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Notebook[{
Cell[TextData[{
StyleBox["Lab 6A - Surface Integrals of Scalar Functions",
FontSize->24,
FontWeight->"Bold",
FontVariations->{"Underline"->True}],
"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n \
questions to: rogness@math.umn.edu"
}], "Text",
CellFrame->True,
TextAlignment->Center,
FontColor->GrayLevel[1],
Background->RGBColor[0, 0, 1]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Introduction",
FontSize->16]], "Section"],
Cell[TextData[{
"All semester long we've been taking the concepts you learned during the \
first year of calculus and generalizing them to higher dimensions. For \
example,\n\n\[FilledSmallSquare] Instead of one derivative, now we have many \
different ",
StyleBox["partial",
FontSlant->"Italic"],
" derivatives, one for each variable in the function. (We also have a \
\"total derivative,\" or \"derivative matrix,\" which is a matrix made up of \
all the partial derivatives.)\n\n\[FilledSmallSquare] A single definite \
integral, such as ",
Cell[BoxData[
\(TraditionalForm\`\[Integral]\_0\%1 x \[DifferentialD]x\)],
FontSize->14],
", could be thought of as an integral along the line segment on the x-axis \
where x ranges from 0 to 1. We've generalized this idea to come up with ",
StyleBox["line",
FontSlant->"Italic"],
" or ",
StyleBox["path integrals",
FontSlant->"Italic"],
", where we integrate over a curve in space instead of a line segment on \
the x-axis.\n\n\[FilledSmallSquare] We also have ",
StyleBox["iterated integrals",
FontSlant->"Italic"],
", which allow us to compute double and triple integrals. Double integrals \
let us integrate a function over a region R in the xy-plane.\n\nOver the next \
two weeks in lab, we're going to generalize this last idea. Instead of \
integrating over a region R in the xy-plane, we're going to integrate over a \
two-dimensional region which is sitting in space, i.e. a surface. This week \
we'll work with integrals of scalar functions on a surface; next week we'll \
integrate vector fields on a surface.\n\nThe concepts in this lab can be \
difficult to understand, so it's very important that you read through this \
whole lab; we'll try to give you lots of hints along the way, but they won't \
be very helpful hints if you miss them! Half of the work in any of these \
problems is finding a parametrization for a surface, and many people have \
trouble with this part. We've tried to make it easier on you by getting you \
started on parametrizing surfaces way back in Lab 2B. You might find it \
useful to open Lab 2B in another window so you can refer back to it.\n\nOne \
reminder before we move on. Many of the formulas involve the length (or ",
StyleBox["norm",
FontSlant->"Italic"],
") of a vector, i.e. ",
Cell[BoxData[
\(TraditionalForm\`\(\(||\)\(x\&\[RightVector]\)\(||\)\)\)]],
". You might recall from an earlier lab that we can compute this in ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" using the ",
StyleBox["Norm",
FontWeight->"Bold"],
" function. Evaluate the next two cells before moving on and make sure the \
output is what you would expect."
}], "Text"],
Cell[BoxData[
\(Norm[{1, 2, 3}]\)], "Input"],
Cell[BoxData[
\(Norm[{x, y, z}]\)], "Input"],
Cell[TextData[{
StyleBox["Mathematica",
FontSlant->"Italic"],
" has a built-in function called ",
StyleBox["Norm",
FontWeight->"Bold"],
", but it assumes that it's working with complex numbers (things like ",
Cell[BoxData[
\(TraditionalForm\`a + b\ i\)]],
"), so it doesn't work quite right for us. Instead ",
StyleBox["Norm",
FontWeight->"Bold"],
" is redefined in math2374.nb like this:\n\n",
Cell[BoxData[
\(\(Norm[v_] = Sqrt[v . v];\)\)], "Input"],
"\n\n(Here ",
StyleBox["v",
FontWeight->"Bold"],
" is a vector.) Why is this the correct definition? Talk to your TA if \
you're not sure."
}], "Text",
CellFrame->True,
Background->GrayLevel[0.833326]],
Cell[TextData[{
"We're also going to use the command ",
StyleBox["Cross",
FontWeight->"Bold"],
" to find the cross product of two vectors. If you aren't sure how to use \
this command, look it up in the help browser."
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Definition",
FontSize->16]], "Section"],
Cell[TextData[{
"Just for your reference, we'll repeat the definition of a surface integral \
of a scalar function. \n\nLet ",
Cell[BoxData[
\(TraditionalForm\`g :
U \[Subset] \[DoubleStruckCapitalR]\^3\[LongRightArrow]\
\[DoubleStruckCapitalR]\)]],
"be a continuous function, and let M be a smooth surface lying in U that is \
parametrized by ",
StyleBox["f",
FontWeight->"Bold"],
"(s,t), where ",
Cell[BoxData[
\(TraditionalForm\`\((s,
t)\) \[Epsilon]R \[Subset] \[DoubleStruckCapitalR]\^2\)]],
". The ",
StyleBox["surface integral of g over M",
FontWeight->"Bold"],
" is:\n"
}], "Text"],
Cell[BoxData[
RowBox[{\(\[Integral]\(\[Integral]\_M g\ \[DifferentialD]\[Sigma]\)\),
"=",
RowBox[{"\[Integral]",
RowBox[{\(\[Integral]\_R\),
RowBox[{"g",
RowBox[{"(",
RowBox[{
StyleBox["f",
FontWeight->"Bold"], \((s, t)\)}], ")"}],
RowBox[{"\[LeftDoubleBracketingBar]",
FormBox[\(\[PartialD]f\/\[DifferentialD]s\[Times]\[PartialD]f\/\
\[PartialD]t\),
"TraditionalForm"], "\[RightDoubleBracketingBar]"}], "ds",
" ", "dt"}]}]}]}]], "DisplayFormula",
FontSize->14]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Example",
FontSize->16]], "Section"],
Cell[TextData[{
"(This is based on similar examples in most textbooks. Here we'll \
concentrate on the concepts and let ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" do all the hard work.)\n\n",
StyleBox["Evaluate the surface integral of ",
FontSlant->"Italic"],
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = \((x\^2 + y\^2)\) z\)],
FontSlant->"Italic"],
" ",
StyleBox["over the upper half of the sphere of radius 1 centered at the \
origin.",
FontSlant->"Italic"],
"\n\nThe first step in any surface integral is generally to find a \
parametrization for the surface. In this case we need to parametrize the \
upper half of the unit sphere. There are a few ways to do this, but because \
this is a sphere, it's probably best to use spherical coordinates.\n\nRecall \
from Lab 2B that one of the best ways to parametrize a surface is often to \
describe it in cylindrical or spherical coordinates, and then convert the \
description back into rectangular coordinates. (Remember, our final \
parametrization has to be in rectangular coordinates!) In this particular \
case, the description of the upper half of the sphere is simply \[Rho]=1, \
where 0\[LessEqual]\[Theta]\[LessEqual]2\[Pi] and 0\[LessEqual]\[Phi]\
\[LessEqual]\[Pi]/2. (If this doesn't make sense, you should open your \
textbook and re-read the section on spherical coordinates.)\n\nSo \[Rho] is \
constant, and the angles \[Theta] and \[Phi] will be our parameters. Using \
the change of coordinate formulas, we get the following parametrization (with \
the same bounds on \[Theta] and \[Phi] as before):"
}], "Text"],
Cell[BoxData[{
\(\(f[theta_, phi_] = {Sin[phi] Cos[theta], Sin[phi] Sin[theta],
Cos[phi]};\)\), "\[IndentingNewLine]",
\(ParametricPlot3DLive[
f[theta, phi], {theta, 0, 2 Pi}, {phi, 0, Pi/2}]\)}], "Input"],
Cell[TextData[{
"Now we're asked to integrate the function ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = \((x\^2 + y\^2)\) z\)]],
" over the sphere. Sometimes people have a hard time visualizing what this \
actually means. Suppose our hemisphere is made up of a mixture of different \
kinds of metals, and different parts of the sphere have a different density, \
represented by the function g. Here's a different graph of our surface, \
where the hemisphere is colored according to the function g; lighter shades \
indicate larger values of g, while dark areas on the sphere represent areas \
where the value of g is lower. "
}], "Text"],
Cell[TextData[{
StyleBox["FYI",
FontWeight->"Bold"],
": You do ",
StyleBox["not",
FontSlant->"Italic"],
" have to understand the following commands; just evaluate them to see the \
picture! If you'd rather have a plain old graph instead of an interactive \
picture, change ",
StyleBox["ParametricPlot3DLive",
FontWeight->"Bold"],
" to ",
StyleBox["ParametricPlot3D",
FontWeight->"Bold"],
". I used a ",
StyleBox["Live",
FontWeight->"Bold"],
" command here because we're going to refer to the picture again in a few \
minutes, so you can keep it off to the side in the ",
StyleBox["Live",
FontWeight->"Bold"],
" window."
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell[BoxData[{
\(\(g[t_, p_] = Cos[p] Sin[p]^2;\)\), "\[IndentingNewLine]",
\(\(graph[t_, p_] = {Sin[p] Cos[t], Sin[p] Sin[t], Cos[p],
GrayLevel[g[t, p]/ .4]};\)\), "\[IndentingNewLine]",
\(ParametricPlot3DLive[graph[t, p], {p, 0, Pi/2}, {t, 0, 2 Pi},
Lighting \[Rule] False]\)}], "Input"],
Cell["\<\
So, for example, if g represents the density of the material making \
up the hemisphere, the lighter areas are constructed of much denser material \
than the darker areas. The integral of g over the surface will give us the \
total mass of the hemisphere.
The first step to compute the integral, as you can see by looking at the \
definition above or in your book, is to find the partial derivatives of our \
parametrization f(\[Theta],\[Phi]). To save some typing and reading time, \
let's simply write \"t\" and \"p\" instead of \"theta\" and \"phi\" for our \
variable names.\
\>", "Text"],
Cell[BoxData[{
\(dt = D[f[t, p], t]\), "\[IndentingNewLine]",
\(dp = D[f[t, p], p]\)}], "Input"],
Cell[TextData[{
"Now we have to cross the partials and find the length of the resulting \
cross product. Remember that we've defined a function ",
StyleBox["Norm",
FontWeight->"Bold"],
" to do this for us."
}], "Text"],
Cell[BoxData[{
\(\(v = Cross[dp, dt];\)\), "\[IndentingNewLine]",
\(Norm[v]\)}], "Input"],
Cell[TextData[{
"This looks truly horrendous, but if you use ",
StyleBox["Simplify",
FontWeight->"Bold"],
" you'll see that it's actually much better.\n\nThe other piece of our \
integrand is g(f(\[Theta],\[Phi])). Recall from Lab 4A that the correct way \
to do this in ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is to use the ",
StyleBox["Apply",
FontWeight->"Bold"],
" command. If you don't remember this, go look at Lab 4A again."
}], "Text"],
Cell[BoxData[{
\(\(g[x_, y_, z_] = \((x^2 + y^2)\) z;\)\), "\[IndentingNewLine]",
\(Apply[g, \ f[t, p]]\)}], "Input"],
Cell["\<\
Again, this can be simplified if you like.
Now that you've seen how we can compute each piece of the integrand \
separately, we can do it all at once. (If we had simply used this command \
right away it probably would have confused everybody)\
\>", "Text"],
Cell[BoxData[{
\(integrand =
Simplify[Apply[g, f[t, p]]*
Norm[Cross[dp, dt]]]\), "\[IndentingNewLine]",
\(Integrate[integrand, {p, 0, Pi/2}, {t, 0, 2 Pi}]\)}], "Input"],
Cell[TextData[{
"So if g represents the density of the material of the hemisphere, the \
total mass of the hemisphere is ",
Cell[BoxData[
\(TraditionalForm\`\[Pi]/2 \[TildeTilde] 1.571\)]],
"."
}], "Text"],
Cell[TextData[{
StyleBox["Careful",
FontWeight->"Bold"],
": we're being a bit sloppy here. If we were actually working with \
densities and mass, we really ought to include units! In fact, it's even \
worse; we usually think of our surfaces as being two-dimensional, whereas you \
might only think a three-dimensional object could have any mass. If you \
prefer, you could imagine the surface as a very, very thin piece of paper or \
rubber."
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell["\<\
Here's an interesting little afterthought. If you look at the \
earlier picture which is shaded in black and white according to g, you can \
see that the larger values of g are concentrated in the middle of the \
hemisphere. Close to the north pole, the shading is much darker, which means \
the values of g are much smaller. If g is a density function, that means \
there's not much mass in this part of the hemisphere. To verify this with \
integrals, you can evaluate the following commands. These are the same \
integrals as before, with one difference; instead of doing the whole \
hemisphere at once, the first integral is restricted to the area near the \
north pole (\[Phi] goes to \[Pi]/6 instead of \[Pi]/2=3\[Pi]/6), the second \
integral does the \"middle\" of the hemisphere (\[Pi]/6\[LessEqual]\[Phi]\
\[LessEqual]2\[Pi]/6), and the third does the section near the equator \
(2\[Pi]/6\[LessEqual]\[Phi]\[LessEqual]3\[Pi]/6).\
\>", "Text"],
Cell[BoxData[{
\(mass1 =
Integrate[
integrand, {p, 0, Pi/6}, {t, 0, 2 Pi}]\), "\[IndentingNewLine]",
\(mass2 =
Integrate[
integrand, {p, Pi/6, 2 Pi/6}, {t, 0,
2 Pi}]\), "\[IndentingNewLine]",
\(mass3 =
Integrate[integrand, {p, 2 Pi/6, 3 Pi/6}, {t, 0, 2 Pi}]\)}], "Input"],
Cell["\<\
You might remember that average density is equal to mass divided by \
surface area. We can find surface area of these pieces by integrating the \
function 1 over the surface. Remember that if we're integrating 1, then the \
integrand is actually 1 times the length of the normal vector:\
\>", "Text"],
Cell[BoxData[{
\(area1 =
Integrate[
Norm[Cross[dp, dt]], {p, 0, Pi/6}, {t, 0,
2 Pi}]\), "\[IndentingNewLine]",
\(area2 =
Integrate[
Norm[Cross[dp, dt]], {p, Pi/6, 2 Pi/6}, {t, 0,
2 Pi}]\), "\[IndentingNewLine]",
\(area3 =
Integrate[
Norm[Cross[dp, dt]], {p, 2 Pi/6, 3 Pi/6}, {t, 0,
2 Pi}]\)}], "Input"],
Cell[BoxData[{
\(avgdens1 = N[mass1/area1]\), "\[IndentingNewLine]",
\(avgdens2 = N[mass2/area2]\), "\[IndentingNewLine]",
\(avgdens3 = N[mass3/area3]\)}], "Input"],
Cell["\<\
So you can see that the density really is concentrated in the \
middle of the sphere, with somewhat less along the equator, and significantly \
less near the north pole!\
\>", "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Donuts and Chocolate Frosting",
FontSize->16]], "Section"],
Cell[TextData[{
"Now we come to the important, real world application: measuring chocolate.\
\n\nIn the homework you may have been asked to parametrize a \"torus,\" which \
is the mathematical term for the surface of a donut shape. This is a very \
famous and widely used surface in mathematica. ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" even includes commands for drawing tori (",
StyleBox["not toruses!)",
FontSlant->"Italic"],
". Note that you don't have to understand these commands, but you can \
evaluate them if you'd like to see a picture."
}], "Text"],
Cell[BoxData[
\(ShowLive[Graphics3D[Torus[3, 1]]]\)], "Input"],
Cell["\<\
In case you didn't manage to do the homework problem in question, \
here's the parametrization for a certain torus. If you graph this \
parametrization with both \[Theta] and \[Phi] going from 0 to 2\[Pi], you'll \
see the whole torus. (Try it if you'd like to.) Note that I'm typing \"t\" \
and \"p\" instead of \"theta\" and \"phi,\" or \"\[Theta]\" and \"\[Phi].\"\
\
\>", "Text"],
Cell[BoxData[
\(\(f[t_, p_] = {\((Sin[t] + 3)\) Sin[p], \((Sin[t] + 3)\) Cos[p],
Cos[t]};\)\)], "Input"],
Cell[TextData[{
"Here's the setup for our example. Suppose we bake a donut, and then we \
dip it upside down in a vat of dark chocolate, so we coat the upper half of \
the donut with gooey chocolate frosting. The chocolate in the vat isn't \
perfectly mixed, so the denser chocolate is settling at the bottom of the \
vat. What that means is that the chocolate which ends up on the ",
StyleBox["top",
FontSlant->"Italic"],
" of the donut -- remember, it's dipped upside down -- is denser than the \
chocolate on the sides. We want to know how much chocolate is actually \
there. We're going to do that by estimating the density of the chocolate \
coating at each point on the top half of the donut, and then integrating the \
density function over this half.\n\nThe way we've chosen the parametrization, \
the chocolate is on the top half of the torus, which just happens to be \
precisely those points on or above the xy-plane, where z=0. Evaluate the \
next two commands to see pictures of the surface we're going to integrate \
over. We use the same parametrization, but now 0\[LessEqual]\[Phi]\
\[LessEqual]2\[Pi] and -\[Pi]/2\[LessEqual]\[Theta]\[LessEqual]\[Pi]/2."
}], "Text"],
Cell[BoxData[
\(\(ParametricPlot3D[f[t, p], {t, \(-Pi\)/2, Pi/2}, {p, 0, 2 Pi},
PlotPoints \[Rule] {15, 30}];\)\)], "Input"],
Cell[BoxData[
\(ParametricPlot3D[f[t, p], {t, \(-Pi\)/2, Pi/2}, {p, 0, 2 Pi},
PlotPoints \[Rule] {15, 30}, ViewPoint -> {2, 2, \(-2\)}]\)], "Input"],
Cell["\<\
Alternatively, you could evaluate the next command to show the \
surface in a LiveGraphics3D window:\
\>", "Text"],
Cell[BoxData[
\(\(ParametricPlot3DLive[f[t, p], {t, \(-Pi\)/2, Pi/2}, {p, 0, 2 Pi},
PlotPoints \[Rule] {15, 30}];\)\)], "Input"],
Cell[TextData[{
"Suppose the density function is ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = \((z + 1)\)\)]],
" ",
Cell[BoxData[
\(TraditionalForm\`grams/cm\^2\)]],
". A graph of the chocolate surface looks like this:"
}], "Text"],
Cell[BoxData[{
\(\(g[t_, p_] = {\((Sin[t] + 3)\) Sin[p], \((Sin[t] + 3)\) Cos[p],
Cos[t], \[IndentingNewLine]\t\t\tGrayLevel[\((1.25 - Cos[t])\)/
1.25]};\)\), "\[IndentingNewLine]",
\(ParametricPlot3DLive[g[t, p], {t, \(-Pi\)/2, Pi/2}, {p, 0, 2 Pi},
PlotPoints \[Rule] {15, 30}, Lighting \[Rule] False]\)}], "Input"],
Cell[TextData[{
"(Again, you do ",
StyleBox["not",
FontSlant->"Italic"],
" need to understand these commands; they're just here so that you can see \
a picture which illustrates our situation.)\n\nUnlike in the previous \
example, the darker shades here represent the denser chocolate (because our \
chocolate is dark, after all). Now that we've seen all of the pretty \
pictures, we're ready to calculate the mass of the chocolate by integrating g \
over the top half of the torus. I'm not going to explain the commands much, \
so if you don't understand what's going on, scroll back up and look at the \
first example again. I will make a few comments; they are in between the (* \
and *) symbols."
}], "Text"],
Cell[BoxData[
\(\(\( (*\
density\ function\ *) \)\(\[IndentingNewLine]\)\(\(g[x_, y_, z_] =
z + 1\ ;\)\[IndentingNewLine]\[IndentingNewLine] (*\ \ \
parametrization\ f[t,
p]\ already\ defined\ *) \[IndentingNewLine]\[IndentingNewLine] (*\
partials\ *) \[IndentingNewLine]
\(dp = D[f[t, p], p];\)\[IndentingNewLine]
\(dt = D[f[t, p], t];\)\[IndentingNewLine]\[IndentingNewLine] (*\
define\ integrand\ and\ integrate\ *) \[IndentingNewLine]
\(integrand =
Simplify[Apply[g, f[t, p]]]*Norm[Cross[dp, dt]];\)\[IndentingNewLine]
\(ans =
Integrate[
integrand, {t, \(-Pi\)/2, Pi/2}, {p, 0,
2 Pi}];\)\[IndentingNewLine]
Simplify[ans]\[IndentingNewLine]
N[ans]\)\)\)], "Input"],
Cell["\<\
There are semicolons after most of these commands, so the only \
outputs are the final answer, in simplified form, and a numerical \
approximation. If you'd like to see the partials, or the integrand, you can \
remove the appropriate semicolons and evaluate the cell again.
So with our given setup, there are almost 97 grams of chocolate on this \
surface. (Sounds like a good donut!)\
\>", "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Exercises",
FontSize->16]], "Section"],
Cell[TextData[{
StyleBox["Note",
FontWeight->"Bold"],
": in the examples above, you were shown graphs which were colored \
according to the scalar function g(x,y,z), instead of ",
StyleBox["Mathematica",
FontSlant->"Italic"],
"'s normal internal coloring scheme. ",
StyleBox["You do not have to produce these kinds of graphs in your lab \
report!",
FontWeight->"Bold"]
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell[TextData[{
StyleBox["Exercise 1",
FontSize->14,
FontWeight->"Bold"],
"\n\nFind the surface area of the surface parametrized (and graphed) by the \
following commands. (You will need to cut and paste before you can evaluate \
them.)\n\n",
Cell[BoxData[{
\(f[s_, t_] = {t\ Cos[2 Pi\ t], \ t\ Sin[2 Pi\ t], \
s\ *\((4 - t)\)}\), "\n",
\(ParametricPlot3D[f[s, t], {s, 0, 1}, {t, 0, 4},
PlotPoints \[Rule] {5, 200}]\)}], "Input"],
"\n\nYour answer should include a sketch of the surface, your exact answer \
-- which will be ugly -- and a numeric approximation, which you can find \
using the ",
StyleBox["N",
FontWeight->"Bold"],
" command in ",
StyleBox["Mathematica",
FontSlant->"Italic"],
". Alternatively, you could use the command ",
StyleBox["NIntegrate",
FontWeight->"Bold"],
" to get your numeric approximation. ",
StyleBox["NIntegrate",
FontWeight->"Bold"],
" will not try to give you an exact answer, but will only give you the \
approximation.\n\n",
StyleBox["Exercise 2",
FontSize->14,
FontWeight->"Bold"],
"\n\nFind the surface area of the surface parametrized (and graphed) by the \
following commands. (You will need to cut and paste before you can evaluate \
them.)\n\n",
Cell[BoxData[{
\(f[s_, t_] = {t\ Cos[2 Pi\ t], \ t\ Sin[2 Pi\ t],
s*\((\ 2 + Sin[7 Pi\ t])\)}\), "\n",
\(ParametricPlot3D[f[s, t], {s, 0, 1}, {t, 0, 4},
PlotPoints \[Rule] {5, 200}]\)}], "Input"],
"\n\nYour answer should include a sketch of the surface, and a numeric \
approximation to the final answer. You'll have to use ",
StyleBox["NIntegrate",
FontWeight->"Bold"],
" instead of ",
StyleBox["Integrate",
FontWeight->"Bold"],
" (see exercise 1). ",
StyleBox["Integrate",
FontWeight->"Bold"],
" will think for a long, long time and probably not give you an answer.",
"\n\n",
StyleBox["Exercise 3\n",
FontSize->14,
FontWeight->"Bold"],
"\nEvaluate the surface integral of the function ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = x\)]],
" over the surface which is the graph of ",
Cell[BoxData[
\(TraditionalForm\`z = Sin[x] Cos[y]\)]],
", where 0\[LessEqual]x\[LessEqual]2\[Pi] and -\[Pi]\[LessEqual]y\
\[LessEqual]\[Pi]. Note that you will have to use ",
StyleBox["NIntegrate",
FontWeight->"Bold"],
" with this exercise; once again ",
StyleBox["Integrate",
FontWeight->"Bold"],
" can't quite get the answer, although it gives it a good try.\n\n",
StyleBox["Exercise 4\n",
FontSize->14,
FontWeight->"Bold"],
"\nEvaluate the surface integral of the function ",
Cell[BoxData[
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with exercises 2 and 3, you'll have to use ",
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Cell["Credits", "Subsection"],
Cell["\<\
This lab was written from scratch in February/March 2002. \
Previously there was no lab which dealt with surface integrals of scalar \
functions. I intetionally kept this lab short (only two exercises to hand \
in, generally) to provide some relief from earlier labs, which were much \
longer. I made minor updates in April 2004 -- new exercises, a bit of \
LiveGraphics3D stuff, etc.
This lab is copyright 2002, 2004 by Jonathan Rogness (rogness@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 if you use parts of our labs, just so I can keep track of it. \
Please send me any questions or comments!\
\>", "Text"]
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