1
00:00:00,000 --> 00:00:00,000

2
00:00:00,000 --> 00:00:08,750
PROFESSOR: Welcome back
to recitation.

3
00:00:08,750 --> 00:00:11,320
In this video I want us to look
at the following problem.

4
00:00:11,320 --> 00:00:14,010
We're going to let f of
x equal 1 over x.

5
00:00:14,010 --> 00:00:17,300
And I want to consider the solid
generated by rotating f

6
00:00:17,300 --> 00:00:19,990
of x about the x-axis
between x equal

7
00:00:19,990 --> 00:00:22,090
1 and x equal infinity.

8
00:00:22,090 --> 00:00:25,000
I want you to find the area of
a cross sectional slice, and

9
00:00:25,000 --> 00:00:29,080
then I want you to find the
volume of the solid.

10
00:00:29,080 --> 00:00:31,360
In the cross sectional slice,
the one I'm interested in is

11
00:00:31,360 --> 00:00:33,570
the one that you see
in the xy plane.

12
00:00:33,570 --> 00:00:36,170
So my suggestion to you is that
right away draw yourself

13
00:00:36,170 --> 00:00:41,090
a picture, just at least a rough
sketch of the curve y

14
00:00:41,090 --> 00:00:43,770
equals f of x and then what
that cross sectional slice

15
00:00:43,770 --> 00:00:46,150
will look like and start
off from there.

16
00:00:46,150 --> 00:00:48,670
And take a while to work on that
and then I'll be back and

17
00:00:48,670 --> 00:00:49,920
show you what I did.

18
00:00:49,920 --> 00:00:58,450

19
00:00:58,450 --> 00:00:59,600
OK welcome back.

20
00:00:59,600 --> 00:01:02,590
Well hopefully you were able to
make some headway on this

21
00:01:02,590 --> 00:01:05,090
problem, and maybe you found
some interesting things.

22
00:01:05,090 --> 00:01:06,150
Hopefully you did.

23
00:01:06,150 --> 00:01:09,370
And we will see what I find and
if they are interesting,

24
00:01:09,370 --> 00:01:10,930
which I think they are.

25
00:01:10,930 --> 00:01:12,140
So let's start off.

26
00:01:12,140 --> 00:01:13,670
I told you the first
thing you should do

27
00:01:13,670 --> 00:01:14,710
is get a rough picture.

28
00:01:14,710 --> 00:01:17,540
So I'm going to draw a rough
picture and make sure that

29
00:01:17,540 --> 00:01:21,170
those match up well, if my
picture matches up with yours.

30
00:01:21,170 --> 00:01:26,600
So let me give you
the xy plane.

31
00:01:26,600 --> 00:01:31,670
And then y equals 1 over x is a
curve that at x equal 1 I'm

32
00:01:31,670 --> 00:01:34,110
going to make y equal
1 up here.

33
00:01:34,110 --> 00:01:35,510
y equal 1 up there. x equal 1.

34
00:01:35,510 --> 00:01:38,590
So when x is 1, y is 1.

35
00:01:38,590 --> 00:01:42,400
And then it's going
to decay down.

36
00:01:42,400 --> 00:01:44,740
As x goes to infinity
it decays

37
00:01:44,740 --> 00:01:46,190
So it looks something
like that.

38
00:01:46,190 --> 00:01:48,650
And then I actually should have
moved this 1 because I'm

39
00:01:48,650 --> 00:01:52,370
going to need that side to get
my cross sectional slice.

40
00:01:52,370 --> 00:01:56,590
So my cross sectional slice, is
going to come down here and

41
00:01:56,590 --> 00:01:58,060
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 --> 00:02:18,030
You saw I believe that the
integral from 1 to infinity of

48
00:02:18,030 --> 00:02:26,230
1 over x dx diverges, because
you end up with log evaluated

49
00:02:26,230 --> 00:02:29,550
at infinity and that is
infinite and so you

50
00:02:29,550 --> 00:02:30,420
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 --> 00:02:37,830
So of course if I double
that, I'm still going

54
00:02:37,830 --> 00:02:38,950
to get that it diverges.

55
00:02:38,950 --> 00:02:42,130
So in fact the cross sectional
slice area,

56
00:02:42,130 --> 00:02:46,393
CSS area, is infinite.

57
00:02:46,393 --> 00:02:51,020

58
00:02:51,020 --> 00:02:53,400
OK, so that's one part.

59
00:02:53,400 --> 00:02:56,170
The other part is to find the
volume of the solid of

60
00:02:56,170 --> 00:02:57,360
revolution.

61
00:02:57,360 --> 00:03:01,980
So the second part was take this
piece from 0 up to f of

62
00:03:01,980 --> 00:03:06,280
x, rotate it around the x-axis
and compute the volume.

63
00:03:06,280 --> 00:03:08,380
Now we know how to compute
the volume.

64
00:03:08,380 --> 00:03:12,090
This is the disc method that
I'm going to use here.

65
00:03:12,090 --> 00:03:14,400
And I will write out what we
need to do and then we'll look

66
00:03:14,400 --> 00:03:16,310
at what the integral gives us.

67
00:03:16,310 --> 00:03:18,930
So for the disc method,
so let's say this

68
00:03:18,930 --> 00:03:22,300
is the volume part.

69
00:03:22,300 --> 00:03:24,300
So the volume, it's going
to be the disc method.

70
00:03:24,300 --> 00:03:26,460
So I need to do pi r squared.

71
00:03:26,460 --> 00:03:28,190
And in this case I'm integrating
from 1 to

72
00:03:28,190 --> 00:03:31,690
infinity, so I need
pi r squared dx.

73
00:03:31,690 --> 00:03:33,160
I'm going to be integrating
something that's

74
00:03:33,160 --> 00:03:35,900
pi r squared dx.

75
00:03:35,900 --> 00:03:38,344
And I know the bounds are 1 to
infinity, and so I need r in

76
00:03:38,344 --> 00:03:41,100
terms of, as a function of x.

77
00:03:41,100 --> 00:03:43,100
That's fairly easy. r is
just the measure from

78
00:03:43,100 --> 00:03:44,870
0 up to f of x.

79
00:03:44,870 --> 00:03:47,060
So that's just 1 over x.

80
00:03:47,060 --> 00:03:48,230
So this is actually
the integral--

81
00:03:48,230 --> 00:03:52,680
I'm going to pull the pi out
--of 1 over x squared from 1

82
00:03:52,680 --> 00:03:57,040
to infinity dx.

83
00:03:57,040 --> 00:03:59,100
OK so how do I do
this integral?

84
00:03:59,100 --> 00:04:03,660
Well you were shown in class
that in actuality this is the

85
00:04:03,660 --> 00:04:07,880
limit-- as some value up here
goes to infinity --of this

86
00:04:07,880 --> 00:04:11,780
integral, but we're just going
to do it the shorthand way

87
00:04:11,780 --> 00:04:13,270
that he also mentioned
in class and keep

88
00:04:13,270 --> 00:04:14,650
the infinity around.

89
00:04:14,650 --> 00:04:17,110
So we can keep that as our
bounds, knowing we're taking

90
00:04:17,110 --> 00:04:21,120
limits as this thing
goes to infinity.

91
00:04:21,120 --> 00:04:25,550
So the integral of 1 over x
squared is negative 1 over x.

92
00:04:25,550 --> 00:04:32,670
So I get minus pi times 1 over
x, evaluated at 1 in infinity.

93
00:04:32,670 --> 00:04:34,950
I can just check that if I need
to, but this is x to the

94
00:04:34,950 --> 00:04:37,950
minus 1 and it's derivative is
negative x to the minus 2.

95
00:04:37,950 --> 00:04:40,070
So I needed that
negative there.

96
00:04:40,070 --> 00:04:41,000
And then I evaluate this.

97
00:04:41,000 --> 00:04:44,520
Well at infinity 1 over x as x
goes to infinity, this first

98
00:04:44,520 --> 00:04:45,620
part goes to 0.

99
00:04:45,620 --> 00:04:51,700
So this is 0 minus negative
pi times 1 over 1.

100
00:04:51,700 --> 00:04:52,830
So negative pi times 1.

101
00:04:52,830 --> 00:04:56,960
So this is just pi.

102
00:04:56,960 --> 00:05:00,440
OK, so hopefully that kind of
blows your mind a little bit,

103
00:05:00,440 --> 00:05:03,170
that you could have something
where this cross sectional

104
00:05:03,170 --> 00:05:06,430
slice is infinite, but in fact
if you look at it the way we

105
00:05:06,430 --> 00:05:11,260
computed the volume, we have
in fact a finite volume.

106
00:05:11,260 --> 00:05:13,770
So I don't know what else I'm
going to say about that except

107
00:05:13,770 --> 00:05:14,750
I think it's really cool.

108
00:05:14,750 --> 00:05:16,410
And you can think about
why that is.

109
00:05:16,410 --> 00:05:20,290
And in fact you might want to
notice that we had to, we

110
00:05:20,290 --> 00:05:22,930
computed things in terms of
cross sectional slices coming

111
00:05:22,930 --> 00:05:23,730
from the other direction.

112
00:05:23,730 --> 00:05:29,480
So we looked at these cross
sectional slices and we got

113
00:05:29,480 --> 00:05:31,860
our, we showed our volume
was finite there.

114
00:05:31,860 --> 00:05:34,610
So we had a sum of a bunch of
finite things and so it made

115
00:05:34,610 --> 00:05:36,870
sense that you were going to
get, and the finite things

116
00:05:36,870 --> 00:05:38,210
were getting small
fast enough.

117
00:05:38,210 --> 00:05:39,200
That's the point.

118
00:05:39,200 --> 00:05:41,070
That when you add up these
finite things that are getting

119
00:05:41,070 --> 00:05:43,320
small fast enough, you
can still end up

120
00:05:43,320 --> 00:05:45,300
with a finite number.

121
00:05:45,300 --> 00:05:46,560
But I guess, yeah, that's
where I'll stop.

122
00:05:46,560 --> 00:05:49,390
So just to say that we were
looking at this sort of solid

123
00:05:49,390 --> 00:05:52,470
of revolution problem again,
but we were dealing with

124
00:05:52,470 --> 00:05:55,020
improper integrals, and we were
showing how you can do

125
00:05:55,020 --> 00:05:57,950
these kinds of problems with
improper integrals.

126
00:05:57,950 --> 00:05:59,200
So now I really will
stop there.

127
00:05:59,200 --> 00:06:00,360