Math 0230
Chapter 7
Calculus 2
Lectures
Applications of Integration
Numeration of sections corresponds to the text
James Stewart, Essential Calculus, Early Transcendentals, Second edition.
Section 7.1
Areas Between Curves
Two approaches:
1. We consider a region S that lies between two curves y = f (x) and y = g(x) and between
the vertical lines x = a and x = b, where f and g are continuous and f ≥ g on [a, b]. Then we
divide the region into small vertical rectangles with base ∆x and heights f (x∗i ) − g(x∗i ). The
corresponding Riemann sum is
n
X
[f (x∗i ) − g(x∗i )]∆x
i=1
and the area A of S is
A = lim
n→∞
n
X
[f (x∗i ) − g(x∗i )]∆x =
i=1
Zb
[f (x) − g(x)] dx
a
To solve problems on finding areas between two curves always draw a picture.
Example 1.
Find the area that lies between two curves y =
√
2x + 3 and y =
between the vertical lines x = 1 and x = 3.
Solution:
First, draw the picture. f (x) =
√
2x + 3, g(x) =
are continuous and f ≥ g on [a, b]. Then
Z3 A=
1
√
1
and
x+1
1
and (important!) f and g
x+1
3
1
1
3/2
2x + 3 −
dx =
(2x + 3) − ln(x + 1)
x+1
3
1
1
= · 93/2 − ln 4 −
3
√
1 3/2
5 5
· 5 − ln 2 = 9 −
− ln 2
3
3
2. We consider a region S that lies between two curves x = f (y) and x = g(y) and between the
horizontal lines y = c and y = d, where f and g are continuous and f (y) ≥ g(y) on [c, d]. Then
1
we divide the region into small horizontal rectangles with height ∆y and bases f (yi∗ ) − g(yi∗ ).
The corresponding Riemann sum is
n
X
[f (yi∗ ) − g(yi∗ )]∆y
i=1
and the area A of S is
A = lim
n
X
n→∞
[f (yi∗ ) − g(yi∗ )]∆y =
i=1
Zd
[f (y) − g(y)] dy
c
To solve problems on finding areas between two curves always draw a picture.
Example 2.
Find the area enclosed by the parabola 4x + y 2 = 12 and the line y = x.
Solution:
The parabola has the vertex at (3, 0). Points of intersection: 4x = 12 − y 2 = 4y,
y 2 + 4y − 12 = 0, y = −6 and y = 2. Points are (−6, −6) and (2, 2). Consider functions
12 − y 2
y2
x = f (y) =
=3−
and x = g(y) = y on [−6, 2], f (y) ≥ g(y). Then the area is
4
4
2
Z2 y3 y2
y2
− y dy = 3y −
−
A=
3−
4
12
2 −6
−6
8
6·6·6
2
1
=6−
− 2 − −18 +
− 18 = 4 − + 18 = 21
12
2·6
3
3
Explain why horizontal rectangles work better here comparing to vertical rectangles (with ∆x).
Example 3.
Sketch the region enclosed by the curves y 2 − x − 2 = 0 and x = ey and lines
y = −1 and y = 1.
Solution:
The parabola x = y 2 − 2 has the vertex at (−2, 0). Two curves do not intersect
inside −1 ≤ y ≤ 1. Consider functions x = f (y) = ey and x = g(y) = y 2 − 2 on [−1, 1],
f (y) ≥ g(y). Then the area is
Z1
A=
−1
1
y
y3
2
y
e − y + 2 dy = e −
+ 2y
3
−1
1
1
1
−1
= e − + 2 − e + − 2 = 3 + e − e−1
3
3
3
2
Section 7.2
Volumes
Let S be a 3d solid placed in the xy-plane. Let Px be a plane perpendicular to the x-axis and
Py be a plane perpendicular to the y-axis. First consider Px type of planes. They slice S. The
intersection of S and Px is called a cross-section and let A(x) be its area, a ≤ x ≤ b. If a slice is
∆x = (b − a)/n thick then its volume is A(x∗i )∆x, where x∗i lies in the interval [xi−1 , xi ]. Then
the volume of S is approximately
n
X
V ≈
A(x∗i )∆x
i=1
or exactly
V = lim
n→∞
Example 1.
n
X
A(x∗i )∆x
Zb
=
A(x) dx
i=1
a
Find the volume of a sphere of radius r.
Solution:
−r ≤ x ≤ r. A cross-section at x is a circle of radius y =
2
A(x) = πy = π(r2 − x2 ) and
Zr
2
2
Zr
π(r − x ) dx = 2π
V =
−r
x3
(r − x ) dx = 2π r x −
3
2
2
2
0
r
0
√
r2 − x2 . Then
= 2π(r3 − r3 /3) =
4π 3
r
3
Now assume we have a solid obtained by rotating about the x-axis the region bounded by
curves y = f (x) and y = 0, when a ≤ x ≤ b. Then each cross-section is a circle with radius
f (x) and the volume of the solid is
Zb
V =π
f 2 (x) dx
a
If a solid is obtained by rotating about the y-axis the region bounded by the curve x = f (y),
when c ≤ y ≤ d, then each cross-section is a circle with radius f (y) and the volume of the solid
is
Zd
V = π f 2 (y) dy
c
Example 2.
Find the volume of the solid obtained by rotating about the y-axis the region
bounded by the curves x = y 2 and x = 0, when 0 ≤ y ≤ 2.
3
Solution:
Z2
V =π
Z2
2 2
(y ) dy = π
0
0
2
y 5 32π
y dy = π =
5 0
5
4
Now assume a region is bounded by the curves y = f (x) and y = g(x), a ≤ x ≤ b, where
f (x) ≥ g(x) [note, the second curve is not y = 0 anymore]. In this case a cross-section is
actually a washer with the outer (larger) radius f (x) and the inner (smaller) radius g(x). The
area of the washer is π[f 2 (x) − g 2 (x)] and the volume of the solid obtained by rotation is
Zb
V =π
[f 2 (x) − g 2 (x)] dx
a
In the case of rotation about the y-axis the coressponding volume is
Zd
V =π
[f 2 (y) − g 2 (y)] dy
c
Example 3.
Find the volume of the solid obtained by rotating about the y-axis the region
bounded by the curves x = y 2 and x = 2y.
Solution: First, we have to find c and d as y coordinates of points of intersection of the curves:
the equation y 2 = 2y gives two solutions c = 0 and d = 2. Also, 2y ≥ y 2 when 0 ≤ y ≤ 2.
Hence
2
3
Z2
Z2
y5
64π
4y
2
2 2
2
4
−
=
V = π [(2y) − (y ) ] dy = π [4y − y ] dy = π
3
5 0
15
0
0
Solids that we have considered so far are called solids of revolution. Now consider an example
of not a solid of revolution
Example 4.
Find the volume of a pyramid whose base is a square with side a and whose
height is h.
Solution: We place the origin O at the vertex of the pyramid and the x-axis along its central
ax
axis. Every cross-section at x (0 ≤ x ≤ h) is a square with the side b =
(it can be found
h
a2
from a corresponding proportion). The cross-sectional area is A(x) = b2 = 2 x2 . Then
h
h
Zh
Zh
a2
a2 h
a2 x3 a2 h3
2
V = A(x) dx = 2
x dx = 2
= 2·
=
h
h 3 0
h
3
3
0
0
4
Section 7.3
Volumes by Cylindrical Shells
Some volumes can be calculated by the method of cylindrical shells. Picture. Let there be a
cylindrical shell with inner radius r1 and outer radius r2 . Then its volume ∆V is the difference
of the volume V2 of the outer cylinder and the volume V1 of the inner cylinder:
∆V = V2 − V1 = πr22 h − πr12 h = π(r22 − r12 )h = π(r2 + r1 )(r2 − r1 )h = 2π
We put r =
r2 + r1
(r2 − r1 )h
2
r2 + r1
(the average radius of the shell) and ∆r = r2 − r1 , then
2
∆V = 2πrh∆r
Another way:
∆V = π(r + ∆r)2 h − πr2 h = 2πrh∆r
where r is the inner radius and r + ∆r is the outer radius.
Let’s apply that to a calculation of the volume V of a solid S obtained by rotating about y-axis
the region bounded by y = f (x) (f (x) must be continuous and nonnegative f (x) ≥ 0), y = 0,
x = a, and x = b, where 0 < a < b. As usual, we divide the interval [a, b] into subintervals
[xi−1 , xi ] each of the length ∆x. Let x̄i be a midpoint of ith subinterval. If the rectangle with
base [xi−1 , xi ] and height f (x̄i ) is rotated about the y-axis, then the result is a cylindrical shell
with average radius x̄i , height f (x̄i ), and thickness ∆x. Its volume is
Vi = 2πx̄i f (x̄i )∆x
and the volume of the solid is approximately
V ≈
n
X
Vi =
i=1
n
X
2πx̄i f (x̄i )∆x
i=1
or exactly
V = lim
n→∞
n
X
Zb
2πx̄i f (x̄i )∆x = 2π
i=1
xf (x) dx
a
If a solid S is obtained by rotating about x-axis the region bounded by x = f (y) (f (y) is
continuous and nonnegative f (y) ≥ 0), x = 0, y = c, and y = d, where 0 < c < d, then its
volume is
Zd
V = 2π yf (y) dy
c
If a solid S is obtained by rotating about the line x = l (which is parallel to the y-axis) the
region bounded by y = f (x) (f (x) is continuous and nonnegative f (x) ≥ 0), y = 0, x = a, and
x = b, where 0 < a < b < l, then its volume is
Zb
(l − x)f (x) dx
V = 2π
a
5
Example 1.
Find the volume of a sphere of radius R.
Solution: √The sphere is obtained
√ by rotating about the y-axis the region bounded by the
2
2
curves y = R − x and y = − R2 − x2 when 0 ≤ x ≤ R.
√
√
√
Then r = x, h = R2 − x2 − (− R2 − x2 ) = 2 R2 − x2 and
ZR √
V = 4π x R2 − x2 dx
√
∆V = 2πx · 2 R2 − x2 ∆x,
0
Substitution: u = R2 − x2 , du = −2xdx, u(0) = R2 , u(R) = 0
Z0
V = −2π
1/2
u
ZR2
du = 2π
1/2
u
2
du = 2π u3/2
3
0
R2
R 2
0
4
= πR3
3
Example 2.
Find the volume of the solid obtained by rotating about the y-axis the region
bounded by the curves x = y 2 and x = 2y. [It is the example 3 in the section 7.2]
√
√
Solution: The region is bounded by the curves y = x and y = x/2 with x/2 ≤ √
x. Before
we found that 0 ≤ x ≤ √
4. Here we apply the method of cylindrical shells: r = x, h = x − x/2,
∆V = 2πrh∆r = 2πx( x − x/2)∆x = 2π(x3/2 − x2 /2)∆x. Then
4
Z4 x2
2
2 5/2 x3
32
2 1
64π
3/2
= 2π
· 32 −
−
V = 2π
dx = 2π x −
= 64π
=
x −
2
5
6 0
5
3
5 3
15
0
Example 3.
Find the volume
√ of the solid obtained by rotating about the y-axis the region
bounded by the curves y = x and y = 0, and x = 4.
Solution:
Draw a picture. a = 0, b = 4, f (x) =
Z4
V = 2π
√
x x dx = 2π
0
Z4
x
√
x. Then
3/2
0
4
2 5/2 128π
dx = 2π x =
5
5
0
Example 4.
Find the volume
√ of the solid obtained by rotating about the x-axis the region
bounded by the curves y = x, y = 0, and x = 4.
Solution:
x = y 2 , 0 ≤ y ≤ 2.
Z2
2
Z2
y · y dy = 2π
V = 2π
0
0
6
2
y 4 y dy = 2π = 8π
4 0
3
Example 5.
Find the volume of the solid obtained by rotating about the line x = −2 the
region bounded by the curves y = 4x − x2 , y = 8x − 2x2 .
Solution:
Draw a picture(!). Vertices of both parabolas are at x = 2: (2, 4) and (2, 8).
Intesections of parabolas: 4x − x2 = 8x − 2x2 , x2 − 4x = 0, x = 0, x = 4. We divide [0, 4] into
subintervals ov width ∆x. Let x be in one of the subinervals. Average radius of a cylindrical
shell is the distance between x and −2 which is x − (−2) = x + 2. The height is the difference
between parabolas: h(x) = 8x − 2x2 − (4x − x2 ) = 4x − x2 . Then
Z4
Z4
2
(x + 2)(4x − x ) dx = 2π
V = 2π
0
2 3 x4
(8x + 2x − x ) dx = 2π 4x + x −
3
4
2
3
0
8
256
= 2π 4 · 16 + · 16 − 4 · 16 =
π
3
3
7
2
4
0
Section 7.4
Arc Length
Picture. We want to find the length L of a curve given by y = f (x), where f 0 (x) is continuous
and x runs through the interval [a, b]. A curve is made of small arcs. Consider the ith arc. Its
length by the Pythagorean theorem is
s
2
q
∆yi
2
2
∆xi
∆si ≈ (∆xi ) + (∆yi ) = 1 +
∆xi
Then
n
X
L=
∆si =
n
X
i=1
s
1+
i=1
∆yi
∆xi
2
∆xi
or exactly
L = lim
n→∞
n
X
s
1+
i=1
∆yi
∆xi
Zb
2
s
1+
∆xi =
Solution:
dy
dx
2
Zb q
1 + (f 0 (x))2 dx
dx =
a
a
because ∆xi → dx and ∆yi → dy as n → ∞ and
Example 1.
dy
= f 0 (x).
dx
Find the length L of the curve given by y = 2(x − 1)3/2 , when 1 ≤ x ≤ 2.
s
2
p
√
dy
dy
1/2
= 3(x − 1) , 1 +
= 1 + 9(x − 1) = 9x − 8. Hence,
dx
dx
Z2
L=
√
9x − 8 dx [u = 9x − 8, du = 9xdx, u(1) = 1, u(2) = 10]
1
1
=
9
Z10
u
1
Example 2.
Solution:
1/2
10
√
1 2 3/2 2
du = · u =
(10 10 − 1)
9 3
27
1
√
Find the length L of the curve given by y = 1 − x2 , when 0 ≤ x ≤ 1.
s
2 r
1
dy
x
dy
x2
= −√
, 1+
= 1+
=√
. Hence,
2
2
dx
dx
1−x
1−x
1 − x2
Z1
L=
√
dx
1 − x2
0
Trig substitution: x = sin θ, 0 ≤ θ ≤ π/2, dx = cos θ dθ,
Zπ/2
L=
cos θ dθ
=
cos θ
0
1 − x2 =
Zπ/2
dθ = π/2
0
8
√
√
cos2 θ = cos θ. Then
Notice, that the curve is a quarter of the unit circle whose length is 2π.
Example 3.
Solution:
Find the length L of the curve given by y = ln(cos x), when 0 ≤ x ≤ π/3.
s
2
√
√
dy
dy
= − tan x, 1 +
= 1 + tan2 x = sec2 x = sec x. Hence,
dx
dx
Zπ/3
√
√
L=
3
−
1
−
0)
=
ln(1
+
3)
sec x dx = ln | sec x + tan x||π/3
=
ln(2
+
0
0
9
Section 7.6
Applications to Physics and Engineering
Work
W = F d work = force x distance (here the force is constant)
F =m
d2 s
dt2
force = mass x acceleration
If the force is variable F = f (x), then the work done in moving an object from a to b is
Zb
W =
f (x) dx
a
√
Example 1. When a particle is located x meters from the origin, a force of 3x newtons acts
on it. How much work is done in moving the particle from x = 3 to x = 12?
Solution:
12
r 12
Z12 √
√ Z 1/2
√ 2 √ 12
x 3x dx = 3 x dx = 3 · x x = 2x
= 42
W =
3
3 3
3
3
Hook’s Law
3
The force is proportional to x: f (x) = kx, k > 0 is the spring constant.
Example 2.
A spring has a natural length of 50 cm. If a 60-N force is required to keep it
stretched to a length of 80 cm, how much work is required to stretch it from 50 cm to 60 cm?
Solution:
80 − 50 = 30 cm = 0.3 m. Then f (0.3) = 60, k · 0.3 = 60, k = 200. Hence
f (x) = 200x and
0.1
Z0.1
2
W = 200x dx = 100x = 1J
0
0
Hydrostatic Pressure and Force Suppose that a thin horizontal plate with area A sq.
meters is submerged in a fluid of density ρ kg per m3 at a depth d meters below the surface.
The fluid directly above the plate has volume V = Ad and the mass m = ρV = ρAd. The
force exerted by the fluid on the plate is F = mg = ρgAd. The pressure on the plate is
P = F/A = ρgd.
Example 3.
A plate in an irrigation canal is in the form of a trapezoid 4 feet wide at the
bottom, 8 feet wide at the top, with the height equal to 3 feet. It is placed vertically in the
canal and is submerged in water 2 feet deep. Find the hydrostatic force in pounds on the plate.
The weight density of water is 62.5 = 125/2 lb/ft3 .
Solution 1: Let’s place the coordinate axes as it is shown on the picture. We use horizontal
strips of equal width ∆y and consider a strip on the depth d = −y with −5 ≤ y ≤ −2 according
10
to the picture.
a
5−d
5+y
2
=
=
, a = (5 + y).
2
3
3
3
4
4
4
The length of the strip is l = 4 + 2a = 4 + (5 + y) = (8 + y) and its area is ∆A = (8 + y) ∆y.
3
3
3
By similar triangles:
The pressure on the strip is P = 62.5d = −62.5y (note that d = −y since y is negative) and
the force is
250
125 4
y · (8 + y) ∆y = −
(8y + y 2 ).
∆F = P ∆A = −
2
3
3
Then the total force is
Z−2
F =−
−5
−2
250
250
y3
250
8
125
2
2
(8y + y ) dy = −
4y +
=−
16 − − 100 +
3
3
3 −5
3
3
3
250
135
=−
−
= 250 · 15 = 3750 lb
3
3
Solution 2:
Now we place the coordinate axes differently. See the picture below. As before
we use horizontal strips of equal width δy and consider a strip on the depth d = 2 − y with
−3 ≤ y ≤ 0 according to the picture.
5−d
5−2+y
2
a
=
=
, a = (3 + y).
2
3
3
3
4
4
4
The length of the strip is l = 4 + 2a = 4 + (3 + y) = (6 + y) and its area is ∆A = (6 + y) ∆y.
3
3
3
By similar triangles:
The pressure on the strip is P = 62.5d = 62.5(2 − y) and the force is
125
4
250
(2 − y) · (6 + y) ∆y =
(12 − 4y − y 2 ).
2
3
3
Then the total force is
∆F = P ∆A =
Z0
F =
−3
0
250
250
y3
250
2
2
(12 − 4y − y ) dy =
12y − 2y −
[0 − (−36 − 18 + 9)]
=
3
3
3 −3
3
=
250
· 45 = 250 · 15 = 3750 lb
3
11
Section 7.7
Differential Equations
Examples. Order of a DE = order of the highest derivative. A solution.
Separable Equations
A separable equation is a 1st order DE that can be written in the form
dy
= g(x) f (y)
dx
Assuming that f (y) neq0 we denote h(y) =
1
. Then the equation becomes
f (y)
h(y) dy = g(x) dx
and we integrate both sides to get
Z
Z
h(y) dy =
g(x) dx
and solve the last for y.
Initial-Value Problem
Example 1.
Solution:
y dy = x dx,
IVP = DE together with the initial condition y(x0 ) = y0
Solve the IVP:
dy
x
= .
dx
y
R
y dy =
where C = 2 (C2 − C1 ).
y0 =
x
,
y
y(0) = 2.
It is a separable equation.
R
x2
y2
+ C1 =
+ C2 ,
2
2
√
Then y = ± x2 + C.
x dx,
y 2 = x2 + C,
Initial condition: y(0) = 2 > 0. We use the solution with ”+” sign. y(0) =
C = 4.
√
Answer: y = x2 + 4.
√
2x
x
x
Checking: y 0 = √
=√
= , y(0) = 4 = 2.
y
2 x2 + 4
x2 + 4
Example 2.
Solve the IVP:
y 0 = e−y (2x − 5),
√
0+C =
√
C = 2,
y(3) = 0.
dy
2x − 5
=
. It is a separable equation.
dx
ey
R y
R
ey dy = (2x − 5) dx,
e dy = (2x − 5) dx, ey = x2 − 5x + C, y = ln(x2 − 5x + C),
Solution:
Initial condition: y(3) = ln(−6 + C) = 0, −6 + C = 1, C = 7.
12
Answer: y = ln(x2 − 5x + 7).
Logistic Growth
dy
= ky (M − y) (logistic DE), 0 < y < M , y(0) = y0 .
dt
The logistic equation models world population growth.
Z
Z
dy
Solution: It is a separable equation.
= k dt.
y (M − y)
1
1 1
1
=
+
. So,
y (M − y)
M y M −y
Z
Z
1
1
1
1
dy +
dy = kt + C,
[ ln y − ln (M − y) ] = kt + C
M
y
M −y
M
y
y
[recall that 0 < y < M ], ln
= M (kt + C),
= eM (kt+C) = AekM t , A = eM C .
M −y
M −y
Example 3.
Solve the IVP:
Initial condition gives
y0
y
y0
y0 M
= A. Hence,
=
ekM t or y(t) =
.
M − y0
M −y
M − y0
y0 + (M − y0 )e−kM t
Notice that lim y(t) = M .
t→∞
Direction Field
y 0 = F (x, y)
Example 4.
Solution:
Sketch the direction field for the DE:
y 0 = 1 − xy.
Notice that on the lines x = 0 and y = 0 we have y 0 = 1.
Mixing Problems
y 0 = F (x, y)
Example 4.
Consider a tank with volume 100 liters containing a salt solution. Suppose a
solution with .1 kg/liter of salt flows into the tank at a rate of 5 liters/min. The solution in the
tank is well-mixed. Solution flows out of the tank at a rate of 5 liters/min. If initially there is
20 kg of salt in the tank, how much salt will be in the tank as a function of time?
13
Solution:
Let y(t) denote the amount of salt in kg in the tank after t minutes. We use a
fundamental property of rates:
Total Rate = Rate in - Rate out
To find the rate in we use
kg
kg
liters
(.1)
= .5
min
liter
min
The rate at which salt leaves the tank is equal to the rate of flow of solution out of the tank
times the concentration of salt in the solution. Thus, the rate out is
5
5
y kg
liters y kg
·
=
min 100 liter
20 min
The differential equation for the amount of salt is
y
y 0 = .5 − , y(0) = 20.
20
Using the method of separation of variable we find
Z
Z
dy
10 − y
dy
1
dy
1
t
=
,
= dt,
=
dt, − ln |10 − y| =
+ C1
dt
20
10 − y
20
10 − y
20
20
10 − y = C2 e−0.05t , (C = e−C1 ), y = 10 − Ce−0.05t .
The initial condition gives y(0) = 10 − C = 20, C = −10.
Hence, the amount of salt in the tank after t minutes is given by the formula
y(t) = 10 + 10e−0.05t .
Notice that lim y(t) = 10 + 0 = 10
t→∞
14