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Here's \ a quick review: as usual, vector-valued objects are in ", StyleBox["boldface", FontWeight->"Bold"], "." }], "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["s", FontSize->14, FontWeight->"Bold"]}]}], StyleBox[" ", FontWeight->"Bold"], "or", " ", RowBox[{\(\[ContourIntegral]\_C\), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["s", FontSize->14, 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]\_S\%\ \)f \ \[DifferentialD]S\ \ \ or\ \ \ \(\[DoubleContourIntegral]\_S f \[DifferentialD]S\)\)\), \(S, a\ surface\ with\[IndentingNewLine] parametrization\ \[CapitalPhi] \((u, v)\)\[IndentingNewLine] and\ parameter\ domain\ R\[IndentingNewLine] \), \(scalar\ function\ f\), \(\[Integral]\(\(\[Integral]\_R\%\ \)f \ \((\[CapitalPhi] \((u, v)\))\)\) \[DoubleVerticalBar] \[PartialD]\ \[CapitalPhi]\/\[PartialD]u\[Times]\[PartialD]\[CapitalPhi]\/\[PartialD]v \ \[DoubleVerticalBar] \[DifferentialD]u \[DifferentialD]v\)}, {\(surface\ \((flux)\)\), RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_S\), RowBox[{ RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["S", FontSize->14, FontWeight->"Bold"]}]}], " ", "or", " ", RowBox[{\(\[DoubleContourIntegral]\_S\), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["S", FontSize->14, FontWeight->"Bold"], " "}]}]}]}]}]}], RowBox[{\(S, a\ surface\ with\), "\[IndentingNewLine]", \(parametrization\ \ \[CapitalPhi] \((u, v)\)\), "\[IndentingNewLine]", StyleBox[\(and\ parameter\ domain\ R\), FontWeight->"Plain"], "\[IndentingNewLine]"}], RowBox[{"vector", " ", "field", " ", StyleBox["F", FontWeight->"Bold"]}], RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_R\%\ \), RowBox[{ StyleBox["F", FontWeight-> "Bold"], \(\((\ \[CapitalPhi] \((u, v)\))\)\[CenterDot]\((\[PartialD]\[CapitalPhi]\/\ \[PartialD]u\[Times]\[PartialD]\[CapitalPhi]\/\[PartialD]v)\)\), \(\ \[DifferentialD]u\), \(\(\[DifferentialD]\)\(v\)\(\ \)\)}]}]}]} }]], "Text", CellFrame->True, TextAlignment->Left, FontSize->13, Background->GrayLevel[0.849989]], Cell["\<\ Notice that each kind of integral has its own definition, and there \ is no obvious way to convert one kind to another. Stokes's Theorem provides \ the missing link between surface integrals of vector fields and line \ integrals of vector fields.\ \>", "Text"], Cell[TextData[{ StyleBox["Stokes's Theorem", FontWeight->"Bold"], ": Suppose M is a smooth surface with a chosen orientation, and let ", StyleBox["F", FontWeight->"Bold"], " be a vector field which has continuous partial derivatives on and \ \"near\" M. Then ", Cell[BoxData[ RowBox[{\(\[Integral]\_\(\[PartialD]S\)\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["s", FontWeight->"Bold"]}]}]}]], "Text"], " = ", Cell[BoxData[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_S\%\ \), RowBox[{ RowBox[{"(", RowBox[{"curl", " ", StyleBox["F", FontWeight->"Bold"]}], StyleBox[")", FontWeight->"Bold"]}], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}]}]}]}]], "Text"], ", where \[PartialD]S is the positively-oriented boundary of S." }], "Text", CellFrame->True, FontWeight->"Plain", Background->GrayLevel[0.849989]] }, Closed]], Cell[CellGroupData[{ Cell["Using Stokes's Theorem: Example 1", "Section", Background->None], Cell[TextData[{ "Applying Stokes's Theorem can be tricky: each problem has its own kind of \ challenge. For example, we might want to calculate the surface integral of \ some vector field ", StyleBox["G", FontWeight->"Bold"], " over some complicated surface with a \"nice\" boundary. We can use \ Stokes's Theorem to avoid working with the surface, if we can find a vector \ field ", StyleBox["F", FontWeight->"Bold"], " whose curl is ", StyleBox["G", FontWeight->"Bold"], ". This can be hard." }], "Text"], Cell[CellGroupData[{ Cell["Example 1", "Subsection"], Cell[TextData[{ "Calculate ", Cell[BoxData[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_M\%\ \), RowBox[{ StyleBox["G", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}]}]}]}]], "Text"], ", where ", StyleBox["G", FontWeight->"Bold"], "(x,y,z)=(0,0,2), and M is the surface parametrized by ", StyleBox["\[CapitalPhi]", FontWeight->"Bold"], "(r,\[Theta])=(r cos \[Theta], r sin \[Theta], sin(4\[Pi]r), with 0\ \[LessEqual]r\[LessEqual]1, 0\[LessEqual]\[Theta]\[LessEqual]2\[Pi]. [Use \ the upward-pointing normal vector.]" }], "Text"], Cell[BoxData[{ \(\[CapitalPhi][r_, t_] = {r*Cos[t], r*Sin[t], Sin[4 Pi*r]}\), "\[IndentingNewLine]", \(\(ParametricPlot3D[\[CapitalPhi][r, t], {r, 0, 1}, {t, 0, 2 Pi}, PlotPoints \[Rule] 60, AxesLabel \[Rule] {X, Y, Z}];\)\), "\[IndentingNewLine]", \(\(ShowLive[%];\)\)}], "Input"], Cell[TextData[{ "This surface is intimidating, but its boundary is very nice. The boundary \ is just the part of the surface where r=1; we get its parametrization by \ substituting r=1 in ", StyleBox["\[CapitalPhi]", FontWeight->"Bold"], "." }], "Text"], Cell[BoxData[{ \(g[t_] = {Cos[t], Sin[t], Sin[4 Pi]}\), "\[IndentingNewLine]", \(ParametricPlot3D[g[t], {t, 0, 2 Pi}]\)}], "Input"], Cell[TextData[{ "We need ", StyleBox["F", FontWeight->"Bold"], " so that curl ", StyleBox["F", FontWeight->"Bold"], "=", StyleBox["G", FontWeight->"Bold"], "; then our integral will be the right-hand side of the equation in \ Stokes's Theorem. One possible choice is ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(-y,x,0). (Check this! Also note that, given ", StyleBox["G", FontWeight->"Bold"], ", there is no specific way for you to find an ", StyleBox["F", FontWeight->"Bold"], " such that curl ", StyleBox["F", FontWeight->"Bold"], "=", StyleBox["G", FontWeight->"Bold"], ". In this case we just \"pulled it out of the hat.\")", " If we use this ", StyleBox["F", FontWeight->"Bold"], ", then the left-hand side of the equation in Stokes's Theorem is almost \ (not quite) easy enough to do in your head:" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_S\%\ \), RowBox[{ StyleBox["G", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}]}]}]}], "=", RowBox[{ RowBox[{ SubsuperscriptBox["\[Integral]", StyleBox[\(\[PartialD]S\), FontSize->14], " "], RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["s", FontWeight->"Bold"]}]}]}], StyleBox["=", FontWeight->"Bold"], RowBox[{ RowBox[{\(\[Integral]\_0\%\(2 \[Pi]\)\), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["g", FontWeight->"Bold"], StyleBox[\((t)\), FontWeight->"Plain"]}], ")"}], "\[CenterDot]", RowBox[{ StyleBox["g", FontWeight->"Bold"], StyleBox["'", FontWeight-> "Plain"]}]}], \((t)\), \(\[DifferentialD]t\)}]}], "=", \(\[Integral]\_0\%\(2 \[Pi]\)\((\(-sin\)\ t, cos\ t, \ 0)\)\[CenterDot]\((\(-sin\)\ t, cos\ t, \ 0)\) \[DifferentialD]t = \(\[Integral]\_0\%\(2 \[Pi]\)1 \ \[DifferentialD]t = 2 \( \(\[Pi]\)\(.\)\)\)\)}]}]}]], "Text"], Cell["\<\ Re-read this example to make sure you followed the process, and ask \ your TA if you have any questions about what we did or why. These problems \ can be tricky!\ \>", "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Using Stokes's Theorem: Example 2", "Section", Background->None], Cell[TextData[{ "More often, we use Stokes's Theorem in the other direction: we have a \ tricky line integral, and try to replace it with an easy surface integral. \ This means we don't have to invent a new vector field with a given curl; \ instead, we have to invent a surface with a given boundary. Any surface with \ the right boundary will be good enough, just like in the last example, where \ any ", StyleBox["F ", FontWeight->"Bold"], "with the right curl was good enough.\n\nIf you'd like to see an example of \ many different surfaces which have the same boundary, copy the following \ address and paste it into your web browser:\n\n\ http://www.math.umn.edu/~rogness/multivar/multiple_surfaces.shtml" }], "Text"], Cell[CellGroupData[{ Cell["Example 2", "Subsection"], Cell[TextData[{ "Calculate ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["s", FontWeight->"Bold"]}]}]}]], "Text"], " , where ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(", Cell[BoxData[ \(TraditionalForm\`\(x\^2\) z\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(\(xy\^2\)\(,\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`z\^2\)]], "), and C is the intersection of the plane ", StyleBox["x+y+z", FontSlant->"Italic"], "=4 with the cylinder ", Cell[BoxData[ \(TraditionalForm\`x\^2 + y\^2 = 16\)]], "." }], "Text"], Cell[TextData[{ "Here is a plot of the two surfaces intersecting. Replace ", StyleBox["ShowLive", FontWeight->"Bold"], " with ", StyleBox["Show", FontWeight->"Bold"], " if you'd prefer not to have an interactive version." }], "Text"], Cell[BoxData[{ \(plane = Plot3D[4 - x - y, {x, \(-5\), 5}, {y, \(-5\), 5}]\), "\n", \(cyl = ContourPlot3D[ x^2 + y^2 - 16, {x, \(-4\), 4}, {y, \(-4\), 4}, {z, \(-4\), 12}]\), "\n", \(ShowLive[{plane, cyl}, BoxRatios \[Rule] {1, 1, 1.6}]\)}], "Input"], Cell["\<\ Let's parametrize the curve C, so we can graph it. Thinking in \ cylindrical coordinates, every point on C is on the cylinder, so has r=4. \ Also, every point on C is on the plane, so has z=4-x-y. Letting \[Theta] \ range from 0 to 2\[Pi], we get the following parametrization.\ \>", "Text"], Cell[BoxData[{ \(\(f[t_] = {4 Cos[t], 4 Sin[t], 4 - 4 Cos[t] - 4 Sin[t]};\)\), "\[IndentingNewLine]", \(curve = ParametricPlot3D[f[t], {t, 0, 2 Pi}]\)}], "Input"], Cell[TextData[{ "What is the orientation of this curve -- \"clockwise\" or \ \"counterclockwise\"? Check your answer using ", StyleBox["PathAnimate3D", FontFamily->"Courier", FontWeight->"Bold"], " or ", StyleBox["PathTangentAnimate3D", FontFamily->"Courier", FontWeight->"Bold"], " from Lab 4." }], "Text", CellFrame->True, Background->GrayLevel[0.849989]], Cell[TextData[{ "To use Stokes's Theorem, we need a surface whose boundary is C. We have \ an ideal surface available: it's just the part of the plane which lies inside \ the cylinder. [", StyleBox["Check", FontWeight->"Bold"], ": does C have the right orientation to be the boundary of this surface?] \ We can parametrize this surface just as we did C; the only difference is that \ instead of r=4, we have 0\[LessEqual]r\[LessEqual]4. " }], "Text"], Cell[BoxData[{ \(\(\[CapitalPhi][r_, t_] = {r*Cos[t], r*Sin[t], 4 - r*Cos[t] - r*Sin[t]};\)\), "\[IndentingNewLine]", \(surface = ParametricPlot3D[\[CapitalPhi][r, t], {r, 0, 4}, {t, 0, 2 Pi}]\)}], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " gives us an oblique view of this surface; you should evaluate the \ following cell to get an interactive version of it. Rotate it and compare to \ the earlier picture (with the plane and the cylinder) and convince yourself \ that our surface is exactly the part of the plane contained within the \ cylinder." }], "Text"], Cell[BoxData[ \(ShowLive[surface]\)], "Input"], Cell[TextData[{ "Finally, we can use Mathematica to calculate our integral: ", Cell[BoxData[ RowBox[{ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["s", FontWeight->"Bold"]}]}]}], "=", RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_S\%\ \), RowBox[{"curl", " ", RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["S", FontWeight->"Bold"]}]}]}]}]}]}]], "Text"], " = ", Cell[BoxData[ RowBox[{"\[Integral]", RowBox[{\(\[Integral]\_R\%\ \), RowBox[{ RowBox[{"(", RowBox[{"curl", " ", StyleBox["F", FontWeight->"Bold"]}], ")"}], \(\((\[CapitalPhi] \((r, t)\))\)\[CenterDot]\((\[PartialD]\[CapitalPhi]\/\ \[PartialD]r\[Times]\[PartialD]\[CapitalPhi]\/\[PartialD]t)\)\), \(\ \[DifferentialD]A\)}]}]}]], TextAlignment->Left, FontSize->13], ", so we need the following calculations. First, we define ", StyleBox["F", FontWeight->"Bold"], ", ", "take its curl,", " and plug in the parametrization of our surface:" }], "Text"], Cell[BoxData[{ \(F[x_, y_, z_] = {x^2*z, x*y^2, z^2}\), "\[IndentingNewLine]", \(curlF[x_, y_, z_] = Curl[F[x, y, z]]\), "\[IndentingNewLine]", \(v1[r_, t_] = Apply[curlF, \[CapitalPhi][r, t]]\)}], "Input"], Cell[TextData[{ "Remember the ", StyleBox["Apply", FontFamily->"Courier", FontWeight->"Bold"], " command? We've used it a few times now; ask your TA for help if you \ don't remember it." }], "Text", CellFrame->True, Background->GrayLevel[0.849989]], Cell["\<\ Next, we take the partial derivatives of our parametrization and \ take the cross product:\ \>", "Text"], Cell[BoxData[{ \(d\[CapitalPhi]dr[r_, t_] = D[\[CapitalPhi][r, t], r]\), "\[IndentingNewLine]", \(d\[CapitalPhi]dt[r_, t_] = D[\[CapitalPhi][r, t], t]\), "\[IndentingNewLine]", \(v2[r_, t_] = Cross[d\[CapitalPhi]dr[r, t], d\[CapitalPhi]dt[r, t]]\), "\[IndentingNewLine]", \(Simplify[%]\)}], "Input"], Cell["\<\ Finally, we take the dot product of our results, and integrate the \ result.\ \>", "Text"], Cell[BoxData[{ \(v1[r, t] . v2[r, t]\), "\n", \(Integrate[%, {r, 0, 4}, {t, 0, 2 Pi}]\)}], "Input"], Cell[TextData[{ "It's probably good to check our answer by calculating the line integral \ directly: ", Cell[BoxData[ RowBox[{\(\[Integral]\_C\%\ \), RowBox[{ StyleBox["F", FontWeight->"Bold"], "\[CenterDot]", RowBox[{"\[DifferentialD]", StyleBox["s", FontWeight->"Bold"]}]}]}]]], " = ", Cell[BoxData[ RowBox[{\(\[Integral]\_0\%\(2 \[Pi]\)\), RowBox[{ StyleBox["F", FontWeight->"Bold"], RowBox[{ RowBox[{"(", RowBox[{ StyleBox["f", FontWeight->"Bold"], \((t)\)}], ")"}], "\[CenterDot]", RowBox[{ StyleBox["f", FontWeight->"Bold"], "'"}]}], \((t)\), \(\[DifferentialD]t\)}]}]], "Text"], ", so we get:" }], "Text"], Cell[BoxData[{ \(w1[t_] = Apply[F, f[t]]\), "\[IndentingNewLine]", \(w2[t_] = D[f[t], t]\), "\[IndentingNewLine]", \(w1[t] . w2[t]\), "\[IndentingNewLine]", \(Simplify[%]\), "\[IndentingNewLine]", \(Integrate[%, {t, 0, 2 Pi}]\)}], "Input"], Cell["\<\ If this looks easier to you than the surface integral, take a \ careful look at the functions you're integrating. If the line integral still \ looks easier, you should try doing both by hand!\ \>", "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Exercises", "Section", Background->None], Cell[TextData[{ "1) Graph the cylinder ", Cell[BoxData[ \(TraditionalForm\`x\^2 + y\^2 = 1\)]], "and the hyperbolic paraboloid ", Cell[BoxData[ \(TraditionalForm\`z = y\^2 - x\^2\)]], "on the same set of axes, and parametrize the curve C where they intersect. \ Verify Stokes's Theorem for the vector field ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(", Cell[BoxData[ \(TraditionalForm\`\(x\^2\) y\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(1\/3\) x\^3\)]], ", ", StyleBox["xy", FontSlant->"Italic"], ") and the curve C. [i.e. Show that the line integral of ", StyleBox["F", FontWeight->"Bold"], " along C equals the appropriate surface integral. You will need to \ produce your own surface.]\n\n2) Now consider the vector field ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(", Cell[BoxData[ \(TraditionalForm\`\(x\^2\) y\)]], ", ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", ", StyleBox["xy", FontSlant->"Italic"], ").\na) Calculate the line integral of ", StyleBox["F", FontWeight->"Bold"], " along the circle parametrized by ", Cell[BoxData[ \(TraditionalForm\`\(f\_1\)(t) = \ \((cos(t), \ sin(t), \ 1)\), \ for\ 0 < t < 2 \( \(\[Pi]\)\(.\)\)\)]], "\nb) Calculate the line integral of ", StyleBox["F", FontWeight->"Bold"], " along the circle parametrized by ", Cell[BoxData[ \(TraditionalForm\`\(f\_2\)(t) = \ \((cos(t), \ sin(t), \ 0)\), \ for\ 0 < t < 2 \( \(\[Pi]\)\(.\)\)\)]], "\nc) Calculate the surface integral of curl(", StyleBox["F", FontWeight->"Bold"], ") over the surface parametrized by ", StyleBox["\[CapitalPhi]", FontSlant->"Italic"], "(", StyleBox["s", FontSlant->"Italic"], ",", StyleBox["t", FontSlant->"Italic"], ")=(cos(", StyleBox["t", FontSlant->"Italic"], "),sin(", StyleBox["t", FontSlant->"Italic"], "),", StyleBox["s), ", FontSlant->"Italic"], "for", StyleBox[" ", FontSlant->"Italic"], "0", StyleBox[""Italic"], "2 \[Pi]", StyleBox[" ", FontSlant->"Italic"], "and 0<", StyleBox["s", FontSlant->"Italic"], "<1.\nd) Explain your answers in terms of Stokes's Theorem.\n\n3) Use \ Stokes's Theorem to evaluate the line integral of the vector field ", StyleBox["F", FontWeight->"Bold"], "(x,y,z)=(", Cell[BoxData[ \(TraditionalForm\`y + sin(x), z\^2 + cos(y), \ x\^3\)]], "), where C is the curve parametrized by ", StyleBox["f", FontWeight->"Bold"], "(t)=(sin t, cos t, sin(2t)), for 0\[LessEqual]t\[LessEqual]2\[Pi].\n\n4) \ Consider the vector field ", StyleBox["G", FontWeight->"Bold"], "(x,y,z)=(x,y,z). Use Stokes's Theorem to prove, by contradiction, that ", StyleBox["G", FontWeight->"Bold"], " is not the curl of any smooth vector field ", StyleBox["F", FontWeight->"Bold"], ". [Hint: Let M be the unit sphere centered at the origin. The path \ integral over \[PartialD]M of every smooth vector field ", StyleBox["F", FontWeight->"Bold"], " equals 0 -- explain why!]" }], "Text", CellFrame->True, Background->RGBColor[1, 0.501961, 0.501961]], Cell[CellGroupData[{ Cell["Credits", "Subsection"], Cell["\<\ This lab was written by James Swenson in 2002. In Spring 2004 \ Jonathan Rogness went through and updated a few minor things to reflect the \ use of our math2374.nb file. This lab is copyright 2002 by James Swenson (swenson@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/. Although it's not specifically required by the license, I'd appreciate it if \ you let me know at rogness@math.umn.edu if you use parts of our labs, just so \ I can keep track of it. Please send me any questions or comments!\ \>", \ "Text"] }, Closed]] }, Closed]] }, FrontEndVersion->"5.1 for Microsoft Windows", ScreenRectangle->{{0, 1280}, {0, 717}}, WindowToolbars->{}, WindowSize->{959, 594}, WindowMargins->{{16, Automatic}, {Automatic, -1}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "PostScriptOutputFile":>FrontEnd`FileName[{"user002", "rogness"}, \ "Newlab.nb.ps", CharacterEncoding -> "WindowsANSI"], "Magnification"->1}, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions", "Subtitle"], Cell["\<\ Modify the definitions below to change the default appearance of \ all cells in a given style. 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