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{SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 266 0 "" }{TEXT 
267 23 "Calculus IV with Maple\n" }{TEXT 268 32 "Copyright 2002, Dr. J
ack Wagner\n" }{TEXT -1 30 "j.wagner@intelligentsearch.com" }}}{EXCHG 
{PARA 4 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 60 "Lesson 11
: Volumes, Volume Integrals and Change of Variables" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 156 "Somewhere arou
nd the middle of the first year Calculus course we learn how to evalua
te integrals by a \"change of variables\" technique.  An integral such
 as " }{XPPEDIT 18 0 "Int(1/(1+x^2),x);" "6#-%$IntG6$*&\"\"\"F',&F'F'*
$%\"xG\"\"#F'!\"\"F*" }{TEXT -1 37 " is transformed by the substitutio
n, " }{XPPEDIT 18 0 "x = tan(u);" "6#/%\"xG-%$tanG6#%\"uG" }{TEXT -1 
17 " to the integral " }{XPPEDIT 18 0 "Int(1,u);" "6#-%$IntG6$\"\"\"%
\"uG" }{TEXT -1 33 ", from which we easily find that " }{XPPEDIT 18 0 
"Int(1/(1+x^2),x) = arctan(x)+K;" "6#/-%$IntG6$*&\"\"\"F(,&F(F(*$%\"xG
\"\"#F(!\"\"F+,&-%'arctanG6#F+F(%\"KGF(" }{TEXT -1 288 " .  The same i
dea may be applied to integrals in higher dimensions.  We may regard t
his procedure as a change of coordinates or as a reparametrization, wh
ichever viewpoint is more convenient.  Look at the problem of finding \+
the area of a circle from the perspective of surface integrals. " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 12 "E
xample 11.1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 33 "We seek to evaluate the function " }{XPPEDIT 18 0 "F;" 
"6#%\"FG" }{TEXT -1 1 ":" }{XPPEDIT 18 0 "proc (x, y) options operator
, arrow; 1 end;" "6#f*6$%\"xG%\"yG7\"6$%)operatorG%&arrowG6\"\"\"\"F+F
+F+" }{TEXT -1 83 " , over the plane area of the circle, considered as
 a surface, with radius a, i.e. " }{XPPEDIT 18 0 "Int(1,A);" "6#-%$Int
G6$\"\"\"%\"AG" }{TEXT -1 3 ".  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "restart:with(plots): with(LinearAlgebra): with(Vector
Calculus):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "A := Int(Int(
1, x), y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "We reparameterize  \+
the area in polar coordinates." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 34 "S := <r*cos(theta), r*sin(theta)>;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 35 "Now we know exactly how to proceed." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 29 "J := Jacobian(S, [r, theta]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "dA := simplify(Determinant(J));" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Area := Int(Int(dA, r=0..a)
, theta=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "value(
Area);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 447 "Now we extend this to higher dimensions.  Actually, what
 we were doing in the case of areas was specializing the more general \+
technique to the two dimensional case. By taking the parallelogram as \+
the element of area, we were regarding area as the volume of a flat pa
rallelopiped; or, if you will, the volume of a parallelopiped whose a \+
height is one and whose base is the area of the given parallelogram.  \+
Just as the element of area was taken as " }{TEXT 257 5 "dA = " }
{XPPEDIT 18 0 "sqrt(det(J[s]^T*J[s]));" "6#-%%sqrtG6#-%$detG6#*&)&%\"J
G6#%\"sG%\"TG\"\"\"&F,6#F.F0" }{MPLTEXT 1 0 1 " " }{TEXT -1 4 "= ||" }
{XPPEDIT 18 0 "J[s];" "6#&%\"JG6#%\"sG" }{TEXT -1 10 "||, where " }
{XPPEDIT 18 0 "J[s];" "6#&%\"JG6#%\"sG" }{TEXT -1 60 " is the Jacobian
 of the parameterization of the surface, so " }{TEXT 258 3 "dV=" }
{TEXT -1 2 "||" }{XPPEDIT 18 0 "J[v];" "6#&%\"JG6#%\"vG" }{TEXT -1 3 "
||=" }{XPPEDIT 18 0 "sqrt(det(J[v]^T*J[v]));" "6#-%%sqrtG6#-%$detG6#*&
)&%\"JG6#%\"vG%\"TG\"\"\"&F,6#F.F0" }{TEXT -1 8 "  where " }{XPPEDIT 
18 0 "J[v];" "6#&%\"JG6#%\"vG" }{TEXT -1 49 "  is the Jacobian of the \+
volume parameterization." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 259 12 "Example 11.2" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "For the sphere,  " }
{XPPEDIT 18 0 "x^2+y^2+z^2 = a^2;" "6#/,(*$%\"xG\"\"#\"\"\"*$%\"yGF'F(
*$%\"zGF'F(*$%\"aGF'" }{TEXT -1 38 " ,  a convenient coordinate change
 is:" }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "x = r*cos(phi)*cos(thet
a);" "6#/%\"xG*(%\"rG\"\"\"-%$cosG6#%$phiGF'-F)6#%&thetaGF'" }{TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "y = r*cos(phi)*sin(th
eta);" "6#/%\"yG*(%\"rG\"\"\"-%$cosG6#%$phiGF'-%$sinG6#%&thetaGF'" }}}
{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "z = r*sin(phi);" "6#/%\"zG*&%\"r
G\"\"\"-%$sinG6#%$phiGF'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "This \+
is expressed in vector form as: [" }{XPPEDIT 18 0 "r*cos(phi)*cos(thet
a);" "6#*(%\"rG\"\"\"-%$cosG6#%$phiGF%-F'6#%&thetaGF%" }{TEXT -1 2 ", \+
" }{XPPEDIT 18 0 "r*cos(phi)*sin(theta);" "6#*(%\"rG\"\"\"-%$cosG6#%$p
hiGF%-%$sinG6#%&thetaGF%" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "r*sin(phi);
" "6#*&%\"rG\"\"\"-%$sinG6#%$phiGF%" }{TEXT -1 1 "]" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 51 "restart: with(LinearAlgebra): with(Vector
Calculus):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "S := <r*cos(p
hi)*cos(theta), r*cos(phi)*sin(theta), r*sin(phi)>;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 34 "J := Jacobian(S, [r, theta, phi]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "v := simplify(sqrt(Determina
nt(Transpose(J).J)), symbolic); " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 62 "V := Int(Int(Int(v, r=0..a), theta=0..2*Pi), phi=-Pi/
2..Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(V);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "Fortunately, in practice, most of
 the volumes and surfaces we have to deal with have convenient and sym
metrical, parameterizations.  In general, the volume of a solid is:" }
}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "V = Int(Int(Int(` `,x),y),z);
" "6#/%\"VG-%$IntG6$-F&6$-F&6$%\"~G%\"xG%\"yG%\"zG" }{TEXT -1 3 " = " 
}{XPPEDIT 18 0 "Int(Int(Int(`||f'(u,v,w)||`,u),v),w);" "6#-%$IntG6$-F$
6$-F$6$%.|gr|grf'(u,v,w)|gr|grG%\"uG%\"vG%\"wG" }{TEXT -1 9 " , where \+
" }{TEXT 260 1 "f" }{TEXT -1 2 ": " }{XPPEDIT 18 0 "proc (U) options o
perator, arrow; V end;" "6#f*6#%\"UG7\"6$%)operatorG%&arrowG6\"%\"VGF*
F*F*" }{TEXT -1 31 " is the reparametrization map. " }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 637 "         Looked at in this way, the computatio
n of volume is nothing more than the computation of the reparameteriza
tion of the Cartesian coordinate volume element, transformed into new \+
coordinates representing the solid in question, between appropriate li
mits.  Looked at from the opposite perspective, reparameterization dis
torts volume by a factor equal to the determinant of the Jacobian matr
ix; that is, by a factor equal to the ratio of the volume of a transfo
rmed parallelepiped to an original parallelepiped.  Volume, of course,
 is meant in the most general sense and is intended to include area an
d arc length as special cases. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 261 12 "Example 11.3" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 383 "Consider an elli
pse in the plane x=5 with its center at (5, 0, 0), semi-major axis of \+
4 in the z direction and semi-minor axis of 2 in the x direction.  Rot
ate this about the z axis through an ellipse in the x-y plane with sem
i-major axis 10 in the y direction and semi-minor axis 5 in the x dire
ction. What is the volume of the resulting solid?\n\011This figure may
 be parameterized by " }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "[(5+2*
s*cos(theta))*cos(phi), 10+2*s*cos(theta)*sin(phi), 4*s*sin(theta)];" 
"6#7%*&,&\"\"&\"\"\"*(\"\"#F'%\"sGF'-%$cosG6#%&thetaGF'F'F'-F,6#%$phiG
F',&\"#5F'**F)F'F*F'-F,6#F.F'-%$sinG6#F1F'F'*(\"\"%F'F*F'-F86#F.F'" }
{TEXT -1 13 " , s = 0..1, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }
{TEXT -1 6 " = 0.." }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }
{TEXT -1 2 ", " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 6 " = 0.." 
}{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 " " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "        It is an elliptical dough
nut with an elliptical cross section.  Note that we are plotting only \+
the surface, so we leave out the s parameter which provides the volume
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "restart: with(plots): \+
with(LinearAlgebra):with(VectorCalculus):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 75 "t := <(5+2*cos(theta))*cos(phi), (10+2*cos(theta))*
sin(phi), 4*sin(theta)>;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
102 "plot3d(t, theta=0..2*Pi, phi=0..2*Pi, axes=normal, scaling=constr
ained, color=blue, style=\nwireframe);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 90 "Now we calculate.  We must include the s parameter since \+
we are now concerned with volume." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 82 "T := <(5+2*s*cos(theta))*cos(phi), (10+2*s*cos(theta)
)*sin(phi), 4*s*sin(theta)>;\011" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "J := Jacobian(T, [theta, phi, s]);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 60 "dV := sqrt(simplify(Determinant(Transpose
(J).J)), symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "Volu
me := int(int(int(dV, theta=0..2*Pi), phi=0..2*Pi), s=0..1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 238 "The functions commonly integrated
 over volumes are densities or changes in density that vary from point
 to point in the volume.  The result is then the entire quantity or th
e net change in density of the physical parameter being studied. " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 12 "E
xample 11.4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Suppose that a sp
here of radius a contains a material whose density obeys an inverse sq
uare law relative to the center of the sphere: i.e. " }{XPPEDIT 18 0 "
D = K/(r^2);" "6#/%\"DG*&%\"KG\"\"\"*$%\"rG\"\"#!\"\"" }{TEXT -1 139 "
  where r is the distance from the center.  What is the total quantity
 of material in the sphere.  With the methodology in hand we compute \+
" }{XPPEDIT 18 0 "Int(D,V);" "6#-%$IntG6$%\"DG%\"VG" }{TEXT -1 1 "." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "restart: with(plots): with
(LinearAlgebra): with(VectorCalculus):" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "We will use the geographi
c parametrization of the sphere." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "S := <r*cos(phi)*cos(theta), r*cos(phi)*sin(theta), r
*sin(phi)>;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "rho := K/r^2
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "J[v] := Jacobian(S, [r
, phi, theta]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "dV := si
mplify(sqrt(Determinant(Transpose(J[v]).J[v])), symbolic);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "Int(Int(Int(rho*dV, phi=-Pi/2..Pi/2
), theta=0..2*Pi), r=0..a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
9 "value(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "If a number of e
xpressions are to be reparameterized it pays to define a simple functi
on to do the job." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 263 12 "Example 11.5" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "We suppose that we have a measure
ment of some physical quantity, F, in a cylindrical container of radiu
s a, and height h, that depends on the radial distance from the center
 of the cylinder and the height above the cylinder's base. " }
{XPPEDIT 18 0 "F = (x^2+y^2)/(z+1);" "6#/%\"FG*&,&*$%\"xG\"\"#\"\"\"*$
%\"yGF)F*F*,&%\"zGF*F*F*!\"\"" }{TEXT -1 313 "  We want to integrate t
his quantity over the entire cylinder.  We take advantage of the symme
tries present and locate the base of the cylinder on the z=0 plane and
 the center of the base at the origin.  Then, to avoid the problems of
 negative trigonometric functions, we work in the first quadrant of th
e plane. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "restart:with(p
lots):with(LinearAlgebra): with(VectorCalculus):" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "We start by defini
ng a function to do the work of reparameterizing into cylindrical coor
dinates." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f := g->simplif
y(subs(\{x=r*cos(theta), y=r*sin(theta), z=z\}, g)):" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Define the elem
ents of the problem." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "F :
= r/(z+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "cyl := <x, sq
rt(r^2-x^2), z>;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "We have defi
ned only one half of the cylinder.  Now we reparameterize the half-cyl
inder. We need a couple of assume statements so that Maple doesn't pri
nt a bunch of messy csgn symbols." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "interface(showassumed=0):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 33 "assume(sin(theta)>0):assume(r>0):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "C := f(cyl);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 34 "The calculation proceeds as usual." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 32 "J := Jacobian(C, [r, theta, z]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "dV := simplify(sqrt(Determinant(Tra
nspose(J).J)), symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "FdV := F*dV;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "The integrati
on is taken over the part of the cylinder lying over the first quadran
t. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Int(Int(Int(FdV, the
ta=0..Pi/2), z=0..h), r=0..a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 11 "4*value(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "It is alw
ays a good idea to test new methods against known results.  We will fi
nd the volume of a solid of revolution." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 12 "Example 11.5" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Given a c
urve parameterized by " }{XPPEDIT 18 0 "f(t) = [x(t), y(t)];" "6#/-%\"
fG6#%\"tG7$-%\"xG6#F'-%\"yG6#F'" }{TEXT -1 95 " , find the volume of t
he solid formed by rotating the curve around the x axis from x=a to x=
b." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "restart: with(plots):
 with(LinearAlgebra): with(VectorCalculus):" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Transformation to cylin
drical coordinates about the x-axis" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 50 "V := <x(t), s*y(t)*cos(theta), s*y(t)*sin(theta)>;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "J := Jacobian(V, [s, theta
, t]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "dV := simplify(sq
rt(Determinant(Transpose(J).J)), symbolic);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 59 "Volume := Int(Int(Int(dV, s=0..1), t=a..b), theta=
0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "simplify(value
(%));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "A result familiar from t
he calculus of a single variable. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 265 8 "Practice" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 66 "1. Find the volume of the solid whose sur
face is parameterized by:" }}{PARA 0 "" 0 "" {TEXT -1 7 "       " }
{XPPEDIT 18 0 "[(phi+2*cos(theta))*cos(phi), (phi+2*cos(theta))*sin(ph
i), 2*sin(theta)];" "6#7%*&,&%$phiG\"\"\"*&\"\"#F'-%$cosG6#%&thetaGF'F
'F'-F+6#F&F'*&,&F&F'*&F)F'-F+6#F-F'F'F'-%$sinG6#F&F'*&F)F'-F66#F-F'" }
{TEXT -1 4 " ,  " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "Pi .. 4*Pi;" "6#;%#PiG*&\"\"%\"\"\"F$F'" }{TEXT -1 3 " \+
, " }{XPPEDIT 18 0 "theta = 0 .. 2*Pi;" "6#/%&thetaG;\"\"!*&\"\"#\"\"
\"%#PiGF)" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "2. F
ind the volume of the solid enclosed in the surface; " }{XPPEDIT 18 0 
"z = x^4+y^4;" "6#/%\"zG,&*$%\"xG\"\"%\"\"\"*$%\"yGF(F)" }{TEXT -1 3 "
,  " }{XPPEDIT 18 0 "-1 <= x;" "6#1,$\"\"\"!\"\"%\"xG" }{TEXT -1 2 ", \+
" }{XPPEDIT 18 0 "y <= 1;" "6#1%\"yG\"\"\"" }{TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 41 "3. Find the volume of the elliptic cone: \+
" }{XPPEDIT 18 0 "z^2 = 5*x^2+y^2;" "6#/*$%\"zG\"\"#,&*&\"\"&\"\"\"*$%
\"xGF&F*F**$%\"yGF&F*" }{TEXT -1 27 "  whose height is 5 units. " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "4. A cylinder of elliptical cross
 section, 6x10 units wide and 100 units high contains a gas whose dens
ity is " }{XPPEDIT 18 0 "2*exp(-.1*d);" "6#*&\"\"#\"\"\"-%$expG6#,$*&-
%&FloatG6$F%!\"\"F%%\"dGF%F.F%" }{TEXT -1 102 "where d is the distance
 from the base of the cylinder. Find the total mass of the gas in the \+
cylinder." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 243 "5. Given a cylinder
 of radius 2 units and height h, and a material whose concentration  i
s proportional to the square of the distance from the center of the ba
se of the cylinder, find the total amount of the material contained in
 the cylinder." }}}}{MARK "0 0 2" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 
}{PAGENUMBERS 0 1 2 33 1 1 }