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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257
"" 0 "" {TEXT -1 73 "Lesson 5: Applications of Integration 3: Area of
a Surface of Revolution" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 31 "Area
of a Surface of Revolution" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Re
call that a surface of revolution is generated when the graph of a fun
ction is revolved around the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT
-1 6 "-axis:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f := x-> 1 + 4*x^2 - x^3;"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "plot(f, -1..2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "plot3d([x,f(x)*cos(t), f(x)*
sin(t)], x=-1..2, t=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Question 1" }}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 18 "Define a function " }{XPPEDIT 18 0 "g;" "
6#%\"gG" }{TEXT -1 4 " by " }{XPPEDIT 18 0 "g(x) = sqrt(9-x^2);" "6#/-
%\"gG6#%\"xG-%%sqrtG6#,&\"\"*\"\"\"*$F'\"\"#!\"\"" }{TEXT -1 48 " . W
hat surface is generated when the graph of " }{XPPEDIT 18 0 "g;" "6#%
\"gG" }{TEXT -1 9 " between " }{XPPEDIT 18 0 "x = -3;" "6#/%\"xG,$\"\"
$!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x = 3;" "6#/%\"xG\"\"$" }
{TEXT -1 24 " is revolved around the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }
{TEXT -1 37 "-axis? Confirm your answer by using " }{TEXT 256 6 "plot
3d" }{TEXT -1 68 " to draw the surface. (Hint: cut-and-paste from the
example above!)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }{TEXT 262 9 "Solution." }{TEXT -1 0 "" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 24 "g := x-> sqrt(9 - x^2) ;" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 10 "The curve " }{XPPEDIT 18 0 "y = sqrt(9-x^
2);" "6#/%\"yG-%%sqrtG6#,&\"\"*\"\"\"*$%\"xG\"\"#!\"\"" }{TEXT -1 33 "
is the upper half of the circle " }{XPPEDIT 18 0 "x^2+y^2 = 9;" "6#/,
&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(\"\"*" }{TEXT -1 93 ", so it is a semi-c
ircle of radius 3, centred at the origin. When it is revolved around \+
the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 42 "-axis, we should get
a sphere of radius 3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "p
lot3d([x,g(x)*cos(t), g(x)*sin(t)], x=-3..3, t=0..2*Pi, axes=boxed, li
ghtmodel=light1);" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "How should
we proceed in order to find the area of such a surface? Naturally, w
e begin by dividing the interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG
%\"bG" }{TEXT -1 305 " over which we are working into a certain number
of subintervals, and finding an approximation to the area of the part
of the surface corresponding to one subinterval. This is the most co
mplicated approximation we have seen so far, and it cannot be done in \+
one step, but the first step is to approximate " }{XPPEDIT 18 0 "f;" "
6#%\"fG" }{TEXT -1 188 " on each subinterval by a line segment, exactl
y as we did when we computed arclengths. Here is the procedure from t
he arclength worksheet that returns a piecewise-linear approximation t
o " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 19 "pl := proc(f,a,b,n)" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 20 " local u,v,k;" }}{PARA 0 "> " 0 "" {MPLTEXT 1
0 38 " k := 1 + floor(n*(x-a)/(b-a));" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 31 " u := a + (k-1)*(b-a)/n;" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 " v := a + k*(b-a)/n;" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 53 " unapply(f(u) + (f(v) - f(u))/(v-u)*(x-u), x);
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 " end:" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 26 "Working with the function " }{XPPEDIT 18 0 "f;" "6
#%\"fG" }{TEXT -1 78 " we defined above, we can plot the piecewise-lin
ear approximations, and (with " }{TEXT 257 6 "plot3d" }{TEXT -1 56 ") \+
the surfaces we get by revolving these approximations." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plot([f,pl(f,-1,2,5)], -1..2, thick
ness=2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "plot3d([x,pl(f
,-1,2,5)(x)*cos(t),pl(f,-1,2,5)(x)*sin(t)], x=-1..2, t=0..2*Pi, axes=b
oxed, lightmodel=light1);" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Que
stion 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "Experiment with differ
ent numbers of subintervals to see how the piecewise-linear surfaces a
pproach the exact surface of revolution. Repeat with the function " }
{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 17 " from Question 1." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 263 9 "
Solution." }{TEXT -1 51 " Here are the piecewise-linear approximation
s for " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 54 ", with 3, 6, and 1
0 subintervals, revolved around the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }
{TEXT -1 6 "-axis:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "plot
3d([x,pl(f,-1,2,3)(x)*cos(t),pl(f,-1,2,3)(x)*sin(t)], x=-1..2, t=0..2*
Pi, axes=boxed, lightmodel=light1);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 109 "plot3d([x,pl(f,-1,2,6)(x)*cos(t),pl(f,-1,2,6)(x)*sin
(t)], x=-1..2, t=0..2*Pi, axes=boxed, lightmodel=light1);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "plot3d([x,pl(f,-1,2,10)(x)*cos(t),
pl(f,-1,2,10)(x)*sin(t)], x=-1..2, t=0..2*Pi, axes=boxed, lightmodel=l
ight1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 231 "With 10 subintervals,
it is already hard to tell the approximate surface apart from the exa
ct one, except perhaps at the narrow neck. (In a moment, we will comp
ute the relevant areas numerically and see how close they really are.)
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "Here \+
are the same approximations for " }{XPPEDIT 18 0 "g;" "6#%\"gG" }
{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "plot3d([x
,pl(g,-3,3,3)(x)*cos(t),pl(g,-3,3,3)(x)*sin(t)], x=-3..3, t=0..2*Pi, a
xes=boxed, lightmodel=light1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 109 "plot3d([x,pl(g,-3,3,6)(x)*cos(t),pl(g,-3,3,6)(x)*sin(t)], x=-
3..3, t=0..2*Pi, axes=boxed, lightmodel=light1);" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 111 "plot3d([x,pl(g,-3,3,10)(x)*cos(t),pl(g,-3,3,
10)(x)*sin(t)], x=-3..3, t=0..2*Pi, axes=boxed, lightmodel=light1);" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "Once again, 10 subintervals see
ms to give a very good approximation over most of the interval. In th
is case, the approximation is not so good near the endpoints, " }
{XPPEDIT 18 0 "x = 3;" "6#/%\"xG\"\"$" }{TEXT -1 5 " and " }{XPPEDIT
18 0 "x = -3;" "6#/%\"xG,$\"\"$!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 9 "Question
." }{TEXT -1 359 " Suppose you wanted to find a good numerical approx
imation to the area of the sphere with as little compuatation as possi
ble. How would you do it? (Just taking more subintervals is not nece
ssarily the answer: your approximation may already be good enough awa
y from the endpoints, so it could be a waste of time and effort to tak
e more subintervals there.)" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 334 "Now for the hardest part: how do we co
mpute, or at least approximate, the surface areas of the piecewise-l
inear surfaces? It is enough to concentrate on one subinterval; if we
can solve that part of the problem, we will do it for each subinterva
l and sum the results. On each subinterval, we are rotating a line se
gment about the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 145 "-axis, \+
which will give a ring. For example, here is a picture of the ring we
get from the 4th subinterval of the 5-subinterval approximation to "
}{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 94 "plot3d([x,pl(f,-1,2,5)(x)*cos(t),pl(f,-1,2,5)(x)
*sin(t)], x=0.8..1.4, t=0..2*Pi, axes=normal);" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 1 "(" }{TEXT 258 5 "Maple" }{TEXT -1 40 " changes the sc
ale, but notice that the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 97
"-axis in this picture goes from 0.8 to 1.4.) Now imagine cutting thi
s ring open parallel to the " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1
430 "-axis, and laying it out flat. It will be approximately rectangu
lar (especially if the subinterval is very short); the length of the r
ectangle will be the circumference of the ring, and its width will be \+
the length of the line segment we rotated. The length of the line seg
ment can be found from the distance formula, exactly as we did when we
discussed arclength. The radius of the ring is given approximately b
y the value of " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 76 " at the r
ight-hand endpoint of the subinterval, and so its circumference is " }
{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 81 " times th
is. Putting everything together, we approximate the area of the ring \+
by" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "2*Pi*f(a+k*(b-a)/n)*sqrt(((b-a)
/n)^2+(f(a+k*(b-a)/n)-f(a+(k-1)*(b-a)/n))^2);" "6#**\"\"#\"\"\"%#PiGF%
-%\"fG6#,&%\"aGF%*(%\"kGF%,&%\"bGF%F+!\"\"F%%\"nGF0F%F%-%%sqrtG6#,&*$*
&,&F/F%F+F0F%F1F0F$F%*$,&-F(6#,&F+F%*(F-F%,&F/F%F+F0F%F1F0F%F%-F(6#,&F
+F%*(,&F-F%F%F0F%,&F/F%F+F0F%F1F0F%F0F$F%F%" }{TEXT -1 2 " ." }}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Question 3" }}{EXCHG {PARA 0 ""
0 "" {TEXT -1 41 "The following procedure takes a function " }
{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 14 ", an interval " }{XPPEDIT
18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 15 ", and a number " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 18 ", and returns the " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 218 "-subinterval approximation
to the area of the surface of revolution. Use it to find (decimal ap
proximations to) the 5, 10 and 100-subinterval approximations to the a
rea obtained by rotating the graph of the function " }{XPPEDIT 18 0 "
f;" "6#%\"fG" }{TEXT -1 28 " defined earlier around the " }{XPPEDIT
18 0 "x;" "6#%\"xG" }{TEXT -1 33 "-axis. Repeat with the function " }
{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 17 " from Question 1." }}{PARA
0 "> " 0 "" {MPLTEXT 1 0 27 "approxarea := proc(f,a,b,n)" }}{PARA 0 ">
" 0 "" {MPLTEXT 1 0 10 " local l;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
15 " l := (b-a)/n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 " sum( 2*Pi*
f(a + k*l) *" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 " sqrt( l^2 + (f(
a + k*l) - f(a + (k-1)*l))^2 ), k=1..n);" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 6 " end:" }{TEXT -1 0 "" }{TEXT 265 0 "" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 266 9 "Solution." }{TEXT -1 49 " \+
To get the decimal approximations, we nust use " }{TEXT 267 5 "evalf
" }{TEXT -1 25 ". (You can try omitting " }{TEXT 268 5 "evalf" }
{TEXT -1 41 " and seeing the exact number computed by " }{TEXT 269 10
"approxarea" }{TEXT -1 31 "; it is not very illuminating!)" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "evalf(approxarea(f,-1,2,5)); evalf(
approxarea(f,-1,2,10)); evalf(approxarea(f,-1,2,100));" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 475 "Recall that when we drew the pictures, w
e said that 10 subintervals seemed to produce a good approximation to \+
the surface of revolution. That is partially confirmed by the numbers
here: there is quite a big change when we move from 5 subintervals to
10, but a smaller difference when we go from 10 subintervals to 100. \+
Even so, the 2nd significant figure changes in going from 10 to 100 i
ntervals: if you want to be confident that your answer is correct to w
ithin 1% (say)," }}{PARA 0 "" 0 "" {TEXT -1 98 "you should definitely \+
try with more subintervals and see if the approximation continues to c
hange." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88
"Similar comments can be made about the approximations we get starting
with the function " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 1 ":" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "evalf(approxarea(g,-3,3,5));
evalf(approxarea(g,-3,3,10)); evalf(approxarea(g,-3,3,100));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 101 "The final step is to recognise that our approximate sums
are Riemann sums for some integral, and let " }{XPPEDIT 18 0 "proc (n
) options operator, arrow; infinity end;" "6#f*6#%\"nG7\"6$%)operatorG
%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 193 ". The Riemann sums will \+
converge to the integral, and we can conclude that the integral must g
ive the exact surface area. There is a slight catch this time: the su
ms we have written down are " }{TEXT 259 3 "not" }{TEXT -1 112 " obvio
usly Riemann sums. The difficulty is that we approximated the radius \+
of each ring by taking the value of " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 207 " at the right-hand endpoint of the subinterval. This is
good enough for numerical work, but as we will discuss in class, it i
s not good enough for the theory. Nevertheless, our sums do in fact c
onverge as " }{XPPEDIT 18 0 "proc (n) options operator, arrow; infinit
y end;" "6#f*6#%\"nG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }
{TEXT -1 17 ", to the integral" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "int
(2*Pi*f(x)*sqrt(1+D(f)(x)^2),x = a .. b);" "6#-%$intG6$**\"\"#\"\"\"%#
PiGF(-%\"fG6#%\"xGF(-%%sqrtG6#,&F(F(*$--%\"DG6#F+6#F-F'F(F(/F-;%\"aG%
\"bG" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 88 "so we are still \+
justified in concluding that this integral gives the exact surface are
a." }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Question 4" }}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 78 "(a) Compute the exact area of the surfac
e generated by rotating the function " }{XPPEDIT 18 0 "g;" "6#%\"gG" }
{TEXT -1 27 " from Question 1 about the " }{XPPEDIT 18 0 "x;" "6#%\"xG
" }{TEXT -1 46 "-axis, by evaluating the integral above (with " }
{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 13 " replaced by " }{XPPEDIT
18 0 "g;" "6#%\"gG" }{TEXT -1 155 "). Compare with the approximations
you found in Question 3. How many subintervals are necessary for th
e approximation to be within 1% of the exact area." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 272 9 "Solution.
" }{TEXT -1 105 " In principal, the first part of the question just i
nvolves translating the integral formula above into " }{TEXT 273 5 "Ma
ple" }{TEXT -1 76 " syntax. The only possible difficulty is that, acc
ording to its help page, " }{TEXT 274 5 "Maple" }{TEXT -1 3 "'s " }
{TEXT 275 3 "int" }{TEXT -1 107 " command integrates expressions and n
ot functions, so you have to use expression syntax. (You can replace \+
" }{TEXT 276 7 "D(g)(x)" }{TEXT -1 4 " by " }{TEXT 277 12 "diff(g(x),x
)" }{TEXT -1 14 " if you like.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 44 "int(2*Pi*g(x)*sqrt(1 + D(g)(x)^2), x=-3..3);" }}}{EXCHG {PARA
0 "" 0 "" {TEXT -1 0 "" }{TEXT 270 9 "Question." }{TEXT -1 5 " Is " }
{XPPEDIT 18 0 "36*Pi;" "6#*&\"#O\"\"\"%#PiGF%" }{TEXT -1 76 " the corr
ect answer for the surface area obtained by revolving the graph of " }
{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 12 " around the " }{XPPEDIT 18
0 "x;" "6#%\"xG" }{TEXT -1 12 "-axis? Why?" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "To compare this exact area with
our numerical results from Question 3, we should use " }{TEXT 271 5 "
evalf" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "ev
alf(36*Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "If we want our ap
proximation to be within 1% of this value, it must be between the foll
owing two values." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "evalf(
36*Pi - 0.01*36*Pi); evalf(36*Pi + 0.01*36*Pi); " }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 173 "Looking back at Question 3, 10 subintervals are n
ot sufficient, but 100 certainly are. With a bit of trial-and-error, \+
you can find that in fact 13 intervals will just work." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(approxarea(g,-3,3,13));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 4 "(b) " }{TEXT 260 47 "(Save the worksheet before trying thi
s part!!!)" }{TEXT -1 35 " Repeat part (a) for our function " }
{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 3 ". " }{TEXT 261 5 "Maple" }
{TEXT -1 356 " should be able to do this, but on my office machine it \+
took 400 seconds to fail to come up with an answer. Whether or not yo
u choose to try this for yourself, what do you conclude about the usef
ulness of the integral formula for surface area? If you needed to fin
d the surface area of a reasonably complicated surface of revolution, \+
how would you do it?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }{TEXT 278 9 "Solution." }{TEXT -1 67 " If you try th
e solution used in part (a), replacing the function " }{XPPEDIT 18 0 "
g;" "6#%\"gG" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 30 ", you will probably find that " }{TEXT 279 5 "Maple" }
{TEXT -1 58 " cannot compute the integral. You might think that using
" }{TEXT 280 15 "evalf(int(...))" }{TEXT -1 55 " would work, but it d
oesn't, because with this command," }{TEXT 281 5 "Maple" }{TEXT -1
257 " attempts to work out the integral exactly (and fails!) and only \+
then computes the decimal approximation. Your conclusion should be th
at even though we have an exact integral formula for the area, it is u
seless because we can't evaluate the integral. (If " }{TEXT 282 5 "Ma
ple" }{TEXT -1 204 " can't, we certainly can't!) Therefore, to comput
e a surface area (or, more generally, to compute any fairly complicate
d integral) we may have to content ourselves with a suitably accurate \+
approximation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 66 "Actually, we can do more. If you have been suitably impr
essed by " }{TEXT 283 5 "Maple" }{TEXT -1 208 "'s power to this point
, you may be surprised that it cannot apparently evaluate an integral,
even numerically. It can, of course, and one way to force it to do s
o is to use the unevaluated integral command " }{TEXT 284 3 "Int" }
{TEXT -1 82 ". (Note the upper-case I.) You can read about this comm
and on the help page for " }{TEXT 285 3 "int" }{TEXT -1 72 ", but to s
ee what it does, compare the output to the following commands:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "int(x^2, x=0..3);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Int(x^2, x=0..3);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Note that " }{TEXT 286 3 "int" }
{TEXT -1 42 " attempts to evaluate the integral, while " }{TEXT 287 3
"Int" }{TEXT -1 87 " returns the integral unevaluated. This will get \+
around our problem with the function " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 13 ": we can ask " }{TEXT 288 5 "Maple" }{TEXT -1 4 " to " }
{TEXT 289 5 "evalf" }{TEXT -1 54 " the integral without attempting to \+
evaluate it first:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "farea
:= evalf(Int(2*Pi*f(x)*sqrt(1 + D(f)(x)^2), x=-1..2));" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 65 "To be within 1% of this answer, we must b
e in the interval below:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48
"evalf([farea - 0.01*farea, farea + 0.01*farea]);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 87 "Looking back at Question 3, even 10 subintervals s
eems to be sufficient fo 1% accuracy:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 29 "evalf(approxarea(f,-1,2,10));" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 222 "The savings in effort from using fewer intervals is
probably not worth the effort spent in looking for the smallest numbe
r that will work, but if you like you can check that 7 subintervals gi
ves 1% accuracy and 6 does not." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 290 27 "Final comment and Questio
n." }{TEXT -1 7 " Using" }{TEXT 291 0 "" }{TEXT -1 1 " " }{TEXT 292
3 "Int" }{TEXT -1 5 " and " }{TEXT 293 5 "evalf" }{TEXT -1 124 " toget
her as we did here is really something of a kludge: it works, but it s
kirts around the real question: how, exactly is " }{TEXT 294 5 "Maple
" }{TEXT -1 102 " managing to evaluate an integral numerically, if it \+
cannot compute the exact answer? What method do " }{TEXT 295 3 "you"
}{TEXT -1 19 " think it is using?" }}}}}}{MARK "0 0 0" 0 }{VIEWOPTS 1
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }