AREAS &
VOLUMES
www.math.ohiou.edu/~vardges/math263b/slides/chapter7/7.17.2areasvolumes.ppt
Areas Between Curves
To find the area:
• divide the area into n strips of
equal width
• approximate the ith strip by a
rectangle with base Δx and height
f(xi) – g(xi).
• the sum of the rectangle areas is a
good approximation
• the approximation is getting better
as n→∞.
y = f(x)
y = g(x)
The area A of the region bounded by the curves y=f(x), y=g(x), and
the lines x=a, x=b, where f and g are continuous and f(x) ≥ g(x) for
all x in [a,b], is
b
A   [ f ( x)  g ( x)]dx
a
Example

2
1
y1  2  x 2
2  x2  x dx
2
1 3 1 2
2x  x  x
3
2 1
8
1 1

 
 4   2    2   
3
3 2

 
y2   x
8
1 1
6 2 
3
3 2
27
36  16  12  2  3

6
6
9

2
If we try vertical strips, we
have to integrate in two parts:
y x
dx
y  x2
dx
0
2
x   x  2 dx
y x
dy
Since the width of the strip
is dy, we find the length of
the strip by solving for x in
terms of y.
y x
y  x2
y2  x
y2 x
  y  2  y
0
x dx  
4
We can find the same area
using a horizontal strip.
y  x2
2

2
2
2
dy
1 2
1 3
y  2y  y
2
3 0
8
24
3
10

3
General Strategy for Area Between Curves:
1
Sketch the curves.
2
Decide on vertical or horizontal strips. (Pick
whichever is easier to write formulas for the length of
the strip, and/or whichever will let you integrate fewer
times.)
3
Write an expression for the area of the strip.
(If the width is dx, the length must be in terms of x.
If the width is dy, the length must be in terms of y.
4
Find the limits of integration. (If using dx, the limits
are x values; if using dy, the limits are y values.)
5
Integrate to find area.
Volumes
To find the volume of a solid S:
• Divide S into n “slabs” of equal width Δx (think of slicing a loaf of bread)
• Approximate the ith slab by a cylinder with base area A(xi) and “height” Δx. The
volume of the cylinder is A(xi)Δx
• the sum of the cylinder areas is a good approximation for the volume of the solid
• the approximation is getting better as n→∞.
x
Let S be a solid that lies between x=a and x=b. If the cross-sectional area of S in
the plane Px , perpendicular to the x-axis, is A(x), where A is an integrable
n
function, then the volume of S is
b
V
lim
*
A
(
x
 i )xi   A( x)dx
max xi 0 i 1
a
Example of a disk
2
How could we find the volume
of the cone?
y x
1
0
1
2
3
4
One way would be
to cut it into a series of disks
(flat circular cylinders)
and add their volumes.
The volume of each disk is:
 r 2  the thickness

 x
2
dx
In this case:
r= the y value of the function
thickness = a small change
in x = dx

2
The volume of each flat
cylinder (disk) is:
y x
 r 2  the thickness
1

0
1
2
3
 x
2
dx
4
If we add the volumes, we get:
   x
2
4
0
dx
4
   x dx
0


2
4
 8
x2
0
Example of rotating the region about y-axis
1
The region between the curve x 
, 1  y  4 and the
y
y-axis is revolved about the y-axis. Find the volume.
y
x
1
1
2
3
4
1
 .707
2
1
 .577
3
1
2
We use a horizontal disk.
The thickness is dy.
4
3
2
The radius is the x value of the
1
function 
.
dy
y
1
2
 1 
V  
dy
 y 
1


4
 
4
0
1
1
1
dy
y
volume of disk
0
2
  ln y 1    ln 4  ln1   ln 2  2 ln 2
4
y
The natural draft cooling tower
shown at left is about 500 feet
high and its shape can be
approximated by the graph of
this equation revolved about
the y-axis:
500 ft
x
x  .000574 y 2  .439 y  185
The volume can be calculated using the disk method with
a horizontal disk.

500
0
.000574 y
2
 .439 y  185 dy  24, 700, 000 ft 3
2
Example of a washer
4
3
y  2x
2
y  x2
The region bounded by
y  x 2 and y  2 x is
revolved about the y-axis.
Find the volume.
1
If we use a horizontal slice:
yx
y  2x
y
x
2
2
yx
0
1
2
The volume of the washer is:

V  
0


4
 y
2
 y
 
2
2

 dy

1 

V     y  y 2  dy
0
4 

4
V 
4
0
1 2
y  y dy
4
The “disk” now has a hole in
it, making it a “washer”.
 R   r   thickness
  R  r  dy
2
2
2
2
outer
radius
4
1 
1
   y 2  y3 
12  0
2
inner
radius
 16 
  8  
3

8

3

y  x2
4
3
y  2x
2
1
0
1
r
y  2x
y
x
2
y  x2
yx
2
r  2 y
y2
   4  2 y   4  4 y  y dy
0
4
1
4
1 2
   3 y  y  4 y 2 dy
0
4
4
V    R2  r 2 dy
0
2

y

  2   2 y
0
2

 dy
2


4
 3 2 1 3 8 
    y  y  y 
12
3 0
 2
16 64 
8

    24    
3 3
3

3
2

y2 
    4  2 y    4  4 y  y dy
0
4 

4
The outer radius is:
y
R  2
2
The inner radius is:
R
4
4
If the same region is
rotated about the line x=2:
Volumes of Solids of Revolution
The solids we considered are examples of solids of revolution because
they are obtained by revolving a region about a line. In general, we
calculate the volume of a solid of revolution by using the basic defining
formula
b
V   A( x)dx
a
or
d
V   A( y)dy
c
and we find the cross-sectional area A(x) or A(y) in one of the following
ways:
• If the cross-section is a disk, we find the radius of the disk (in terms of
x or y) and use
A = π(radius)2
• If the cross-section is a washer, we find the inner radius rin and outer
radius rout and compute the area of the washer by subtracting the area of
the inner disk from the area of the outer disk:
A = π(outer radius)2 - π(inner radius)2