[DifferentialD]V\), FontSize->14]}]}]}]}], StyleBox["-", FontSize->14], RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\[CapitalDelta]\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight-> "Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}]}]}]], "Text"], "\t\t", StyleBox["[Solving for our original integral; here I've multiplied by -1 to \ get the correct sign]", FontSlant->"Italic"] }], "Text", Background->RGBColor[1, 1, 0.733333]], Cell["\<\ We're interested in the integral over M, not -M, so let's fix that:\ \ \>", "Text"], Cell[TextData[{ "4(a). -", Cell[BoxData[ RowBox[{ StyleBox[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_M\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}], FontSize->14], StyleBox["=", FontSize->14], RowBox[{ RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox[\(\[Integral]\_S\), FontSize->14], RowBox[{ StyleBox["div", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["F", FontSize->14, FontWeight->"Bold"], StyleBox[" ", FontSize->14], StyleBox[\(\[DifferentialD]V\), FontSize->14]}]}]}]}], StyleBox["-", FontSize->14], RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\[CapitalDelta]\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight-> "Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}]}]}]], "Text"], "\t\t", StyleBox["[integral over -M = - (integral over M)]\n\n", FontSlant->"Italic"], "4(b). ", Cell[BoxData[ RowBox[{ StyleBox[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_M\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}], FontSize->14], StyleBox["=", FontSize->14], RowBox[{ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\[CapitalDelta]\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}], StyleBox["-", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox["\[Integral]", FontSize->14], RowBox[{ StyleBox[\(\[Integral]\_S\), FontSize->14], RowBox[{ StyleBox["div", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["F", FontSize->14, FontWeight->"Bold"], StyleBox[" ", FontSize->14], StyleBox[\(\[DifferentialD]V\), FontSize->14]}]}]}]}]}]}]], "Text"], "\t\t\t", StyleBox["[Multiplying by -1]", FontSlant->"Italic"] }], "Text", Background->RGBColor[1, 1, 0.733333]], Cell[TextData[{ "Now there's another step which makes this particular problem much easier. \ It turns out that ", StyleBox["F", FontWeight->"Bold"], " is a ", StyleBox["very", FontSlant->"Italic"], " nice vector field: evaluate the cell below." }], "Text"], Cell[BoxData[{ \(F[x_, y_, z_] = {y\^2 - 2 y\ z, \ z\^2 - 2 x\ z, x\^2 - 2 x\ y}\), "\[IndentingNewLine]", \(Div[F[x, y, z]]\)}], "Input"], Cell[TextData[{ "Hence the triple intregral of div ", StyleBox["F", FontWeight->"Bold"], " is 0, and w", "e can conclude that ", Cell[BoxData[ RowBox[{ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\[CapitalDelta]\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}], "-", RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_M\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight-> "Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}]}]]], " = 0, so the integrals are equal! This means that we can work with the \ unit disk in place of our terrible surface M.\n", "\n", "Let's finish off the calculation on \[CapitalDelta]: ", Cell[BoxData[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_\[CapitalDelta]\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", StyleBox["n", FontWeight->"Bold"]}], \(\[DifferentialD]\[Sigma]\)}]}]}]]], "=", Cell[BoxData[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_R\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["g", FontWeight->"Bold"], \((r, \[Theta])\)}], ")"}], "\[CenterDot]", RowBox[{"(", RowBox[{ FractionBox[ RowBox[{"\[PartialD]", StyleBox["g", FontWeight->"Bold"]}], \(\[PartialD]r\)], "\[Times]", FractionBox[ RowBox[{"\[PartialD]", StyleBox["g", FontWeight->"Bold"]}], \(\[PartialD]\[Theta]\)]}], ")"}]}], \(\[DifferentialD]A\)}]}]}]], "Text"], ", so first, we substitute ", StyleBox["g", FontWeight->"Bold"], " into ", StyleBox["F", FontWeight->"Bold"], ":" }], "Text"], Cell[BoxData[ \(v1[r_, theta_] = Apply[F, g[r, theta]]\)], "Input"], Cell[TextData[{ "Next, we cross the partial derivatives of ", StyleBox["g", FontWeight->"Bold"], "." }], "Text"], Cell[BoxData[{ \(dgdr[r_, theta_] = D[g[r, theta], r]\), "\[IndentingNewLine]", \(dgdtheta[r_, theta_] = D[g[r, theta], theta]\), "\[IndentingNewLine]", \(Cross[dgdr[r, theta], dgdtheta[r, theta]]\), "\[IndentingNewLine]", \(v2[r_, theta_] = Simplify[%]\)}], "Input"], Cell["Lastly, we dot our results together and integrate.", "Text"], Cell[BoxData[{ \(v1[r, theta] . v2[r, theta]\), "\[IndentingNewLine]", \(Integrate[%, {r, 0, 1}, {theta, 0, 2 Pi}]\)}], "Input"], Cell[TextData[{ "So the flux of ", StyleBox["F", FontWeight->"Bold"], " through \[CapitalDelta] is ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], ". Notice that this is the flux ", StyleBox["up", FontSlant->"Italic"], " through \[CapitalDelta]: the normal vector above is (0,0,", StyleBox["r", FontSlant->"Italic"], "), and ", StyleBox["r", FontSlant->"Italic"], " is positive. By the Divergence Theorem, the flux of ", StyleBox["F", FontWeight->"Bold"], " ", StyleBox["down", FontSlant->"Italic"], " through M is ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\)\/4\)]], "; the flux of ", StyleBox["F", FontWeight->"Bold"], " ", StyleBox["up", FontSlant->"Italic"], " through M is ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], ". Remember, when we apply the Divergence Theorem, our combined surface \ integral has to use the outward normal vector; this is the downward normal to \ M and the upward normal to \[CapitalDelta]." }], "Text", FontSize->14] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Exercises", "Section", Background->None], Cell[TextData[{ StyleBox["Exercise 1", FontWeight->"Bold"], "\n\nLet ", StyleBox["r", FontWeight->"Bold"], "(x,y,z)=(x,y,z), and ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=\[DoubleVerticalBar] ", StyleBox["r", FontWeight->"Bold"], "\[DoubleVerticalBar] ", StyleBox["r", FontWeight->"Bold"], ", where the double bars stand for magnitude, and let M be the surface of \ the solid bounded by the paraboloid ", Cell[BoxData[ \(TraditionalForm\`z = 25 - x\^2 - y\^2\)]], " and the ", StyleBox["xy", FontSlant->"Italic"], "-plane. Use the Divergence Theorem to compute the outward flux of ", StyleBox["F", FontWeight->"Bold"], " through M. 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[Hint: ", StyleBox["Mathematica", FontSlant->"Italic"], " can't compute the integral exactly; you should use ", StyleBox["NIntegrate", FontFamily->"Courier", FontWeight->"Bold"], ".]\n\n", StyleBox["Exercise 3", FontWeight->"Bold"], "\n\nUse the Divergence Theorem to calculate the outward flux of ", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{\(F(x, y, z)\), "=", FormBox[\((\(z\^2\) x, \ \(1\/3\) y\^3 + tan(z), \ \(x\^2\) z + y\^2)\), "TraditionalForm"]}], FontSize->14], TraditionalForm]]], " through the top half of the sphere ", Cell[BoxData[ \(TraditionalForm\`x\^2 + y\^2 + z\^2 = 1\)]], ". [Hint: the surface is not closed; you need a closed surface to apply \ the Divergence Theorem. 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For example, \ \"UnmatchedBracket\" style defines how unmatched bracket, curly bracket, and \ parenthesis characters are displayed (typically by coloring them to make them \ stand out).\ \>", "Text"], Cell[StyleData["UnmatchedBracket"], StyleMenuListing->None, FontColor->RGBColor[0.760006, 0.330007, 0.8]] }, Closed]] }, Open ]] }] ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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