Applications of Integration - Unit 7
Area between curves:
Area
□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is a basic problem to practice finding the volume
using the washer method.
Problem:
Find the volume that f(x) and g(x) sweep out from -2 to 4 when they are rotated around the x-axis.
f(x)= (x-2)^2 + 3
g(x)= (1/.3)x
Solution:
v 
 f
v 
 x  2
v 
 x
2
 g 2 dx
4
2
2
4
2
2

2
 3  x /3 dx
2
 4 x  7  x 2 /9 dx now use numerical integration
2

fnint x 2  4 x  7  x 2 /9, x,2,4
2

 406.5333

□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is an intermediate problem designed to help
students practice finding the surface area of conics.
Problem:
Show that the surface area of a half circle with diameter 6 rotated about the x-axis = 4r 2  36
Solution:
First we need to come up with the equation for the circle being described.Since we know the radius, (half the diameter)
y2  x2  9
we can quickly generate or
y  9  x2
Now, we have to apply the equation for finding surface area, S  2

But to use this, we first have to find f x 
1/ 2
1
y  9  x 2
2x 
2
2x

y 
2
2 9 x
x
y 
9  x2



a
 f x 
b

1  f x  dx
2
now we need to square y prime,
y 
2

x2
9  x 2 
now we can plug in everything into the equation

S  2
3

9  x 2 1
3
 2
3
x2
dx
9  x2
 9
 9  x 9  x
2
3
 2

dx
2 

3
 3dx
3
 6x 33
 36

Chap 7: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):
FINDING THE ARC LENGTH:
The equation for finding the arc length, s, of a function is:
b
s=

1   f ' ( x) dx
2
a
PROBLEM: (Hard)
Find the arc length of f(x) = 3x2 – 1 on [-1, 2].
SOLUTION:
f ‘(x) = 6x
2

s=
1  6 x  dx
2
1
2
=

1  36 x 2 dx
1
1
tan( b)
6
1
dx = sec 2 (b) db
6
x=
=

1
6
1
=
6
=
1  tan 2 (b) 
 sec
3
 (sec
1
sec 2 (b) db
6
db
2
x)(sec x)dx
u= sec x
du= sec xtan x
dv= sec 2 xdx
v= tan x
1
[sec xtan x-  (tan x)(sec x tan x)dx ]
6
1
=
[sec xtan x-  (tan 2 x)(sec x)dx ]
6
=
1
6
1
=
6
1
=
6
1
=
6
=
[sec xtan x-
 (sec
[sec xtan x-
 (sec
[sec xtan x-
 (sec
[sec xtan x-
 (sec
2
x  1)(sec x)dx ]
3
x  sec x)dx ]
3
3
xdx   sec xdx ]
xdx   sec xdx ]
sec x  tan x
dx
sec x  tan x
sec 2 x  sec x tan x
7  sec 3 xdx = sec xtan x+ 
dx
sec x  tan x
6  sec 3 xdx = sec xtan x-
 sec
3
xdx   sec x
Using a “u substitution”:
u=sec x + tan x, du= sec x tan x + sec2 x
7  sec 3 xdx = sec xtan x+

du
u
7  sec 3 xdx = sec xtan x+ ln |sec x + tan x|
 sec
3
xdx =
sec x tan x  ln | sec x tan x |
+C
7
Chap 7: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): FINDING THE SURFACE AREA:
PROBLEM: (Medium)
Find the surface area on [-1, 2] if f(x) from the previous problem is revolved around y = -4.
SOLUTION:
2

A = 2 [ f ( x)  4] dx
1
2

= 2 [3x 2  1  4] dx
1
2

2

= 2 3 x 2 dx + 2 3 dx
1
1
2
1
= 2 [ x 3 ] 21 + 2 [3x]
= 2  [2 3  (1) 3 ] + 2 [3(2)  3(1)
= 2  (9) + 2 (9)
= 36 
Chap 7: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): FINDING AREA BETWEEN CURVES:
The rule here is *ALWAYS top – bottom!
 top - bottom
PROBLEM: (Medium)
Find the area between g and h on the interval [-4, 8]
g=
1 2
x
8
h=x+1
SOLUTION:
Because the graphs switch top and bottom positions at x = -0.9, we must do two separate integrals:
0.9
A=

[
4
x2
 ( x  1)] dx +
8
0.9

=
4
[
x2
] dx 8
0.9
8

[( x  1) 
0.9
0.9
x2
] dx
8
8
8
 x dx -  [1] dx +  [ x] dx +  [1] dx -
4
4
 0.9
0.9
8

0.9
[
x2
] dx
8
1 x 3 0.9 x 2 0.9
x2 8
1 x3 8
0.9
8
= [ ]  4 - [ ]  4 - [ x]  4 + [ ] 0.9 + [ x] 0.9 - [ ] 0.9
8 3
8 3
2
2
= 2.63 – (-7.595) – 3.1 + 31.595 + 8.9 – 21.3637
= 26.2563
Chap 7: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):
FINDING VOLUME OF A SOLID USING THE DISC METHOD:
PROBLEM: (Easy)
What is the volume of the solid when 3x2 is revolved around y = -2 on the interval [0,2]?
SOLUTION:
2

V =  (3x 2 ) 2 dx
0
2

=  (9 x 4 ) dx
0
x5 2
]0
5
32
= 9  (  0)
5
288
=
5
= 9[
Chap 7: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):
CENTER OF MASS:
PROBLEM: (Medium)
You have a uniform lamina (sheet) made from f(x) = x – 2 and g(x) = ln x. Find the center of mass.
SOLUTION:
x
 x[ f ( x)  g ( x)]dx
 [ f ( x)  g ( x)]dx
 f ( x)  g ( x) 
[ f ( x)  g ( x)] dx
2

y
 [ f ( x)  g ( x)] dx
 
Equate the two statements: x – 2 = ln x
Solve for x. e x 2  x
Find the zeroes. These will be your interval of integration. [0.159, 3.146]
3.146
x
 x[ x  2  ln x]dx
0.159
3.146
 [ x  2  ln x]dx
0.159
 x  2  ln x 

[ x  2  ln x] dx
2
0.159 
3.146
y

3.146
 [ x  2  ln x] dx
0.159
 2.721439085
=1.3963
x
 1.949090479
.4043936047
y=
 0.2075
 1.949090479
Chap 7: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Labor Day and Volume
find the volume of a cylinder by revolving the line y=2 about the x-axis by using the disc method. Integrate from 0 to 3.
V

Adx
A  r 2
r y 2

V 
 4

3
0
3
0
 2 2 dx
dx  4  x 0  4  3
3
 12
That wasn’t so bad was it?

Chap 7: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Areas Between Curves
Let’s find the area between curves. They are f(x) = x+2 and g(x) = x2 from zero to their intersect point. An intersect point
is where two lines intersect.
First find the intersect point:
f (x)  g(x)
 x  2  x2
 0  x2  x  2
 x 2
Now do the rest:

Atotal  Atop  Abottom
 f (x)dx   g(x)dx
  (x  2)dx   (x )dx
  (x  2)dx   (x )dx

2
2
2
0
0
2
2
2
1
 1 
  x 2  2x   x 3
2
0 3 0
1
 1

  (2  0) 2  2(2  0)  (2  0) 3 
2
 3

8 
8
 2  4     6 
3
3
10

3
Now pat yourself on the back.
 Chap 7: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): The Area of a Kiss
The following question involves the graph of f(x) = -(x-1)3 + 1 which, when revolved about the y-axis, forms a shape
similar to that of a Hershey’s Kiss.
Find the area of the first quadrant.
A

2
0
[(x 1)3 1]dx
A    [(x 1)3 1]dx
2
0
2


A    (x 1) 3 dx 
0

2
0
dx
2
2
1
A   [(x 1) 4 ] [x ]
0
0
4
1
1
A   [14 ]  [14 ]  2
4
4
1 1
A   2
4 4
A2
The Volume of a Kiss
 The following question involves the graph of f(x) = -(x-1)3 + 1
revolved about the y-axis, forms a shape similar to that of a
Kiss.
Find the volume of the Kiss when revolved around the y-axis
method.
which, when
Hershey’s
using the disc
y  (x 1) 3  1
 1 y  (x 1) 3
 (1 y)1 3  1  x
 [(1 y)
V    [(1 y)
2
V 
 1] dy
23
 2(1 y)1 3  1]dy
0
2
0
2
13
2
2
2
3
3
(1 y) 5 3 ] 2 [ (1 y) 4 3 ]  [y ]
0
0
0
5
4
3
3 5 3
V
(1) 5 3 
(1)  2
5
5
6
16
V
2
5
5
V  [

Chap 7: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Lifting and Work
A 20 lbs bag of kittens is being raised by a 100 ft iron chain weighing 0.5 lbs/ft. Determine the work needed to raise the
weight 50 ft (assume a 100% efficient pulley).
Rope=0.5 lbs/ft
y=100 ft
20
lb
s
Let y equal the distance raised.
F = Force of weight + Force of rope = 20 lbs + 0.5(100+y) lbs/ft
W = F x distance, using infinitesimal heights the differential equation is formed:
dW  Fdy
W 

50
0
[20  0.5(100  y)]dy
50
W  [70y  0.25y 2 ]  70(50)  0.25(50) 2
0
W  3500  625
W  2875 ft lbs


Chap 7: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Area and Volume of f(x) = sin x
Consider the graph of sinx from 0 to . Now find the volume when it is revolved about the x-axis using the disc method.
Volume using the Disc Method:
dV  r 2 dx
V 




2

0

0
sin 2 x dx
(1 cos2x)dx

 

1
 x  sin 2x
0
2  2


2
V
 
2
2
units3
Chap 7: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Area of f(x) = sin x
 Consider the graph of sinx from 0 to . Now find the surface area of it when it is revolved about the x-axis using
numerical integration rounded to the nearest hundredth of a unit.
Surface Area:
dS  2rds
r  f (x)  sin x
1
 dy 2 2
ds  
 
1 
 dx
 dx  
1
 dS  2 sin x 1 cos2 x 2 dx
 S  2


0
1
sin x 1 cos 2 x 2 dx
Using numerical integration:

S = 14.42 units2
Chap 7: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an intermediate-level problem. It
deals with finding volume of a solid using cross-sections. This type of problem should be used when students are learning
about volumes and integration, along with the disc/washer method.
The base of a solid is a region in the first quadrant bounded by y = x2+1, x = 2, and the x and y axis. If this base is made
up of square cross-sections perpendicular to the x-axis, what is the volume of the solid?
Solution: To solve this problem, we need to find a general equation for the area of each cross-section. To then find volume
of the entire solid, we can use an integral involving this equation over a certain interval.
Each cross-section has one side which lies in the xy-plane. This side is a vertical line segment, from the x-axis up to the


2
function y=x2+1 . Therefore the area of each cross-section will be x 2  1 . So to find volume of the entire solid, we
simply integrate this expression from 0 to 2.
2


V   x 2  1 dx
2
0
2
  (1  2 x 2  x 4 )dx
0
2
2
1 

 x  x 3  x 5 
5 0
 3
=13.73
Chap 7: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an intermediate-level problem,
involving finding area between curves. In this problem, you have to find area between two different equations.
What is the area of the region in the first quadrant bounded by the y-axis, y=cos2x, and y=2x?
Solution: To solve this problem we first need to find where cos2x and 2x are equivalent. This will be the point to which
we integrate. Since cos2x is above 2x, we subtract 2x when integrating to find area.
cosx=2x
x=.3695
Now integrate:
0.3695
 cos 2 x  2 xdx
0
=0.2
Chap 7: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an intermediate-level problem.
Students need to be comfortable with the washer method.
The image above shows the graph of both y = x and y = cosx. B and A are the regions created when the graphs
intersect. Find the volume created when B is revolved around the line y = 1.
Solution: To solve this problem, we need to know the equation
The volume generated is given by the equation:
b


V     f x   k   g x   k  dx
2
2
a
Since we’re revolving around y = 1, k = 1.
Now we need the intersection of y = x and y = cosx:
cosx = x
x = .74
.74 is our value for b.
So,
V 
 x  1  cos x  1 dx
0.74
2
2
0
Using your calculator, find that V = 0.996
Chap 7: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an advanced-level problem. area
between curves, derivative review
f ( x)  x 2 . Find:
a) The equation of the line tangent to the graph with slope m = 4
b) The area that is bounded by f(x), the tangent line with slope 4, and the x-axis
Solution:
To solve this problem we will use derivative techniques as well as integration.
To solve for part a, we must find the derivative of f(x)
f ’(x) = 2x
To find the point on the graph where the slope = 4, set this equal to the derivative 2x.
2x = 4
x=2
Now, plug this into the original equation to get y = 4
Find the complete equation for the line using point-slope form:
4 = 4(2)+b
b = -4
g(x) = 4x – 4
Now part b:
We already know that the two graphs intersect at x = 2, by definition of a tangent line. So to find the area trapped by these
two graphs and the x-axis we integrate:
2
2
 x dx 
0
x3
3
2

0
8
3
Now, subtract the area under g(x):
g(x) crosses the x-axis at: x = 1
A right triangle is created with a base of 1 and height of 4. Using A = .5bh, we find the area of the triangle to be 2.
So, for our final answer A =
8
2
2 
3
3
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 7: f ( x)   x 2  14 x  45 and g ( x)  x 2  16 x  63 .
a. Find the area between the curves.
Solution: When f(x) = g(x), -x2 + 14x – 45 = x2 – 16x + 63
6 or x = 9
x=
g (x )
 2 3

Area =  [ f ( x)  g ( x)] dx  
x  15 x 2  108 x  = 9
 3
6
6
9
9
b. Find the volume of the shaded region when it is rotated about the
line x  4 .
f (x )
Shell
Method:
Each Shell:
dx A litt
b
Total Volume (adding up all the shells) =
 2 (r )(h) dx
a
Total Volume =
9
2  ( x  4) [ f ( x)  g ( x)] dx  63
6
c. Find the volume when the region is rotated about the line y  3 .
9
A cross-section of the
volume:
f(x) + 3
9


6
6
Volume =  [( f ( x)  3) 2 dx   ( g ( x)  3) 2 ] dx
9

=  [( f ( x)  3) 2  ( g ( x)  3) 2 ] dx  81
g(x) + 3
y = -3
6
dx
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 7: f ( x)  4 x 2  7 x  4 and g ( x)   x 2  4 x  7
Fundamental Theorem of Calculus), how to find the area of a semicircle, and the formula for arc length.
a. Find the area of the region between the curves.
Solution: f(x) = g(x) when x = -1.81327 and when x = 1.21327


 5 3 3 2

 5 x  3 x  11 dx  
x  x  11x 
 23.1027
Area =  g ( x)  f ( x) dx 

2
 3
 1.81327
1.81327
1.81327
1.21327
1.21327
2
1.21327
b. Let the region be the base of a solid with semicircles cross-sections perpendicular to the y-axis, find the volume of the
solid.
dV 
Solution:
1.21327
V
 r2
2
dx 
  g ( x)  f ( x) 

2
  g ( x)  f ( x) 


2

1.81327


2

2
  5x

1.21327
 dx 
8

2
2
2
 3x  11 dx  83.1031
2
1.81327
c. Find the circumference of the region.
Solution: Circumference of region = Arc length of f(x) + Arc length of g(x) from –1.81327 to 1.21327.
b

Arc length =

1  f ' ( x)

2
dx
a

1.21327

Circumference of region =
1.21327

=

1.21327

2
+

1.21327
2

dx
1  g ' ( x)

2
dx
+ 1.81327
1.81327
1  8 x  7 dx
1.81327
1  f ' ( x)
1   2 x  4 dx
2
1.81327
= 21.416 + 14. 306 = 35.722
Chap 7: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): Find the area between two curves
f ( x)  x 4  4 x 2  4 and g(x)  x 2 . (In this problem, you need to find the intercept of the two curves first. Then
you add the parts of the region together).
Solution: To find the intersection of two f ( x)  x 4  4 x 2  4 and g(x)  x 2 ,
x 2  x 4  4x 2  4
0  ( x 2  1)( x 2  4)
So the intersections are 1-,1,2, and -2.
Therefore, the area is
1
1
2
1
A   (g(x) - f(x))dx  
2
(f(x) - g(x))dx   ( g ( x)  f ( x)) dx
1
1
1
  (x 2 - x 4  4x 2 - 4)dx  
2

  15 x 5  53 x 3  4 x
 
1
2
1
5
1
2
(x 4  4x 2  4 - x 2 )dx   (x 2 - x 4  4x 2 - 4)dx
x 5  53 x 3  4 x
 
1
1
1
5
1

2
x 5  53 x 3  4 x 1
 (2.5333  1.0667)  (2.5333  2.5333)  (2.5333  1.0667)
 5.0666
Chap 7: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): Disk & Shell advanced] The graph of the
relation x 2 / 3  y 2 / 3  1 is shown below. Find the volume of the solid which is generated by revolving the region about
the x-axis. (The oval represents the region swept out by a thin, rectangular element of thickness is dx. Imagine the left side
of the oval coming out at you.)
Solution #1 (Disk method): The radius of the cross section is simply y, since we’re revolving around the x-axis.
x 2 / 3  y 2 / 3  1  y  (1  x 2 / 3 ) 3 / 2 . In the first and second quandrants y  (1  x 2 / 3 ) 3 / 2 . So the area of the
cross section is


A( x)   r 2   (1  x 2 / 3 ) 3 / 2 , and the volume element is dV = A(x) dx. Therefore, the
1
volume is given by V 
2
1
 A( x) dx = 2 A( x) dx
1
0
1

 2  (1  x
0
 dx
2 / 3 3/ 2 2
)
1
 2  (1  x 2 / 3 ) 3 dx
0
1
7

9
9
1 3
5
32
3
 9 9 1

 2  (1  3x 2 / 3  3x 4 / 3  x 2 )dx  2  x  x  x  x   2 1     
5
7
3 0
 5 7 3  105

0
1
Solution #2 (Shell method): In the second
picture imagine the thin blue segment of
thickness dy revolving about the x-axis,
sweeping out a cylindrical shell. The volume of
this shell is its area times its thickness; its area
its circumference times its height; and its
circumference is its radius time 2. The radius
is y  (1  x 2 / 3 ) 3 / 2 since we’re revolving
around the x-axis, and by symmetry, the height
y = radius of disk
for disk method.
1
y  (1  x 2 / 3 ) 3 / 2
1
-1

of the cylinder is 2x. V  2 yh dy
0
1
0
1
 2  yh dy
0
-1
Since h = the hight of the shell and y is its
2
3
1
3
2
V  2  y 2(1  y ) dy
0
radius,
2
1
3
 4  y (1  y 3 ) 2 dy
0
1
2
3
2 
Let t  (1  y ) , then dt   y 3 dy .
3
So
3
0
V  4  (t ) 2 y (
1
0
1
2 3
y )dt
3
3
2
 4( )  t 2 (1  t )dt
3 1
1
3
1
y = radius of
cylindrical shell for
shell method.
5
8
   (t 2  t 2 )dt
3 0
dy
}
1
5
7
8 2 2 2 2 
  t  t 
3 5
7 0
-1
8 2 2
 (  )
3 5 7
32


105
0
1
-1
h = 2x = 2 (1  y 2 / 3 ) 3 / 2
= height of shell in shell
method.
Chap 7: □ Sample problem (by Frances Ha,
Lusiana Hadi, & Amity Xu):
Find the length of a smooth curve y  3x 2  4 on the interval [0,1 6] .
(Using the equation of the length of a smooth curve L 
solve the problem).

b
a
1  ( f ' ) 2 dx and you also need to use integration by Parts to
Solution: Since y '  6 x ,

L

b
a
1
6
0

1  ( f ' ) 2 dx
1  (6 x ) 2 dx
1

x  cot  ,


6

dx   1 csc 2  d ,

6

1
6
 16 cot   1  cot   

0  16 cot   0  cot   
4

2

1
csc 2 1  cot 2  d
2
6
1 4
   csc 3  d
6 2
  
4
u  csc 

2
dv  csc  d



Since
L 
 v   cot 


1
 cot  csc     cot  ( csc  cot  d )
6
1
1
cot  csc    cot 2  csc  d
6
6
1
1
cot  csc    (csc 2   1) csc  d
6
6
1
1
1
cot  csc    csc 3  d   csc  d
6
6
6
1
1
1
cot  csc    csc 3  d  ln csc   cot 
6
6
6


 du   csc  cot d
1
csc 3  d ,
6
1
1
cot  csc   L  ln csc   cot 
6
6
cot  csc   ln csc   cot 
 /4
2L  
 C  / 2
6
cot  csc   ln csc   cot 
 /4
L
 C  / 2
12
L
 .1893  2.4849
 2.2956
Chap 7: □ Sample problem (by David Mesri & Jake Mathis): Using the disk method, rotate y  2 x 3  3x  5
about the x-axis and evaluate the volume swept out from x  0 to x  5 .
b
Solution: The disk method is as follows: V 
  yx
dx . In this case, yx  is known so you can plug it in,
2
a
  2 x

2
5
integrate term by term and evaluate the expression. V 
3
 3x  5 dx
0
5
    2 x 3 dx 
0

0 3x dx  50 dx
5
5
5
 1 

3
    x 4    x 2  5 x   300 .
2
 2 
0
Chap 7: □ Sample problem (by David Mesri & Jake Mathis): Find the area between these two curves (from 0 to 5.09):
f x  
x and g x   x 3  5x 2
  f x 
5.09
Solution: The area is as follows

 g x  dx
5.09
 
0
0


x  x 3  5 x 2  dx   59.64 .

□ Sample problem Disk & Shell. (by David Mesri & Jake Mathis): Using both the shell and disk method, find the
VOLUME??? swept out by e(x – 3) from 0 to 7.
Solution: One must be familiar with both the disk and shell method formulas to approach this problem.
Disk: V  
  f x 
2
7

V    e x
 3

dx

2
dx
0
7


1
  e 2 x
2 0
 6

 2 dx
* since du is 2 dx, we must place
1
2
out front
u  2x  6
du  2 dx


1
  e u dx = e u
2
2
= 4682.47
Shell:
 
 
=

 2

 2 x  6  7  27   6 

e
e
 e 2 0   6  = e 8  e 6 
0=
2
2
To use the shell method one needs to understand what exactly they are doing. The shell method works by filling up the
area swept out by the equation with cylinders of infinitesimal thickness. Here’s a run through. The easy part is coming
up with an equation for the circumference of the cylinder (2πr). Obviously, r is equal to y(x). The more difficult part of
using the shell method is finding the varying length of the cylinders. Use the above picture for reference. Obviously,
the light blue cylinder is shorter than the dark blue cylinder. Basically, the easiest way to think about this example is:
Upper x limit – x. For example, if the above picture is from 0 to 10 and the blue cylinder starts at 1. Its length is (101)=9. (Note: One has to think about the exact situation, for example this wouldn’t the case if we were integrating right
to left.) However, one has to remember that the cylinder has an infinitesimal thickness (dy). So we have to get x in
terms of y.
Okay so this is the basic equation (for the given situation)
V  2
upper y bound



y (upper x bound )  xin terms of y 
 dy
lower y bound
In the problem we asked to find the area swept out by y=e(x – 3) from 0 to 7.
found by
plugging in 7 to
e( x -3)
V  2
54.59815



x in terms of y

y  e x
 3

y 7  ln  y   3 dy ln  y   x  3
0.04979707
found by
plugging in 0 to
e( x -3)
 4682.47
ln  y   3  x
Upper x bound
*Used numerical integration
Chap 7: □ Sample problem (by Liz King & Katherine Wallig): This is a complicated problem that deals with the area
between two curves. The two graphs cross each other more than once, so you will have to do to antiderivatives in order
to figure this question out.
Find the area between the two functions y = x2 – 3 and y = 0.17x3.
Solution: First graph these functions just so you can see what they look like and where the area between them is, then
find the points of intersection so that you know from where to where one function is on top of the other, and where it is
below the other
From the graph you know that the functions intersect at -1.54, 2.18, and 5.24, and that first the cubic function is on top,
then the squared function.
2.1845664
3
2
 (.17 x  x  3)dx 
1.541751
5.2395373
2
 (x
 3  .17 x 3 )dx , this you can solve by breaking it up into different parts and
2.1845664
plugging in the boundary numbers, or plug it into your calculator and you should get the numerical answer of 11.45
units2.
Chap 7: □ Sample problem (by Liz King & Katherine Wallig): This problem deals with such a thing called the washer
method. You have probably learned the disk method by now, and that too is a washer method, only the inside part that
you are subtracting is zero so we ignore it. In this problem, since it is more complex, you can not ignore the inside, but
must set up a couple of antiderivatives to find the volume.
Find the volume when the area from -3 to 3 between the functions y = .5x2-2
and y = .15x4 is rotated around y = -3.
Solution: We know that area is equal to pi times radius squared, and we are creating a bunch of different circles with
thickness dx and holes in the middle. The radius is .15x4 + 3 minus the inside circle of .5x2- 2 + 3. We added the plus
three because we are rotating it around the line y = -3. This gives us the equations
3
3
3
3
4
2
2
2
  (.15x  3) dx    (.5x  2  3) dx
This equation you could workout by hand or plug it into a calculator and
come up with the exact answer of 191.595  or an approximate answer of 601.91 units cubed.
Chap 7: □ Sample problem (by Liz King & Katherine Wallig): This one involves using a shell method, which means
you make a tube around the y axis that looks like a toilet (hee hee it says toilet) paper tube, then with the thickness dx,
many of them to create a volume.
Find the volume of the area between the two functions y = 2x and y = -2x2+5-2x when it is flipped around the y axis.
Solution: Here like a flattened toilet paper tube, the area is length times width. Length is 2 times pi times radius
(because that is circumference) and the radius is equal to absolute value of x. Width is equal to -2x2+5-2x. The volume
1
 2 ( x )(2 x
is
2
 2 x  5)dx which is approximately 161.26 units3.
2
chap 7: □ Sample problem (by Merla Hübler & Lisa
Portis): area between graphs that change top and bottom
If f ( x)  5 sin( 5x ) and g ( x)  5 cos x , find the area of the
shaded region, writing the integration out the way you would
put it directly into a graphing calculator to solve and writing
it out the way you would if you were to integrate it by hand.
Plug both integrations into the calculator to find the answer
the two different ways.
Solution: The two graphs should be graphed on a graphing
calculator to find the needed intersection points. On the
calculator, the middle intersection point is not needed, but
when doing the problem without a calculator for the
integration, you must use the middle intersection point to
split it up.
With the help of the graphing calculator, the first and last intersection points are found to occur at x = -3.927 and x =
1.309.
With a graphing calculator to do the integration, you can simply use these to points and type in
1.309

3.927
5 sin x5  5 cos x dx .
However, when doing the integration by hand, you must find the intervals of x in which each graph is on top. Seeing,
with the help of the calculator, that f(x) is on top from x=-3.927 to the middle intersection point, which occurs at x= 1.963, the first integration should be

1.963
3.927
1.309
the second integration should be

1.963
(5 sin x5  5 cos x )dx , and seeing that g(x) is on top for the rest of the way,
(5 cos x  5 sin x5 )dx. Have the calculator solve both, and add them together.
The answer obtained through both methods should be the same: 13.2358.
chap 7: □ Sample problem (by Merla Hübler & Lisa Portis):
Advanced problem for practice with volumes and surface areas of revolution. Find the volume and surface area of the
object if the shaded region were hollowed out (the surface surrounding it would remain there). The diameter of the base of
the two cones (including the shaded region) is 10 inches, the diameter of the shaded region is 2 inches, and the length of
every diagonal side of the cones is 7 inches. Don’t use a calculator for the integrations.
Solution:
A 2-D graph should be set up so that the volume and surface area can be found by using their respective revolution
equations. Centering the shape around the origin, the graph set up to find the volume should only go from x=1 to x=5
since a shape with diameter of 2 in. is hollowed out.
Using the washer method, an equation for the y-coordinates is needed to use for f(x), so use two points to find the
equation for the line.
One of the points is already known to be (5,0), but another one is needed. Using the Pythagorean theorem, the other
endpoint can easily be found to use as the other point:
x 2  y 2  (diagonal length) 2
42  y 2  72
16  y 2  49
y 2  33
y  33
So, two points to use in the equation y  y1  m( x  x1 ) , which
will be used to find the slope of the line, are (5,0) and (1, 33 ).
0  33  m(5  1)
 33  4m
m
 33
4
Now, use the equation y  mx  b and one of the points to find the equation of the line.
0

33
4
b 
5
33
4
y 

33
4
(5)  b
x
5
33
4
Now that the equation for the y-coordinates, or f(x), in the washer method equation for
finding the volume of the object, is known, the washer method can be used.
b
V    [ f ( x)]2 dx
a
5
V    [  433 x  5 433 ] 2 dx
1
5
33 2
   ( 16
x  3316(10 ) ) x 
1

33
16
25 ( 33 )
16
)dx 
33
16
33
1
 [ x3  5 x 2  25 x]15  16
 [ 125
3  5( 25)  125  ( 3  5  25)]
3
 ( 543 )  44
Because this rotation only created half of the figure, the answer should be multiplied by 2 to get the whole figure’s
volume:
44  2  88π in3 (don’t forget your units!)
To get the figure’s surface area, the equation S  2
b
 r ( x)
a
1  [ f ' ( x)] 2 dx should be used. In this equation, a should
be 0 instead of 1, since only the inside of the figure is hollowed out. Also, r(x) should be what f(x) was in the volume
equation, since the axis of rotation is still the x-axis, and f’(x) is the same as r’(x).
d
r’(x)= dx
(
 33
4
x  5 433 )=  433
5
S  2  (  433 x  5 433 ) 1  (  433 ) 2 dx
0
5
5
0
0
 144   (  433 x  5 433 )dx  141633   ( x  5)dx 
7 33
8
[ x2  5 x]50 
2
7 33
8
[ 252  25]  175
16  33
Once again, this number only accounts for half of the total surface area,
so the answer is
175
8
π 33 in2.
chap 7: □ Sample problem (by Merla Hübler & Lisa Portis): work There is an octagonal-cylinder-shaped tank laying
on its side filled up halfway with a liquid that has a weight density of
exit
  25 lb ft 3 . The tank has sides of 5 2 ft, a length of l = 5 ft, and the
hole is centered at the top. Calculate the work necessary to empty out the
tank.
Solution:
Start out with the simple equation for work: W = Fd, where F is the force,
this case, the weight of the liquid. The weight can be calculated using the
density and area:
in
dV  A  dy
weight    A  dy
H can be defined as the height of the empty part of the
tank, and y as the variable used for the liquid level,
going downwards. By combining the weight and the
height, and equation is set up for the work:
dW    A  ( y  H )  dy
H

W   dW   [   A  ( y  H )]dy
0
The integration for the tank can be split up into two
parts. The first is from 0 to 5 2 2 , the length where the
sides of the tank are vertical.
W1 
5 2
2
 [   r  l  ( y  H )]dy
where r is the diameter at
0
5 2
5 2
2
 y2
 2
 Hy 
y = 0. W1    r  l  [ y  H ]dy    r  l 
2
0
0
  25 lb ft 3
r  5  5  5 2  10  5 2 ft


5 2
2
 y2

W1  (25)(10  5 2 )(5)   5  5 2 2  y 
2
0
H  5  5 2 2 ft
l  5 ft
 5 2 2
2
 (1250  625 2 ) 
 5  5 22 

2



    0 = 77,732.402
5 2
2


ft lb. The second filled part of the tank is the part that has sloping sides. You can find the slope of one of these sides to
be –1 by using the Pythagorean theorem or noticing that 5 2 creates a 45   45  90  triangle. Although the slope
seems to be positive in the picture, if you choose to define y as increasing downwards, as I did, then the width of the
tank is decreasing as y increases. Splitting up the tank vertically, you can just work with half of it, and double it. The
distance from this vertical center to a edge of the tank is defined as x.
Equation of line:
5 2
2
y  mx  b
A point on the line that you know:
y  x  b
 5   5 22  b


y  x  5  5 2

x  y  5  5 2

5 2
2
,
5 2
2

5
b  55 2
Using the above equation, we now know x according to y. This gives us the
A  2  x  l  2l ( y  5  5 2 )
function for the area according to y:
We can now use the original general equation for work:
W2 
5 5 2 2
 [   2l  ( y  5  5
2 )( y  H )]dy  2l
5 2
2

 y3 5 2 2
 2l 
 4 y  50  752 2
 3
5 5 2 2
 [( y  5  5
2 )( y  5  5 2 2 )]dy
5 2
2

5 5
 2
y
 5 22
2
 (5  5 2 ) 3
 
2
 2l  
 5 2 2 (5  5 2 2 ) 2  50  752 2 (5  5 2 2 )    
 
3

 



      50 
3
5 2
2

3
5 2
2
3
75 2
2
 
5 2
2

 2(5)( 25)[929.74  393.74]  134000 ft lb. The total work can now be found by adding W1 and W2 :
W1  W2  77732.402  134000  211732.402 ft lb.
chap 7: Kevin Stanford & Mike Mitchner area between two curves & washer method. difficult Consider the shaded
region shown in the graph bounded by y = x and y = x2 /2. a. Find area of the shaded region.
Solution: Solution Intro To solve the problem it must first be found where the two curves intersect, which by setting
them equal to each other is found to be at x=0
and x=2. After this all that is needed is the
formula for finding area between curves and
the washer method for finding the volume.
1 2
x
2
2x  x 2
x
This simple finds the
x  0orx  2
two points at which the curves intersect
After finding the points of intersection we use
them to find our area through an integral

2
0
2
1
1 
8
2
1

x  x 2 dx   x 2  x 3    2    0 
2
6 0 
6
3
2
b. Find the volume of the shaded region when revolved about the x-axis.
Solution: V  

2
0
2
 2  1 2 2 
 8 32   16
1 3 1 5 
x        0    3.35
 x   x   dx    x 
20  0
 2  
3
 3 20   15

c. Find the volume of the shaded region revolved about the line y = -4.
Solution:
V 
2
0
2

2
1 2
 
1 4

2
2
2
x   4   x   4  dx   0 x  8 x  16   x  4 x  16  dx
2
 
4



2
 1
4
  1

 176
 176
   x 3  4 x 2  16 x    x 5  x 3  16 x    
 0 
  36.86.
3
  20
 0
 15
 15
 3
1 2
x
Here we subtracted a negative four from both x and 2 because the area is revolving over a larger space than in part
two. If you draw a picture with the line y = -4 drawn in, it is easy to see that the distance over which the shaded region
rotates grows four units higher, so we are subtracting a negative four, or in essence adding four to the height of the arc
to solve this equation.
chap 7: Kevin Stanford & Mike Mitchner Disc method
Find the area swept out by f ( x)  2 x 2  x on the closed interval [-1, 2] when
it is rotated about the x-axis.
Solution: The area is equal to a sum of all the infinitely thin “discs” or
cylinders that range from x = -1 to x = 2.
2
Area of a Cylinder  2r h . r  f (x) h  dx
dA  2f ( x) dx . 2
2
2

1
2
f ( x) 2 dx = 2  (2 x 2  x ) 2 dx . Evaluate the
1
2

integral using a calculator: 2 (2 x 2  x ) 2 dx  56.55 .
1
chap 7: Kevin Stanford & Mike Mitchner area between curves , disc method, washer method Little Timmy Tucker
has two functions: f ( x)  e x and g ( x )  ln x . a. Timmy wants to know the area between the two graphs from x = ½
to x = 1.
Solution: Solution Intro: To find the solution to this problem one must first find inner peace. Then you remember that
you take the integral of the upper equation minus the lower equation to find the area between the two graphs. Upper
1
equation = e
x
Lower equation=ln x
 (e
x
 ln x)dx  1.22 .
1/ 2
b. Find the volume generated when this region is revolved around y = 4 using the washer method.
Solution: Intro: The washer method where you find the volume by using infinitely thin donuts or washers if you will.
b

b

Disc Method:  [ f ( x)] dx . Washer Method:  [ y12  y 22 ]dx
2
a
a
1

y1  4  ln x
y2  4  e x
graph since you are revolving it about y = 4.  [( 4  ln x) 2  (4  e x ) 2 ]dx  23.61 .
1/ 2
The 4 is used to adjust the
chap 7: basic Disk method Patrick McCall & Nathan Dornfeld: Find the volume of the line y = 3x rotated about the xaxis from 0 to 3.
3

Solution: dA   (3x) 2 dx  A   9 x 2 dx =  3x 3
= 81  254.47 .
3
0
0
chap 7: basic Patrick McCall & Nathan Dornfeld: Find the area between the curves
y
x and y = x.
Solution: The points of intersection of the two curves define the region of interest. To find these points we equate the
1
x = x  x 2  x  x = 1 or x = 0.
functions and solve for x:
1

x dx   xdx =
0
0
3
2 2
x
3
1
0

1 2
x
2
1
0
=
1
.
6
chap 7: intermediate Patrick McCall & Nathan Dornfeld: Derive the arc length formula.
Solution: First we break the curve into little bits of arc length that are infinitesimally
small. Each of those bits would have a length ds, and each has an x component dx
and a y component dy. Using the Pythagorean Theorem, ds 2  dx 2  dy 2 . So,
 dy 2
ds  dx 2  dy 2  dx 2 1 
2
 dx 
2

dy
  dx 1     1  f ' ( x)2 dx .

 dx 

ds
The total arc length from x = a to x = b is the sum of all the infinitesimally small
dx
bits of arc length. We sum them up via integration. Thus, s 

b
a
dy
ds   1   f ' ( x)2 dx .
b
a
What would happened if you tried to find the arc length of a curve by integrating dx from a to b rather than ds ?
chap 7: intermediate Patrick McCall & Nathan Dornfeld: Disk method Given y1 
swept out by the area between the curves when they are revolved about the x-axis.
Solution: V  
1

x and y2 = x, find the volume
1
x dx    x 2 dx , where the first integral represents the volume swept out by y1, and the second
2
0
0
integral that of y2. We are subtracting because we only want the volume generated by the area between the curves.
(Some people prefer to write a single integral of a difference.) Proceeding with the integration we obtain
1
1
V    x dx    x dx =
2
0
 x2
2
0
1

0
1
 x3
=
3
0

2


3
=

.
6
chap 7: basic Patrick McCall & Nathan Dornfeld: arc length Find the length of the curve y = e x from x = 0 to 10.
Solution: s =

b
a


1  f x 
'
2
dx . f x   e , so s 
'
x
10

1  e x dx  295.75 , using numerical integration
0
chap 7: hard Patrick McCall & Nathan Dornfeld: disk , shell method. Find the volume of the hyperboloid formed by
revolving the right branch of the hyperbola given by x 2  y 2  1 about the x-axis from x  1 to x  10 . Show that the
disk and shell methods yield the same result.
Solution: Disk method: In the first quadrant we have y 
x 2  1 , where y is the radius of each thin disk, since we’re
revolving around the x-axis. For an arbitrary disk, its area is given A   y 2 and its infinitesimal volume is
dV  A dx   y 2 dx (since cross-sectional area times thickness is volume). We find the total volume by adding up the
10

10

volumes of an infinite number of infinitely thin disks, doing so via integration: V   y dx  
1
2


2
x 2  1 dx
1
10
x

 1000
  1 
   x 2  1dx     x    
 10     1  324 .
  3 
 3
3
1
1
10
3
Shell method: This time we use thin, horizontal strips that sweep out cylindrical shells when they’re revolved around the
x-axis. For an arbitrary strip: its thickness is dy, its length is 10 - x; the radius of the cylinder it sweeps out is y; the
circumference of the cylinder is 2 y; the surface area is 2 y(10 - x); and the volume of the cylinder is 2 y(10 - x) dy.
Since our strips are oriented horizontally, we will be integrating with respect to y, and, hence, y limits are needed.
99
2
99

 2

0

y 10  1  y 2 dy  20
0
99
99
 y dy  2 
0
 y2 
20  y dy  20  
 2 0
0
99
 2 y 10  x dy
99 when x = 10. Thus, V 
x  y  1  y = 0 when x = 1, and y =
2
y 1  y 2 dy . For the first integral we have
0
99
 10 (99)  990 . For the second integral we use the substitution u  1  y 2 , implying
du = 2y dy. We must not forget to switch to u limits (or convert back in terms of y after finding an antiderivative).
Opting for the former, u = 1 when y = 0, and u = 100 when y =
2
99

y 1  y 2 dy  
100

u du   
1
0
 
2 3/ 2
u
3
100
1
99 . So,
2
   (1000  1)  666 . Thus, V = 999 - 666 = 324, the same
3
result as we got with the disk method.
chap 7: Shell method Find the volume of the solid that results when the region is bounded by the curve y = 16 - x 2 and
the curve y = 16 - 4x by revolving the region about the y-axis. Use the cylindrical shells method.
Solution: We first determine where the graphs intersect by setting the two
expressions equal to each other:
16  x 2  16  4 x  x 2  4 x  x 2  4 x  0  x( x  4)  0  x = 0 or 4.
These will be our limits of integration. Since each function is 1-1 on the interval [0,
4], we could use thin, rectangular strips oriented either horizontally or vertically
without having the same function marking both endpoints for any strip. This means
we could use the washer method (horizontal strips perpendicular to the axis of
revolution) or the shell method (vertical strips parallel to the axis). Let’s use the shell
method, whose general formula for revolution about the y-axis is
b
V  2  x  f ( x)  g ( x)dx . Here the difference represents the height of the strip
a
(parabola minus line); dx is its thickness; x is the distance between the strip and the
y-axis, so x is the radius when the region is revolved; and 2 x is the circumference
of the cylindrical shell swept out by the strip. This circumference times the height is the surface area of the cylindrical
shell, and this area times dx is the volume of the shell. We add up all these volumes via integration:


4
  x 4 4x3 

V  2  x (16  x )  (16  4 x) dx  2  x( x  4 x) dx  2  ( x  4 x ) dx  2 

3 0
 4
0
0
0
  4 4 4  4 3 

  0  0)   134.04. Show you get the same result with the washer method.
 2 

3 
 4

4
2
4
4
2
3
2
chap 7: (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): volume of a sphere-Disk Method Find the volume of a
sphere with radius r.
Solution: Step 1: x 2  y 2  r 2
Step 2: Find equation for upper semi circle assume y  0 and solve for y: x 2  y 2  r 2 y 2  r 2  x 2 y 
r
r2  x 2
2
  ( r 2  x 2 ) dx =
Step3: 
r
r
 1
1
1
2
1 
2
2
  (r 2  x 2 )dx =  (r 2 x  x 3 )   [r 3  r 3  (r 2  v  (v)3 )] =  [ r 3  (r  r 3 ]   [ r 3  ( r 3 )] =
3
3
3
3
3
3
3
r
r
2
2
4
  [ r 3  ( r 3 )]  r 3 done!
3
3
3



r


Chap 7: (by Shaofeng Sun & Artem Rogachev): intermediate disc method Find the volume of y = 5x2+2x+10
when revolved around y = 2, on the interval of (2,5).
Solution: Interval (2,5) will be the limits of integration. Using washer’s method we will find the volume. Remember
the volume formula. Since the graph is above x-axis in the interval we use (5x2+2x+10-2) for r. So the integral is:
5
v    (5x 2  2 x  8) 2 dx  701015
. do numerical integration.
2
Chap 7: (by Shaofeng Sun & Artem Rogachev): advanced disc method Two functions: f(x)=x2-1, and g(x)=2x+2. You
are to take the region enclosed by the two functions and rotate it about the x-axis, which goes through the region. Find the
volume of the shape.
Solution: This is not an easy problem since we have to know which function is going to stick out more. We can still do
it. So first we are going to find the bounds of the region. Second we will find the interval on which one function’s radius
is bigger then the others. Finally we will use the volume formula, to evaluate the volume. To find the intersections set
the two functions equal to each other: x2 – 1 = x+1, x2-x-2=0, x=2 or x =-1;
At x=2, y=6; and at x=-1, y=0; Now find the interval for which x 2  1  2 x  2 and vice versa for the interval along
which lies the surface. There are no values of x for which x 2  1  2 x  2 , so that means 2x+2 is further from x-axis
the whole time. This means that this volume is equivalent to just rotating 2x+2. So just use the volume formula to with r
=2x+2.
2
 x3

2

2
x

2
dx

4

x

2
x

1
dx

4


x

x








3
 1
1
1
2
2
2
2
8
  13

 4   4  2  
 1  1   36
 3

 3


so the volume is 36  .