4
More Applications of Definite Integrals: to Average
value of a function; Volumes, arclength and other
matters
4.1
During a particularly soggy week in Vancouver, rainfall reached epic proportions. The rainfall
pattern was as follows: A constant 20mm on the first 24 hrs, a steady increase from 20 up
to 50 mm over the next 24 hrs (assume linear increase), 50mm rain over the next 24 hrs, a
steady drop from 50 down to 40 mm over the next 24 hrs, and a flat 30 mm over the next
24 hrs. Determine the total amount of rain during this period and the average daily rainfall
for the same period.
Solution
See Figure 1.
rainfall (in cm)
5
4
3
constant rate
20mm/day
2
A3
A4
A2
1
A5
A1
1
2
3
4
5
Figure 1: For problem 4.1 solution
Total amount of rain = A1 + A2 + A3 + A4 + A5
9
7
= (2 + + 5 + + 3)
2
5
= 18 cm
Average rainfall over 5 days =
18 cm total
= 3.6 cm/day.
5 days
4.2
Consider the periodic function,
f (t) = sin (2t) + cos (2t)
1
days
(a) What is the frequency, the amplitude, and the length of one cycle in this function?
(b) How would you define the average value of this function over one cycle ?
(c) Compute this average value and show that it is zero. Now explain why this is true using
a geometric argument.
Solution
(a) Consider A sin(ωt + ϕ), A = amplitude, ϕ = phase shift, ω = frequency.
A sin(ωt + ϕ) = A(sin(ωt) cos ϕ + cos(ωt) sin ϕ)
If ϕ =
π
4
and ω = 2, then
π
1
1
A sin(2t + ) = A sin(2t) · √ + cos(2t) · √
4
2
2
√
If A = 2, then
√
π
2 sin(2t + ) = sin(2t) + cos(2t) = f (t)
4
!
Therefore, frequency ω = 2 (meaning 2 cycles in length 2π), amplitude A =
one cycle = 2π/2 = π.
√
2, length of
(b) Average value of f (t) over one cycle:
favg =
Z length
1Zπ
1
f (t) dt =
(sin(2t) + cos(2t)) dt
(length of cycle) 0
π 0
(c)
favg
"
#π
1
cos(2t) sin(2t)
=
−
+
π
2
2
0
1
1
1
=
− +0 − − +0
π
2
2
= 0
(cos(2π) = 1, cos 0 = 1, sin(2π) = 0, sin 0 = 0.)
See Figure 2. Over one cycle, there is equal amount of area above the x-axis as below the
x-axis, so the average value of the function over one cycle is 0.
2
1/2
2
−π/8
Figure 2: For problem 4.2
4.3
t days after the Spring Equinox, the length of time from sunrise to sunset (in hours) is given
by
πt
l(t) = 12 + 4 sin
182
(a) Explain the meaning of the word “Equinox” and describe what happens on that day
according to the above formula.
(b) What is the length of the shortest and the longest day and when do these occur according
to this formula?
(c) How long is one complete cycle in this expression?
(d) Sketch l(t) as a function of t.
(e) Find the average day-length over the month immediately following the Equinox.
(f) Find the average day length over the whole year. Explain your result with a simple
geometric or intuitive argument.
Solution
πt
l(t) = 12 + 4 sin
182
(a) Equinox is a time of year when the length of “day” (sunrise → sunset) equals the length
of “night” (sunset → sunrise). This happens twice a year, once in spring and once in fall.
On the Equinox, l(t) = 12 hours
πt
⇒ 12 = 12 + 4 sin
182
3
⇒ t = 0 (Spring Equinox) or t = 182 (Fall Equinox)
(b) We know that −1 ≤ sin
πt
182
≤ 1. So the longest day will occur when sin
the shortest day will occur when sin
Longest day
sin
πt
182
= −1.
πt
182
= 1 and
πt
π
πt
= ⇒ t = 91
=1⇒
182
182
2
l(91) = 12 + 4 · 1 = 16 hours, occurring 91 days after the Spring Equinox.
Shortest day
πt
πt
3π
sin
= −1 ⇒
=
⇒ t = 273
182
182
2
l(273) = 12 + 4(−1) = 8 hours, occurring 273 days after the Spring Equinox.
(c) One complete cycle is 2(182) = 364 days, from this expression:
sin(ωt) has frequency ω ⇒ ω cycles in every 2π.
2π
= π/182
= 364.
Therefore the length of one cycle = 2π
ω
(d) See Figure 3.
16
14
12
Fall Equinox
Spring Equinox
l(t)
hours
10
8
6
4
2
0
0
91
182
273
364
t
days
Figure 3: For problem 4.3 (d)
(e) Average day-length over 1st month after Spring Equinox (assume 30 days in month):
lavg =
=
=
=
1 30
πt
1 30
l(t) dt =
12 + 4 sin
dt
30 0
30 0
182
30
1
4 · 182
πt
12t −
cos
30
π
182 0
4 · 182
4 · 182
π · 30
1
12 · 30 −
+
cos
cos 0
30
π
182
π
15π
364
364
cos
+
≈ 13 hours
12 −
15π
91
15π
Z
Z
4
(f) Average day length over whole year: (expect ≈ 12 hours)
lavg =
=
=
=
=
1 Z 365
l(t) dt
365 0
365
1
4 · 182
πt
12t −
cos
365
π
182 0
π · 365
4 · 182
4 · 182
1
cos
12 · 365 −
+
365
π
182
π
365π
728
728
cos
12 −
+
365π
182
365π
12.0009 (≈ 12, as expected)
From the graph in (d), we can see that over 364 days (almost a year) the average value of
l(t) is 12 hours.
4.4
The current in an AC electric circuit is given by
I(t) = A cos(ωt)
The Power in the circuit is defined as P (t) = I 2 (t).
(a) What is meant by one cycle in this situation?
(b) Sketch graphs of I(t) and P (t). Explain why P (t) is always positive, and indicate how
its zeros are related to zeros of I(t). What are the maximal and minimal values of each of
these functions?
(c) Find the average power and the average current over half a cycle. (Note: in computing
the average power, you will need to use the trick cos2 (ωt) = 12 (1 + cos(2ωt))).
Solution
(a) One cycle in this situation is one period (max to max or min to min) in the current
function I(t) = A cos(ωt).
The frequency is ω so the period (i.e. the length of one cycle) is 2π
.
ω
This length is equivalent to two periods (but one cycle) of the power function. The power is
P (t) = I 2 (t) = A2 cos2 (ωt) = A2
P (t) =
1 + cos(2ωt)
2
A2
(1 + cos(2ωt))
2
using double-angle formula.
(b) See Figure 4.
5
!
I(t)
P(t)
A
A
2π / ω
2
t
−A
π/ω
t
Figure 4: For problem 4.4
P (t) ≥ 0 since it is the square of I(t).
I(t) = 0 ⇒ cos(ωt) = 0
π 3π
, ...
⇒ ωt = ,
2 2
π 3π
,
, ...
⇒ t=
2ω 2ω
P (t) = 0 ⇒ I 2 (t) = 0 ⇒ I(t) = 0
π 3π
,
, ...
⇒ t=
2ω 2ω
Therefore, I(t) and P (t) have the same zeroes.
Maximal and minimal values: −A ≤ I(t) ≤ A, 0 ≤ P (t) ≤ A2 .
The amplitude of I(t) is A, and thus it fluctuates between a maximum of A and a minimum
of −A.
Because P (t) is the square of I(t), it must always be greater than 0, which is therefore its
minimum. Its maximum is A2 , the square of the amplitude of I(t).
(c) Average power over a half cycle: (cycle length is
Pavg =
=
=
=
2π
)
ω
ω Z π/ω
ω Z π/ω 2
P (t) dt =
A cos2 (ωt) dt
π 0
π 0
ω Z π/ω A2
(1 + cos(2ωt)) dt
π 0
2
"
#π/ω
ωA2
sin(2ωt)
t+
2π
2ω
0
2 A2
ωA π
+0−0 =
watts
2π ω
2
6
y
y
y=x
2
y=x
x
1
x
1
y
y
3
y=x
4
y=x
x
1
1
x
Figure 5: For problem 4.5 (c) solution
4.5
(a) Find the average value of xn over the interval [0, 1].
(b) What happens as n becomes arbitrarily large (that is, n → ∞)?
(c) Explain your answer to part (b) by considering the graphs of these functions.
(d) Repeat parts (a) - (c) using the functions x1/n .
Solution
(a) Let f (x) = xn ,
favg
1
=
1−0
Z
1
0
1
xn+1 1
, for n 6= −1
x dx =
=
n+1 0 n+1
n
(b) As n → ∞, favg → 0.
(c) See Figure 5.
As n → ∞, the area under the curve in the interval [0, 1] gets smaller and smaller.
(d) Let g(x) = x1/n ,
gavg
1 Z 1 1/n
x1/n+1 1
=
=
x
dx =
1−0 0
1/n + 1 0
1
n
1
, for n 6= −1
+1
As n → ∞, gavg → 1. See Figure 6.
As n → ∞, the area under the curve in the interval [0, 1] gets closer to 1.
7
y
y
y=x
1/2
y=x
1
x
y
1
x
1
x
y
1/3
1/4
y=x
y=x
1
x
Figure 6: For problem 4.5 (d) solution
4.6
Symmetry
(a) Find the average value of the function sin x over the interval [−π, π].
(b) Find the average value of the function x3 over the interval [−1, 1].
(c) Find the average value of the function x3 − x over the interval [−1, 1].
(d) Explain these results graphically.
(e) Find the average value of an odd function f (x) over the interval [−a, a]. (Remember that
f (x) is odd if f (−x) = −f (x).)
(f) Suppose now that f (x) is an even function (that is, f (−x) = f (x)) and its average value
over the interval [0, 1] is 2. Find its average value over the interval [−1, 1].
Solution
(a)
Z π
π
1
1
1
sin x dx =
(− cos x) = − (1 − 1) = 0
π − (−π) −π
2π
2π
−π
(b)
1
1 − (−1)
Z
1 x4 1
1 4
4
=0
1
−
(−1)
x dx =
=
2 4 −1 8
−1
1
3
8
y
y
y = x3
y = sin x
π
−π
1
−1
X
X
y
3
y = x −x
−1
1
X
Figure 7: For problem 4.6 (e) solution
(c)
1
1 − (−1)
Z
x4 x2 1
1 1 1 1 1
=0
−
−
−
+
=
4
2 −1 2 4 2 4 2
!
1
(x − x) dx =
2
−1
1
3
(d) See Figure 7. All these graphs are symmetric about the origin, so that the integral
R0
Rπ
−π sin x dx = − 0 sin x dx and they cancel out (similarly for the other two functions).
(e) Odd function (symmetric about the origin): f (−x) = f (x).
1
a − (−a)
Z
a
f (x) dx =
−a
1
2a
Z
a
f (x) dx =
−a
1
2a
Z
0
f (x) dx +
−a
Z
a
0
f (x) dx
Now, let x = −u, then dx = −du, x = −a ⇒ u = a etc.
Z
0
−a
f (x) dx = −
= −
=
Z
Z
0
−a
Z
0
a
= −
Therefore,
Z
0
a
0
a
f (−u) du
(−f (u)) du ⇒ since f is odd
f (u) du = −
Z
a
0
Z
a
0
f (u) du reversing limits of integration
f (x) dx since u is just a “dummy00 variable of integration
f (x) dx +
Z
a
0
f (x) dx = −
9
Z
a
0
f (x) dx +
Z
a
0
f (x) dx = 0
Thus, the average value of an odd function over a symmetric interval [−a, a] is 0.
(e) We are given that f (x) is an even function (that is, f (−x) = f (x)) and its average value
over the interval [0, 1] is 2. This means that
1 1
f¯[0,1] =
f (x) dx =
1 0
The average value over the interval [−1, 1] is
Z
f¯[−1,1] =
1
(1 − (−1))
Z
1
f (x) dx = · 2
2
−1
1
Z
Z
1
0
1
0
f (x) dx = 2
f (x) dx =
Z
1
0
f (x) dx = f¯[0,1]
(We have used the symmetry of the function here: the definite integral of an even function
over a symmetric interval about the y-axis will be double the value of the definite integral
over half that interval.) Thus
f¯[−1,1] = f¯[0,1] = 2.
4.7
[98 Final]
Find the average value of the function f (x) = sin( πx
) over the interval [0,2].
2
Solution
f¯ =
1
2−0
Z
2
0
πx
1 −2
dx =
sin
2
2 π
πx
cos
2
4.8
2
0
=
−1
2
(−1 − 1) =
π
π
The intensity of light cast by a street lamp at a distance x (in meters) along the street from
the base of the lamp is found to be approximately I(x) = 20 − x2 in arbitrary units for
−20 < x < 20.
(a) Find the average intensity of the light over the interval −5 < x < 5.
(b) Find the average intensity over −7 < x < 7.
(c) Find the value of b such that the average Intensity over [−b, b] is Iav = 10.
Solution
The intervals and the function are symmetric, so
!
Z a
Z a
1
1
x3 a
1
2
2
Iav =
20x −
(20 − x ) dx = (2) (20 − x ) dx =
a − (−a) −a
2a
a
3 0
0
3
1 R5
(20 − x2 ) dx = 51 [100 − 53 ] = 35
.
5 0
3
R
3
Iav = 17 07 (20 − x2 ) dx = 71 [140 − 73 ] = 11
.
3
R
3
b
Solve 10 = 1b 0 (20 − x2 ) dx = 1b [20b − b3 ] =
(a) Iav =
(b)
(c)
Volumes of surfaces of revolution
10
2
20 − b3 . This gives b =
√
30 ≈ 5.477 meters.
4.9
Find the volume of a cone with height h equal to the base radius r, by summing up volumes
of disk-shaped slices. We will place the cone on its side, as shown in the Figure 8, and let x
represent position along its axis.
(a) Using the diagram shown below (Figure 8), explain what kind of a curve in the xy plane
we would use to generate the surface of the cone as a surface of revolution. (Hint: notice
that the “curve” is nice and straight.)
y
generating curve
r
x
h
Figure 8: For problem 4.9
(b) Using the proportions given in the problem, specify the exact function y = f (x) that we
need to describe this “curve”.
(c) Now find the volume enclosed by this surface of revolution for 0 ≤ x ≤ 1.
(d) Show that, in this particular case, we would have got the same geometric object, and
also the same enclosed volume, if we had rotated the “curve” about the y axis. Be careful
how you set up the integration step in this case.
(e) Suppose you are now given a new line, y = ax and told to compute the two volumes as
in (c) and (d) above. By what factor would they differ?
Solution
(a) (b) The curve that generates the cone surface of revolution is the straight line y =
(since r = h).
r
h
x=x
(c) “Disk method” - rotate about the x−axis. Each disk has width dx and radius y. Volume
of each disk is (area of base) × (width) = πy 2 dx. Because the curve used to generate this
surface of revolution is y = x, the volume πy 2 dx = πx2 dx. Summing up the volumes
enclosed by each of the disks, we find the total volume enclosed by this surface of revolution
for 0 < x < 1 is
Z 1
Z 1
2
V =
πy dx =
πx2 dx = π/3 cubic units
0
0
11
(d) Rotate about the y−axis: since r = h, we obtain the same geometrical object. In this
case, each disk has width dy and radius x. The volume of each disk is therefore πx2 dy =
πy 2 dy.
Volume enclosed by surface of revolution is then for 0 < y < 1 and is
Z
1
0
πx2 dy =
Z
1
0
πy 2 dy = π/3 cubic units
(same as in part (c)).
(e) The two volumes are: a2 π/3 and a−2 π/3 respectively. Thus, they differ by a factor a4 .
4.10
Find the volume of the cone generated by revolving the curve y = f (x) = 1−x (for 0 < x < 1)
about the y axis. Use the disk method, with disks stacked up along the y axis.
Solution
Figure 9:
The disks have thickness ∆y → dy and radii in the x direction so that x = f (y) = 1 − y
is the function to use. From the interval 0 < x < 1, we find the interval on the y-axis is
0 < y < 1. Thus,
Z
Z
V =π
V =π
Z
1
0
1
0
[f (y)]2 dy = π
1
0
[1 − y]2 dy
y 2 y 3 1
[1 − 2y + y ] dy = π y − 2 +
2
3 0
!
2
1
π
V = π(1 − 1 + ) = .
3
3
12
4.11
In this problem you are asked to find the volume of a pyramid with a square base of width
w and with a height h. (This is related to the Cheops pyramid problem, but we will use
calculus.) Let the variable z stand for distance down the axis of the pyramid with z = 0
at the top, and consider “slicing” the pyramid along this axis (into horizontal slices). This
will produce square “slices” (having area A(z) and some width ∆z). Calculate the volume
of the pyramid as an integral by figuring out how A(z) depends on z and integrating it.
z
h
0
w
Figure 10: For problem 4.11
Solution
Let the side length of the square cross-section at height z be x. Then by similar triangles,
x/z = w/h, or x = (w/h)z. Now the area of the square slice at height z is A(z) = x2 =
(w/h)2 z 2 and its thickness is ∆z. Thus
V =
Z
h
0
(w/h)2 z 2 dz = (w/h)2 [z 3 /3]|h0 = w 2 h/3.
4.12
Find the volume of the “bowl” obtained by rotating the curve y = 4x2 about the y axis for
0 ≤ x ≤ 1.
Solution
We are rotating about the y axis, so, using the ”disk” method, disk
q width is dy, the radius
of each disk is in the x direction, i.e. it is given by x = f (y) = y/4. The volume will be
given by
Z ymax
V =
π[f (y)]2 dy.
ymin
To find ymax and ymin we use the interval 0 ≤ x ≤ 1. y(0) = 0 and y(1) = 4. Therefore the
interval on the y-axis is 0 ≤ y ≤ 4.
Z
4
0
πZ4
y dy = 2π.
πx dy =
4 0
2
13
4.13
Consider the curve
y = f (x) = 1 − x2
0<x<1
rotated about the y axis. Recall that this will form a shape called a paraboloid. Use the
cylindrical shell method to calculate the volume of this shape.
Solution
The cylindrical shells will be encasing one another with their radii in the direction of the
x axis, their heights along the y axis, and their thickness ∆x → dx along the x axis. The
radius of a given shell is x, the height is f (x) = 1 − x2 . The volume of one shell is
Vshell = 2πrhτ = 2πxf (x) dx
The volume of the entire shape for 0 < x < 1 is
V = 2π
V = 2π
4.14
Z
1
0
Z
1
0
xf (x) dx = 2π
Z
1
0
x(1 − x2 ) dx.
x2 x4 1
1 1
π
(x − x ) dx = 2π
=
−
−
= 2π
2
4 0
2 4
2
!
3
On his wedding day, Kepler wanted to calculate the amount of wine contained inside a wine
barrel whose shape is shown below in Figure 11. Use the disk method to compute this
volume. You may assume that the function that generates the shape of the barrel (as a
surface of revolution) is y = f (x) = R − px2 , for −1 < x < 1 where R is the radius of the
widest part of the barrel. (R and p are both positive constants.)
y
y = R − px 2
1
x
Figure 11: For problem 4.14
Solution
14
radius
r=R−px^2
dx
Figure 12: For problem 4.14 solution
−1 ≤ x ≤ 1, R > 0, p > 0.
Disk method: See Figure 12.
The radius of each disk is y = R − px2 , and the thickness is ∆x → dx. The volume of each
disk is therefore V = π(R − px2 )2 ∆x
Because f (x) = R − px2 is an even function, the volume for −1 < x < 1 is twice that for
0 < x < 1.
Volume
= 2π
= 2π
Z
Z
1
0
1
0
(R − px2 )2 dx
(R2 − 2Rpx2 + p2 x4 ) dx
5 1
x3
2 x
= 2π R x − 2Rp
+p
3
5 0
!
2Rp p2
2
= 2π R −
+
3
5
2
!
2RP
p2
So the amount of wine in the barrel is 2π R −
+
3
5
2
!
cubic units.
4.15
Find the volume of the solid obtained by rotating the region bounded by the given curves
f (x) and g(x) about the specified line. Sketch the region, the solid, and a typical disk or
‘washer’.
√
(a) f (x) = x − 1, g(x) = 0, from x = 2 to x = 5, about the x-axis.
√
(b) f (x) = x, g(x) = x/2, about the y-axis.
(c) f (x) = 1/x, g(x) = x3 , from x = 1/10 to x = 1, about the x-axis.
Solution
15
(a) The volume of the solid is equal to
Z
5
2
π(x − 1)dx = π(x
2
5
/2 − x)
= 7.5π.
2
(b) To calculate the volume of the solid, we have to find the x-coordinates of the 2 curves
as a function of the y-coordinates, and we have to determine the boundaries on the y-axis.
For the first task, we simply take the inverse functions: x = y 2 and x = 2y. Thus, the
solid is obtained by rotating the region between the curve x = 2y and the curve x = y 2
around the y-axis (see sketch). This region is bounded by the two points (x, y) = (0, 0) and
(x, y) = (4, 2) (the latter point can be found by solving f (x) = g(x)). Therefore, we have to
integrate from 0 to 2 on the y-axis. We get the volume of the solid as
V =
Z
2
0
V = π(
π[(2y)2 − (y 2 )2 ]dy
4y 3 y 5 2 64π
− ) =
3
5 0
15
(c) Since f (x) ≥ g(x) in the region considered, the volume of the solid is
V =
Z
1
(πf (x)2 − πg(x)2 )dx = π
1
10
V = π(−1/x − x
7
1
/7)
1
10
Z
1
1
10
(x−2 − x6 )dx
= π(62 + 1/107 )/7 ≈ 8.857π.
Length of a curve and arclength
4.16
Compute the length of the line y = 2x + 1 for −1 < x < 1 using the arc-length formula.
Check your work by using the simple distance formula (or Pythagorean theorem).
Solution
dy/dx = 2 so
L=
Z
√ Z
L= 5
1
−1
1
−1
q
1+
(dy/dx)2 dx
=
Z
1
√
1 + 4 dx
−1
√ 1
√
√
dx = 5x = 5(1 − (−1)) = 2 5.
−1
We can compare this to the simple distance formula: The endpoints of the line are (−1, −1)
and (1, 3) so the length of the line is
q
√
√
√
d = (1 − (−1))2 + (3 − (−1))2 = 22 + 42 = 20 = 2 5.
16
4.17
Set up the integral that represents the length of the following curves: Do not attempt to
calculate the integral in any of these cases
(a)
y = f (x) = sin(x) 0 < x < 2π.
(b)
y = f (x) =
√
x 0 < x < 1.
(c)
y = f (x) = xn
− 1 < x < 1.
Solution
(a)
y = f (x) = sin(x) 0 < x < 2π.
dy/dx = cos(x) so
L=
Z
2π
0
q
(b)
y = f (x) =
1 + cos2 (x) dx
√
x 0 < x < 1.
dy/dx = 12 x−1/2 , and (dy/dx)2 = ( 21 )2 x−1 =
1
4x
1
s
L=
Z
0
so
1+
1
dx
4x
(c)
y = f (x) = xn
dy/dx = nxn−1 so
L=
Z
1
√
− 1 < x < 1.
1 + n2 x2n−2 dx
−1
Other applications (work, energy, etc)
17
4.18
A spring has a natural length of 16 cm. When it is stretched x cm beyond that, Hooke’s Law
states that the spring pulls back with a restoring force F = kx dynes, where the constant k
is called the spring constant, and represents the stiffness of the spring. For the given spring,
8 dynes of force are required to hold it stretched by 2 cm. How much work (dynes-cm) is
done in stretching this spring from its natural length to a length 24 cm?
Solution
Since F = 8 when x = 2 we find that 8 = 2k so that k = 4 and F = 4x is the force. When
the spring is at its natural length, 16 cm, no work is done. To stretch the spring by ∆x
beyond this would require work
W = F ∆x = 4x∆x.
To sum up all the work involved in stretching from 16 to 24 cm, i.e. for the increase in length
x = 0 to x = 8, we would calculate
W =
Z
F dx =
Z
8
0
4xdx = 2x
4.19
8
2
0
= 128 dynes − cm.
Calculate the work done in pumping water out of a parabolic container. Assume that the
container is a surface of revolution generated by rotating the curve y = x2 about the y axis,
that the height of the water in the container is 10 units, that the density of water is 1gm/cm 3
and that the force due to gravity is F = mg where m is mass and g = 9.8m/s2 .
Solution
The paraboloid can be thought of as a bowl full of water. The total work done is equivalent
to the sum of work done to pump each bit of the water up to the height 10, or equivalently,
a sum over “disks” of water, each at some height y. The work done is Work = (force)
(distance). The distance from a disk at height y to height 10 is 10 − y. Therefore, Work =
Force × (10 − y). The force is F = mg where g = 9.8m/s2 = 980cm/s2 . The mass of a disk
at height y is
my = (1 gm/cm3 )(πr 2 ∆y cm3 )
where the radius of the disk is the x coordinate of a point on the parabola y = x2 , giving
√
r = x = y. Thus, the work done to move this disk to height 10 cm is
√
wy = my g(10 − y) = π( y)2 ∆y(980)(10 − y) = 980π(10 − y)y∆y,
and the total work, which is a sum over all disks, is the integration of wy over 0 < y < 10:
W = 980π
Z
10
0
(10 − y)y dy = 980π
18
Z
10
0
(10y − y 2 ) dy
1 1
y 2 y 3 10
−
W = 980π 10 −
|0 = 980π · 103
= 5.131 × 105 ergs.
2
3
2 3
!
(One erg is 1 gm cm2 s−2 or 1 dyne-cm).
4.20
A vertical shear is a transformation of the plane which takes the vertical line x = a and
pushes it upward by a distance sa (the constant s is called the strength of the shear). For
instance, the rectangle shown below in Figure 13 is transformed into a trapezoid.
y
Before shear
y
x=a
After shear
x=a
x
x
Figure 13: For problem 4.20
(a) Set up a definite integral which shows that the area of the trapezoid is the same as the
area of the original rectangle. We say that the shear has preserved the area of the rectangle.
(b) Now consider a more general shape:
Before Shear
After shear
Figure 14: For problem 4.20(b)
Show that the area before the shear is the same as the area after the shear.
Solution
(a) See Figure 15.
19
y
y
Before shear
After shear
line x = a
y = sx + b
y=b
b
b
y = sx
sa
x=a
a
x
x
Figure 15: For problem 4.20 (a) solution
Area of rectangle = ab.
Area of trapezoid =
Ra
0 [(sx + b) − sx] dx =
So shear is area-preserving for a rectangle.
(b) See Figure 16.
Ra
0
a
b dx = bx = ab.
0
a
a
a
−a
a
−a
−a
−a
Circle (before shear)
Ellipse (after shear)
Figure 16: For problem 4.20 (b) solution
Consider the circle as made up of rectangle strips, each of which is undergoing a shear of
strength s. From part (a) we know that this shear is area-preserving. Therefore, the area of
the circle (before shear) is the same as the area of the ellipse (after shear).
20