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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 29 "The Area of Familiar Surfaces"
}}{PARA 257 "" 0 "" {TEXT -1 9 "Jim Herod" }}{PARA 258 "" 0 "" {TEXT
-1 21 "School of Mathematics" }}{PARA 259 "" 0 "" {TEXT -1 12 "Georgia
Tech" }}{PARA 260 "" 0 "" {TEXT -1 21 "herod@math.gatech.edu" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 " In
order to put surface area into a perspective, recall how the arc leng
th of the graph for a function f from R to R was computed:" }}{PARA 0
"" 0 "" {TEXT -1 47 " arclength = "
}{XPPEDIT 18 0 "int(sqrt(1+(diff(f,x)^2)),x=a..b)" "6#-%$intG6$-%%sqrt
G6#,&\"\"\"F**$-%%diffG6$%\"fG%\"xG\"\"#F*/F0;%\"aG%\"bG" }{TEXT -1 1
"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 " \+
It is reasonable to want to compute the arclength of a curve that i
s not the graph of a function, but rather is given parametrically. If \+
the curve is given by" }}{PARA 0 "" 0 "" {TEXT -1 64 " \+
r(t) = [ x(t), y(t), z(t) ], " }}{PARA 0 "" 0 ""
{TEXT -1 4 "then" }}{PARA 0 "" 0 "" {TEXT -1 43 " \+
arclength = " }{XPPEDIT 18 0 "int(sqrt((diff(x,t))^2+(diff(y
,t))^2+(diff(z,t))^2),t=a..b)" "6#-%$intG6$-%%sqrtG6#,(*$-%%diffG6$%\"
xG%\"tG\"\"#\"\"\"*$-F,6$%\"yGF/F0F1*$-F,6$%\"zGF/F0F1/F/;%\"aG%\"bG"
}{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 39 " The subject for this worksheet is " }{TEXT 256 12 "s
urface area" }{TEXT -1 113 ". It is well to understand this concept fo
r the associated ideas are precursors to those of the surface integral
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 " \+
We begin with most familiar surfaces and work toward more complicated
surfaces." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 27 "The area of a parallelogram" }}{PARA 0 "" 0 "" {TEXT -1
217 "\n The simplest area will be a parallelogram in the plane. We
take the parallelogram to have non-parallel sides which are represent
ed by the vector a and b. Recall that the area of such a parallelogram
is precisely" }}{PARA 0 "" 0 "" {TEXT -1 56 " \+
|a| |b| sin(" }{XPPEDIT 18 0 "theta" "6#%&thetaG"
}{TEXT -1 15 ") = | a x b |, " }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }
{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 90 " is the angle between
a and b and | a x b | is the length of the cross product of a and b.
" }}{PARA 0 "" 0 "" {TEXT -1 178 " We find the area of a parallelo
gram with three corners at [1,0,0], [0,2,0], and [0,0,3]. The equation
for the plane that goes through these points can be seen by inspectio
n:" }}{PARA 0 "" 0 "" {TEXT -1 33 " "
}{XPPEDIT 18 0 "x/1+y/2+z/3=1" "6#/,(*&%\"xG\"\"\"F'!\"\"F'*&%\"yGF'\"
\"#F(F'*&%\"zGF'\"\"$F(F'F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}
{PARA 0 "" 0 "" {TEXT -1 56 "We write z as a function of x and y then \+
plot the plane." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solve(x/1
+y/2+z/3=1,z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 173 "J:=plot3
d(-3*x-3/2*y+3,x=0..1,y=0..2,axes=normal,orientation=[-10,70]):\nK:=sp
acecurve([1,t,0],t=0..2,color=BLACK):\nL:=spacecurve([t,2,0],t=0..1,co
lor=BLACK):\ndisplay(J,K,L);" }}}{PARA 0 "" 0 "" {TEXT -1 30 "The two \+
vectors we compute are" }}{PARA 0 "" 0 "" {TEXT -1 76 " (
[0, 2, 0] - [1, 0, 0] ) and ( [0, 0, 3] - [1, 0, 0] ). " }}
{PARA 0 "" 0 "" {TEXT -1 130 "The length of the parallelogram with the
se vectors as sides is given by the norm of the crossproduct. This is \+
computed as follows." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with
(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "norm(crossprod
([-1,2,0],[-1,0,3]),2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 " \+
Suppose that r(s,t) is a surface in " }{XPPEDIT 18 0 "R^3" "6#*$%\"R
G\"\"$" }{TEXT -1 56 " and that the three dimensional vector valued fu
nctions " }{XPPEDIT 18 0 "diff(r,s)" "6#-%%diffG6$%\"rG%\"sG" }{TEXT
-1 6 " and " }{XPPEDIT 18 0 "diff(r,t)" "6#-%%diffG6$%\"rG%\"tG" }
{TEXT -1 221 " are continuous functions of s and t. Suppose also tha
t r(a,b) is a point on the surface. We paste a parallelogram tangent t
o the surface at this point. The length of the sides of the parallelo
gram may be symbolized by" }}{PARA 0 "" 0 "" {TEXT -1 11 " | \+
" }{XPPEDIT 18 0 "diff(r,s)" "6#-%%diffG6$%\"rG%\"sG" }{TEXT -1 21 "(a
,b) | s and | " }{XPPEDIT 18 0 "diff(r,t)" "6#-%%diffG6$%\"rG%\"t
G" }{TEXT -1 10 "(a,b) | t." }}{PARA 0 "" 0 "" {TEXT -1 134 "The area \+
of this tangent parallelogram is, then, the cross product of these two
vectors. This leads to a representation of the area as" }}{PARA 0 ""
0 "" {TEXT -1 23 " | " }{XPPEDIT 18 0 "diff(r,s)"
"6#-%%diffG6$%\"rG%\"sG" }{TEXT -1 8 "(s,t) x " }{XPPEDIT 18 0 "diff(r
,t)" "6#-%%diffG6$%\"rG%\"tG" }{TEXT -1 11 "(s,t) | dA" }}{PARA 0 ""
0 "" {TEXT -1 76 "where the double integral is taken over the appropri
ate region in the plane." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 25 "The area of a hemisphere." }}{PARA 0 ""
0 "" {TEXT -1 55 "We expect the area of a hemisphere with radius a to \+
be " }{XPPEDIT 18 0 "4*Pi*a^2" "6#*(\"\"%\"\"\"%#PiGF%%\"aG\"\"#" }
{TEXT -1 84 ". If we take the northern hemisphere, we can display it \+
as the graph of a function." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot3d(1-(s^2+t^2),s
=-1..1,t=-sqrt(1-s^2)..sqrt(1-s^2));" }}}{PARA 0 "" 0 "" {TEXT -1 114
"For this example, we use the computational procedure for surface area
given above by defining the function r(s,t)." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 34 "r:=(s,t)->[s,t,sqrt(1-(s^2+t^2))];" }}}{PARA 0 "
" 0 "" {TEXT -1 61 "We need to have the derivatives of r with respect \+
to s and t." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "drds:=diff(r(
s,t),s);\ndrdt:=diff(r(s,t),t);" }}}{PARA 0 "" 0 "" {TEXT -1 55 "The c
omputational procedure asks for the cross product." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 83 "term:=crossprod(drds,drdt);\ncrss:=sqrt(term
[1]^2+term[2]^2+term[3]^2);\nsimplify(%);" }}}{PARA 0 "" 0 "" {TEXT
-1 47 "It remains only to compute the double integral." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "int(int(sqrt(1/(1-s^2-t^2)),t=-sqrt
(1-s^2)..sqrt(1-s^2)),s=-1..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 35 "The area of the surface of a sphere" }}{PARA 0 "" 0 ""
{TEXT -1 169 "Of course, this area can be computed as simply twice the
answer to the above problem. You learn nothing from doing the problem
that way. Rather, parameterize the sphere." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 67 "x:=(s,t)->cos(s)*sin(t);\ny:=(s,t)->sin(s)*sin(t);
\nz:=(s,t)->cos(t);" }}}{PARA 0 "" 0 "" {TEXT -1 48 "To verify that th
is is a sphere, check this sum." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 27 "x(s,t)^2+y(s,t)^2+z(s,t)^2;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 12 "simplify(%);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 68 "We compute the derivatives of r = [x, y
, z] with respect to s and t." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 107 "drds:=[diff(x(s,t),s),diff(y(s,t),s),diff(z(s,t),s)];\ndrdt:=[d
iff(x(s,t),t),diff(y(s,t),t),diff(z(s,t),t)];" }}}{PARA 0 "" 0 ""
{TEXT -1 93 "Next, take the crossproduct of these terms, compute the l
ength of that vector, and integrate." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 "term:=crossprod(drds,drdt);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 43 "cross:=sqrt(term[1]^2+term[2]^2+term[3]^2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "int(int(%,t=0..Pi),s=0..2*Pi);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 47 "T
he surface area of a portion of a Revere Bowl." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 22 "restart; with(linalg):" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 24 "r:=(s,t)->[s,t,s^2+t^2];" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 64 "plot3d(r(s,t),s=-1..1,t=-sqrt(1-s^2)..sqrt(1-s
^2),axes=NORMAL); " }}}{PARA 0 "" 0 "" {TEXT -1 45 "We make the setup \+
for computing the integral." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
43 "drds:=diff(r(s,t),s);\ndrdt:=diff(r(s,t),t);" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 71 "term:=crossprod(drds,drdt);\ncross:=sqrt(term
[1]^2+term[2]^2+term[3]^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
52 "int(int(cross,t=-sqrt(1-s^2)..sqrt(1-s^2)),s=-1..1);" }}}{PARA 0 "
" 0 "" {TEXT -1 195 "If the above answer is not satisfactory, recall h
ow to integrate numerically. An alternate strategy is to draw this fig
ure using polar coordinates. That is, we parameterize the curve differ
ently." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart; with(lina
lg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "R:=(r,s)->[r,s,r^2]
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot3d([r,t,r^2],r=0..
1,t=0..2*Pi,coords=cylindrical);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 43 "dRdr:=diff(R(r,s),r);\ndRds:=diff(R(r,s),s);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "term:=crossprod(dRdr,dRds);
\ncross:=sqrt(term[1]^2+term[2]^2+term[3]^2);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 35 "int(int(cross*r,r=0..1),t=0..2*Pi);" }}}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Exercise for the student" }}
{PARA 0 "" 0 "" {TEXT -1 134 "You job, should you decide to take it, i
s to compute the surface area of a torus. In case you forget, here is \+
what a torus looks like." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "
restart: with(linalg): with(plots):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 75 "x:=(s,t)->(3+cos(s))*cos(t);\ny:=(s,t)->(3+cos(s))*si
n(t);\nz:=(s,t)->sin(s);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33
"r:=(s,t)->[x(s,t),y(s,t),z(s,t)];" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 85 "plot3d(r(s,t),s=-Pi..Pi,t=0..2*Pi,axes=NORMAL,color=B
LUE,\n orientation=[43,30]);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 0 }
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