1
00:00:00,000 --> 00:00:08,750
PROFESSOR: Welcome
back to recitation.

2
00:00:08,750 --> 00:00:11,320
In this video I want us to
look at the following problem.

3
00:00:11,320 --> 00:00:14,010
We're going to let f
of x equal 1 over x.

4
00:00:14,010 --> 00:00:16,290
And I want to consider
the solid generated

5
00:00:16,290 --> 00:00:19,860
by rotating f of x about
the x-axis between x

6
00:00:19,860 --> 00:00:22,090
equal 1 and x equal infinity.

7
00:00:22,090 --> 00:00:24,840
I want you to find the area
of a cross-sectional slice,

8
00:00:24,840 --> 00:00:28,889
and then I want you to find
the volume of the solid.

9
00:00:28,889 --> 00:00:31,180
In the cross-sectional slice,
the one I'm interested in

10
00:00:31,180 --> 00:00:33,570
is the one that you
see in the xy-plane.

11
00:00:33,570 --> 00:00:36,170
So my suggestion to you is
that right away draw yourself

12
00:00:36,170 --> 00:00:41,090
a picture, just at least a
rough sketch of the curve y

13
00:00:41,090 --> 00:00:43,770
equals f of x, and then what
that cross-sectional slice

14
00:00:43,770 --> 00:00:46,144
will look like and
start off from there.

15
00:00:46,144 --> 00:00:48,060
And take a while to work
on that and then I'll

16
00:00:48,060 --> 00:00:49,510
be back and show you what I did.

17
00:00:58,450 --> 00:00:59,600
OK welcome back.

18
00:00:59,600 --> 00:01:02,290
Well, hopefully you were
able to make some headway

19
00:01:02,290 --> 00:01:05,090
on this problem, and maybe you
found some interesting things.

20
00:01:05,090 --> 00:01:06,150
Hopefully you did.

21
00:01:06,150 --> 00:01:09,370
And we will see what I find
and if they are interesting,

22
00:01:09,370 --> 00:01:10,930
which I think they are.

23
00:01:10,930 --> 00:01:12,004
So let's start off.

24
00:01:12,004 --> 00:01:13,670
I told you the first
thing you should do

25
00:01:13,670 --> 00:01:14,710
is get a rough picture.

26
00:01:14,710 --> 00:01:16,280
So I'm going to
draw a rough picture

27
00:01:16,280 --> 00:01:18,590
and make sure that
those match up well,

28
00:01:18,590 --> 00:01:21,170
that my picture
matches up with yours.

29
00:01:21,170 --> 00:01:26,600
So let me give you the xy-plane.

30
00:01:26,600 --> 00:01:31,530
And then y equals 1 over x is
a curve that at x equal 1--

31
00:01:31,530 --> 00:01:34,110
I'm going to make
y equal 1 up here.

32
00:01:34,110 --> 00:01:35,510
y equal 1 up there. x equal 1.

33
00:01:35,510 --> 00:01:38,590
So when x is 1, y is 1.

34
00:01:38,590 --> 00:01:42,400
And then it's going
to decay down.

35
00:01:42,400 --> 00:01:44,740
As x goes to infinity it decays

36
00:01:44,740 --> 00:01:46,190
So it looks something like that.

37
00:01:46,190 --> 00:01:48,260
And then I actually
should have moved this 1

38
00:01:48,260 --> 00:01:50,820
because I'm going to
need that side to get

39
00:01:50,820 --> 00:01:52,370
my cross-sectional slice.

40
00:01:52,370 --> 00:01:54,610
So my cross sectional
slice is going

41
00:01:54,610 --> 00:01:58,060
to come down here and
look something like this.

42
00:01:58,060 --> 00:02:00,160
Oops, that's supposed
to be symmetric.

43
00:02:00,160 --> 00:02:02,290
That's still supposed to
be something like this.

44
00:02:02,290 --> 00:02:05,030
It's supposed to be
symmetric about the x-axis.

45
00:02:05,030 --> 00:02:10,250
And the cross-sectional
slice is the area of that.

46
00:02:10,250 --> 00:02:13,950
Now, if you remember
what you saw in lecture.

47
00:02:13,950 --> 00:02:16,750
You saw, I believe,
that the integral from 1

48
00:02:16,750 --> 00:02:23,270
to infinity of 1
over x dx diverges,

49
00:02:23,270 --> 00:02:28,150
because you end up with
log evaluated at infinity

50
00:02:28,150 --> 00:02:30,420
and that is infinite and
so you wind up with this.

51
00:02:30,420 --> 00:02:31,960
This actually diverges.

52
00:02:31,960 --> 00:02:35,570
That was the area of the
top part from 0 to f of x.

53
00:02:35,570 --> 00:02:37,620
So of course if I
double that, I'm still

54
00:02:37,620 --> 00:02:38,950
going to get that it diverges.

55
00:02:38,950 --> 00:02:45,570
So in fact the cross sectional
slice area, CSS area,

56
00:02:45,570 --> 00:02:46,393
is infinite.

57
00:02:51,020 --> 00:02:53,400
OK, so that's one part.

58
00:02:53,400 --> 00:02:55,330
The other part is
to find the volume

59
00:02:55,330 --> 00:02:57,360
of the solid of revolution.

60
00:02:57,360 --> 00:03:02,370
So the second part was take
this piece from 0 up to f of x,

61
00:03:02,370 --> 00:03:06,280
rotate it around the x-axis
and compute the volume.

62
00:03:06,280 --> 00:03:08,380
Now we know how to
compute the volume.

63
00:03:08,380 --> 00:03:12,090
This is the disc method
that I'm going to use here.

64
00:03:12,090 --> 00:03:13,880
And I will write out
what we need to do

65
00:03:13,880 --> 00:03:16,310
and then we'll look at
what the integral gives us.

66
00:03:16,310 --> 00:03:22,300
So for the disc method, so let's
say this is the volume part.

67
00:03:22,300 --> 00:03:24,300
So the volume, it's going
to be the disc method.

68
00:03:24,300 --> 00:03:26,460
So I need to do pi r squared.

69
00:03:26,460 --> 00:03:28,680
And in this case I'm
integrating from 1 to infinity,

70
00:03:28,680 --> 00:03:31,439
so I need pi r squared dx.

71
00:03:31,439 --> 00:03:32,980
I'm going to be
integrating something

72
00:03:32,980 --> 00:03:35,894
that's pi r squared dx.

73
00:03:35,894 --> 00:03:37,560
And I know the bounds
are 1 to infinity,

74
00:03:37,560 --> 00:03:41,100
and so I need r in terms
of, as a function of x.

75
00:03:41,100 --> 00:03:43,730
That's fairly easy. r is
just the measure from 0 up

76
00:03:43,730 --> 00:03:44,870
to f of x.

77
00:03:44,870 --> 00:03:46,814
So that's just 1 over x.

78
00:03:46,814 --> 00:03:48,230
So this is actually
the integral--

79
00:03:48,230 --> 00:03:50,830
I'm going to pull
the pi out-- of 1

80
00:03:50,830 --> 00:03:57,040
over x squared from
1 to infinity dx.

81
00:03:57,040 --> 00:03:59,100
OK so how do I do this integral?

82
00:03:59,100 --> 00:04:03,420
Well you were shown in
class that in actuality this

83
00:04:03,420 --> 00:04:07,270
is the limit, as some value
up here goes to infinity,

84
00:04:07,270 --> 00:04:10,574
of this integral,
but we're just going

85
00:04:10,574 --> 00:04:12,990
to do it the shorthand way
that he also mentioned in class

86
00:04:12,990 --> 00:04:14,650
and keep the infinity around.

87
00:04:14,650 --> 00:04:16,180
So we can keep
that as our bounds,

88
00:04:16,180 --> 00:04:21,120
knowing we're taking limits as
this thing goes to infinity.

89
00:04:21,120 --> 00:04:25,550
So the integral of 1 over x
squared is negative 1 over x.

90
00:04:25,550 --> 00:04:32,612
So I get minus pi times 1 over
x, evaluated at 1 in infinity.

91
00:04:32,612 --> 00:04:34,070
I can just check
that if I need to,

92
00:04:34,070 --> 00:04:36,460
but this is x to the minus
1 and its derivative is

93
00:04:36,460 --> 00:04:37,950
negative x to the minus 2.

94
00:04:37,950 --> 00:04:39,959
So I needed that negative there.

95
00:04:39,959 --> 00:04:41,000
And then I evaluate this.

96
00:04:41,000 --> 00:04:43,900
Well at infinity, 1 over
x-- as x goes to infinity,

97
00:04:43,900 --> 00:04:45,620
this first part goes to 0.

98
00:04:45,620 --> 00:04:51,700
So this is 0 minus
negative pi times 1 over 1.

99
00:04:51,700 --> 00:04:52,830
So negative pi times 1.

100
00:04:52,830 --> 00:04:56,960
So this is just pi.

101
00:04:56,960 --> 00:05:00,440
OK, so hopefully that kind of
blows your mind a little bit,

102
00:05:00,440 --> 00:05:02,200
that you could have
something where

103
00:05:02,200 --> 00:05:05,360
this cross-sectional slice
is infinite, but in fact

104
00:05:05,360 --> 00:05:07,710
if you look at it the way
we computed the volume,

105
00:05:07,710 --> 00:05:11,260
we have in fact a finite volume.

106
00:05:11,260 --> 00:05:12,719
So I don't know
what else I'm going

107
00:05:12,719 --> 00:05:14,801
to say about that except
I think it's really cool.

108
00:05:14,801 --> 00:05:16,410
And you can think
about why that is.

109
00:05:16,410 --> 00:05:18,455
And in fact you
might want to notice

110
00:05:18,455 --> 00:05:21,140
that we had to--
we computed things

111
00:05:21,140 --> 00:05:22,670
in terms of
cross-sectional slices

112
00:05:22,670 --> 00:05:24,003
coming from the other direction.

113
00:05:24,003 --> 00:05:28,080
So we looked at these
cross-sectional slices

114
00:05:28,080 --> 00:05:31,860
and we got our-- we showed
our volume was finite there.

115
00:05:31,860 --> 00:05:34,030
So we had a sum of a
bunch of finite things

116
00:05:34,030 --> 00:05:36,072
and so it made sense that
you were going to get--

117
00:05:36,072 --> 00:05:38,280
and the finite things were
getting small fast enough.

118
00:05:38,280 --> 00:05:39,200
That's the point.

119
00:05:39,200 --> 00:05:40,800
That when you add up
these finite things that

120
00:05:40,800 --> 00:05:42,633
are getting small fast
enough, you can still

121
00:05:42,633 --> 00:05:44,810
end up with a finite number.

122
00:05:44,810 --> 00:05:46,560
But I guess, yeah,
that's where I'll stop.

123
00:05:46,560 --> 00:05:49,232
So just to say that we were
looking at this sort of solid

124
00:05:49,232 --> 00:05:50,690
of revolution
problem again, but we

125
00:05:50,690 --> 00:05:53,460
were dealing with
improper integrals,

126
00:05:53,460 --> 00:05:56,260
and we were showing how you
can do these kinds of problems

127
00:05:56,260 --> 00:05:57,950
with improper integrals.

128
00:05:57,950 --> 00:05:59,861
So now I really will stop there.