# Finding Volumes Using Integrals

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In this section, we will learn about: Using integration to find out the volume of a solid.
1. 6
APPLICATIONS OF INTEGRATION
2. APPLICATIONS OF INTEGRATION
6.2
Volumes
In this section, we will learn about:
Using integration to find out
the volume of a solid.
3. In trying to find the volume of a solid,
we face the same type of problem as
in finding areas.
4. We have an intuitive idea of what volume
However, we must make this idea precise
by using calculus to give an exact definition
of volume.
5. We start with a simple type of solid
called a cylinder or, more precisely,
a right cylinder.
6. As illustrated, a cylinder is bounded by
a plane region B1, called the base, and
a congruent region B2 in a parallel plane.
 The cylinder consists of
all points on line segments
perpendicular to the base
and join B1 to B2.
7. If the area of the base is A and the height of
the cylinder (the distance from B1 to B2) is h,
then the volume V of the cylinder is defined
V = Ah
8. In particular, if the base is a circle with
radius r, then the cylinder is a circular
cylinder with volume V = πr2h.
9. RECTANGULAR PARALLELEPIPEDS
If the base is a rectangle with length l and
width w, then the cylinder is a rectangular box
(also called a rectangular parallelepiped) with
volume V = lwh.
10. IRREGULAR SOLIDS
For a solid S that isn’t a cylinder, we first
‘cut’ S into pieces and approximate each
piece by a cylinder.
 We estimate the volume of S by adding the volumes
of the cylinders.
 We arrive at the exact volume of S through a limiting
process in which the number of pieces becomes large.
11. IRREGULAR SOLIDS
We start by intersecting S with a plane
and obtaining a plane region that is called
a cross-section of S.
12. IRREGULAR SOLIDS
Let A(x) be the area of the cross-section of S
in a plane Px perpendicular to the x-axis and
passing through the point x, where a ≤ x ≤ b.
 Think of slicing S
with a knife
through x and
computing the
area of this slice.
13. IRREGULAR SOLIDS
The cross-sectional area A(x) will vary
as x increases from a to b.
14. IRREGULAR SOLIDS
We divide S into n ‘slabs’ of equal width ∆x
using the planes Px1, Px2, . . . to slice the solid.
 Think of slicing a loaf of bread.
15. IRREGULAR SOLIDS
If we choose sample points xi* in [xi - 1, xi], we
can approximate the i th slab Si (the part of S
that lies between the planes Pxi  1 and Pxi ) by a
cylinder with base area A(xi*) and ‘height’ ∆x.
16. IRREGULAR SOLIDS
The volume of this cylinder is A(xi*).
So, an approximation to our intuitive
conception of the volume of the i th slab Si
V ( Si )  A( xi*)x
17. IRREGULAR SOLIDS
Adding the volumes of these slabs, we get an
approximation to the total volume (that is,
what we think of intuitively as the volume):
n
V  A( xi *) x
i 1
 This approximation appears to become
better and better as n → ∞.
 Think of the slices as becoming thinner and thinner.
18. IRREGULAR SOLIDS
Therefore, we define the volume as the limit
of these sums as n → ∞).
However, we recognize the limit of Riemann
sums as a definite integral and so we have
the following definition.
19. DEFINITION OF VOLUME
Let S be a solid that lies between x = a
and x = b.
If the cross-sectional area of S in the plane Px,
through x and perpendicular to the x-axis,
is A(x), where A is a continuous function, then
the volume of S is:
n b
V lim  A( xi *) x  A( x) dx
x  a
i 1
20. When we use the volume formula
b
V   A( x)dx , it is important to remember
a
that A(x) is the area of a moving
cross-section obtained by slicing through
x perpendicular to the x-axis.
21. Notice that, for a cylinder, the cross-sectional
area is constant: A(x) = A for all x.
 So, our definition of volume gives:
b
V  A dx  A  b  a 
a
 This agrees with the formula V = Ah.
22. SPHERES Example 1
Show that the volume of a sphere
of radius r is
3
V  r .
4
3
23. SPHERES Example 1
If we place the sphere so that its center is
at the origin, then the plane Px intersects
the sphere in a circle whose radius, from the
Pythagorean Theorem,
2 2
y r  x
24. SPHERES Example 1
So, the cross-sectional area is:
2 2 2
A( x)  y  (r  x )
25. SPHERES Example 1
Using the definition of volume with a = -r and
b = r, we have:
r r
V  A( x) dx    r  x  dx 2 2
r r
r
2 2
2  (r  x ) dx (The integrand is even.)
0
3 r 3
 2 x   3 r 
2  r x   2  r  
 3 0  3
3
 r
4
3
26. The figure illustrates the definition of volume
when the solid is a sphere with radius r = 1.
 From the example, we know that the volume of
the sphere is 43  4.18879
 The slabs are circular cylinders, or disks.
27. The three parts show the geometric
n n
of the Riemann sums
2 2
 A( xi )x   (1  xi )x when n = 5, 10,
i 1 i 1
and 20 if we choose the sample points xi*
to be the midpoints xi .
28. Notice that as we increase the number
of approximating cylinders, the corresponding
Riemann sums become closer to the true
29. VOLUMES Example 2
Find the volume of the solid obtained by
rotating about the x-axis the region under
the curve y  x from 0 to 1.
Illustrate the definition of volume by sketching
a typical approximating cylinder.
30. VOLUMES Example 2
The region is shown in the first figure.
If we rotate about the x-axis, we get the solid
shown in the next figure.
 When we slice through the point x, we get a disk
with radius x .
31. VOLUMES Example 2
The area of the cross-section is:
A( x)  ( x ) 2  x
The volume of the approximating cylinder
(a disk with thickness ∆x) is:
A( x)x  xx
32. VOLUMES Example 2
The solid lies between x = 0 and x = 1.
1
So, its volume is: V A( x)dx
0
1
 xdx
0
2 1
x  
  
2 0 2
33. VOLUMES Example 3
Find the volume of the solid obtained
by rotating the region bounded by y = x3,
Y = 8, and x = 0 about the y-axis.
34. VOLUMES Example 3
As the region is rotated about the y-axis, it
makes sense to slice the solid perpendicular
to the y-axis and thus to integrate with
respect to y.
 Slicing at height y,
we get a circular
disk with radius x,
where x  3 y
35. VOLUMES Example 3
So, the area of a cross-section through y is:
2 2 2/3
A( y )  x  ( y )  y
3
The volume of the approximating
cylinder is:
2/3
A( y )y  y y
36. VOLUMES Example 3
Since the solid lies between y = 0 and
y = 8, its volume is:
8
V  A( y ) dy
0
8
23
 y dy
0
8
 5
3
96
 5 y
3

  0 5
37. VOLUMES Example 4
The region R enclosed by the curves y = x
and y = x2 is rotated about the x-axis.
Find the volume of the resulting solid.
38. VOLUMES Example 4
The curves y = x and y = x2 intersect at
the points (0, 0) and (1, 1).
 The region between them, the solid of rotation, and
cross-section perpendicular to the x-axis are shown.
39. VOLUMES Example 4
A cross-section in the plane Px has the shape
of a washer (an annular ring) with inner
radius x2 and outer radius x.
40. VOLUMES Example 4
Thus, we find the cross-sectional area by
subtracting the area of the inner circle from
the area of the outer circle:
2 2 2
A( x)  x   ( x )
2 4
 ( x  x )
41. VOLUMES Example 4
1
Thus, we have: V  A( x) dx
0
1
2 4
  ( x  x ) dx
0
3 5 1
x x 
   
 3 5 0
2

15
42. VOLUMES Example 5
Find the volume of the solid obtained
by rotating the region in Example 4
about the line y = 2.
43. VOLUMES Example 5
Again, the cross-section is a washer.
This time, though, the inner radius is 2 – x
and the outer radius is 2 – x2.
44. VOLUMES Example 5
The cross-sectional area is:
2 2 2
A( x)  (2  x )   (2  x)
45. VOLUMES Example 5
So, the volume is:
1
V A( x) dx
0
1
2 2
   2  x   (2  x)  dx
2
0 
1
  x  5 x  4 x  dx
4 2
0
5 3 2 1
x x x  8
   5  4  
 5 3 2 0 5
46. SOLIDS OF REVOLUTION
The solids in Examples 1–5 are all
called solids of revolution because
they are obtained by revolving a region
about a line.
47. SOLIDS OF REVOLUTION
In general, we calculate the volume of
a solid of revolution by using the basic
defining formula
b d
V  A( x) dx or V  A  y  dy
a c
48. SOLIDS OF REVOLUTION
We find the cross-sectional area
A(x) or A(y) in one of the following
two ways.
49. WAY 1
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
50. WAY 2
If the cross-section is a washer, we first find
the inner radius rin and outer radius rout from
a sketch.
 Then, we subtract the area of the inner disk from
the area of the outer disk to obtain:
A = π(outer radius)2 – π(outer radius)2
51. SOLIDS OF REVOLUTION Example 6
Find the volume of the solid obtained
by rotating the region in Example 4
about the line x = -1.
52. SOLIDS OF REVOLUTION Example 6
The figure shows the horizontal cross-section.
It is a washer with inner radius 1 + y and
outer radius 1  y .
53. SOLIDS OF REVOLUTION Example 6
So, the cross-sectional area is:
2 2
A( y )  (outer radius)   (inner radius)
2 2

 1  y    1  y 
54. SOLIDS OF REVOLUTION Example 6
The volume is:
1
V A( y )dy
0
1 2
  2
  1  y
0 
    1  y  dy

1

  2 y  y  y dy
0
2

3 1
 4 y 2 y 2 y3  
     
 3 2 3
0
2
55. In the following examples, we find
the volumes of three solids that are
not solids of revolution.
56. VOLUMES Example 7
The figure shows a solid with a circular base
of radius 1. Parallel cross-sections
perpendicular to the base are equilateral
Find the volume of the solid.
57. VOLUMES Example 7
Let’s take the circle to be x2 + y2 = 1.
The solid, its base, and a typical cross-section
at a distance x from the origin are shown.
58. VOLUMES Example 7
2
As B lies on the circle, we have y  1  x
So, the base of the triangle ABC is
2
|AB| = 2 1  x
59. VOLUMES Example 7
Since the triangle is equilateral, we see
that its height is 3 y  3 1  x 2
60. VOLUMES Example 7
Thus, the cross-sectional area is :
2 2
A( x)  2 1  x  3 1  x
1
2
2
 3(1  x )
61. VOLUMES Example 7
The volume of the solid is:
1
V  A( x) dx
1
1 1
2 2
 3(1  x ) dx 2  3(1  x ) dx
1 0
3 1
 x  4 3
2 3  x   
 3 0 3
62. VOLUMES Example 8
Find the volume of a pyramid
whose base is a square with side L
and whose height is h.
63. VOLUMES Example 8
We place the origin O at the vertex
of the pyramid and the x-axis along its
central axis.
 Any plane Px that
passes through x and
is perpendicular to
the x-axis intersects
the pyramid in a
square with side of
length s.
64. VOLUMES Example 8
We can express s in terms of x by observing
x s 2 s
from the similar triangles that  
h L 2 L
Therefore, s = Lx/h
 Another method is
to observe that the
line OP has slope
L/(2h)
 So, its equation is
y = Lx/(2h)
65. VOLUMES Example 8
Thus, the cross-sectional area is:
2
2 L 2
A( x) s  2 x
h
66. VOLUMES Example 8
The pyramid lies between x = 0 and x = h.
h
So, its volume is: V   A( x) dx
0
2
L 2
h
  2 x dx
0 h
2 3 h
L x  L2 h
 2  
h 3 0 3
67. In the example, we didn’t need to place
the vertex of the pyramid at the origin.
 We did so merely to make the equations
simple.
68. Instead, if we had placed the center of
the base at the origin and the vertex on
the positive y-axis, as in the figure, you can
verify that we would have
obtained the integral:
2
L
h
2
V   2 (h  y ) dy
0 h
2
Lh

3
69. VOLUMES Example 9
A wedge is cut out of a circular cylinder of
radius 4 by two planes. One plane is
perpendicular to the axis of the cylinder.
The other intersects the first at an angle of 30°
along a diameter of the cylinder.
Find the volume of the wedge.
70. VOLUMES Example 9
If we place the x-axis along the diameter
where the planes meet, then the base of
the solid is a semicircle
with equation
2
y  16  x , -4 ≤ x ≤ 4
71. VOLUMES Example 9
A cross-section perpendicular to the x-axis at
a distance x from the origin is a triangle ABC,
2
whose base is y  16  x and whose height
2
is |BC| = y tan 30 =
° 16  x 3.
72. VOLUMES Example 9
Thus, the cross-sectional area is:
21 2
A( x)  16  x 
1
2 16  x
3
2
16  x

2 3
73. VOLUMES Example 9
The volume is:
4
V   A( x) dx
4
2
4 16  x 1 4

4
2 3
dx 
3
 0
16  x  2
dx
3 4
1  x  128
  16 x  3  
3 0 3 3