Ch. 8 – Applications of Definite
Integrals
8.3 – Volumes
• The volume of a solid with a known, integrable cross-sectional area
b
of A(x) from x=a to x=b is
V   A( x) dx
a
• Ex: Find the volume of the solid formed by revolving the region of
the curve between the x-axis and y = 2 + cosx around the x-axis
over [0, 2π].
– Graph the function from [0, 2π]. Picture the shape visually.
– The revolution of this shape will produce an hourglass-like shape
– The formula above says we need a cross-sectional area formula, and since the
cross-section is a circle, here’s our formula:
A( x)   r 2   y 2
  (2  cos x) 2
r
– Now integrate over the proper interval.
Use your calculator.
2
V    (2  cos x) 2 dx
0
 88.826u
3
• Ex: The region enclosed by the y-axis and the graphs of y = x + 2
and y = 4 – x2 is revolved around the x-axis to form a solid. Find its
This is called the
volume.
washer method
A( x) because
r r
the crosssection
 (4  x ) is
 ( xa 2)
washer
– Find a cross-sectional area formula:
2
2
1
2
2 2
2
– What are the limits of integration?
x  2  4  x2
( x  2)( x  1)  0
x2  x  2  0
x  1 or  2
– Now integrate over the proper interval.
Use your calculator.
1
V    (4  x 2 ) 2   ( x  2) 2  dx
0
 22.619u
3
r2
r1
• Ex: The region enclosed by the graphs of y  x  3 and y = x + 3 is
revolved around the x-axis to form a solid. Find its volume. NO
CALCULATOR!
– Find a cross-sectional area formula:
A( x)   r2 2   r12
  ( x  3) 2   ( x  3) 2
– What are the limits of integration?
( x  3)2  x  3
x3 x3
x2  6 x  9  x  3
x2  5x  6  0
x  2 or  3
– Now integrate over the proper interval.
V 
2
 ( x  3)   ( x  3) 2  dx
3
 x3 5 x 2
 2

  
 6x 

2
 3
 3
6
r2
r1
2
y

1

x
• Ex: The region enclosed by the x-axis and the graph of
is revolved around the line y = 3 to form a solid. Find its volume.
NO CALCULATOR!
– Use the washer method, but the washer radii will be trickier to find:
A( x)   r2 2   r12
  (3)2   (3  y)2   9  (3  (1  x 2 )) 2 


  9  (2  x 2 ) 2    9  (4  4 x 2  x 4 )    5  4x 2  x 4 
– Whew! Limits of integration:
1  x2  0
x  1
– Now integrate over the proper interval.
1
V    (5  4 x 2  x 4 ) dx
1

4 x3 x5  1
   5x 
 
3
5  1

 
4 1 
  2  5   
3 5 
 
104

15
• Ex: The region enclosed by the x-axis and the graph of
y   x 2  5 x  4 over [1, 4] is revolved around the y-axis to
form a solid. Find its volume.
– When the radius of the cross-section is parallel to the axis of revolution, use
the shell method!
b
Shell Method : V  2  r ( x)h( x) dx
a
• r(x) is the radius as a function of x
• h(x) is the height as a function of x
– Find radius and height formulas:
r ( x)  x
h( x )  y   x 2  5 x  4
– Now integrate over the proper interval.
4
V  2  x( x 2  5 x  4) dx
1
4
 2  ( x3  5 x 2  4 x) dx
1
 x 4 5 x3
45
2 4
 2   
 2x 

3
 4
1
2
• Ex: The region enclosed by the line x = 1, the x-axis, and the graph
of y  x is revolved around the line x = 3 to form a solid. Find
its volume. NO CALCULATOR!
– Use the shell method, but watch out for the tricky radius!
– Find radius and height formulas:
h( x )  y
r ( x)  3  x
 x
– Now integrate over the proper interval.
1
V  2  (3  x) x dx
0
1
3
V  2  (3 x  x 2 ) dx
0
 32 2 52  1
 2  2 x  x 
5

0
2  16

 2  2   
5

5
x
3–x
y  x . This
• Ex: A region is bounded by the x-axis, x = 3, and
region is cut into square cross-sections that sit perpendicular to the
x-axis. Find the volume of the solid generated by these crosssections without a calculator.
– I’ll cheat and let you look at my calculator for the graph.
– We need a cross-sectional area formula for a square:
2
x

(
x
)
A( x)  s  y
2
2
– Now integrate over the proper interval.
V 
3
0
x2 3
x dx 
2 0
 4.5u 3
y  x . This
• Ex: A region is bounded by the x-axis, x = 3, and
region is cut into equilateral triangle cross-sections that sit
perpendicular to the x-axis. Find the volume of the solid generated
by these cross-sections without a calculator.
– I’ll cheat and let you look at my calculator for the graph.
– Know this formula for equilateral triangle cross-sections:
s2 3
y2 3
x 3
A( x) 


4
4
4
– Now integrate over the proper interval.
V 
3
0
x 3
x2 3 3
dx 
4
8 0
9 3 3

u
8