1. APPLICATIONS OF INTEGRATION6
APPLICATIONS OF INTEGRATION

2. In general, we calculate the volume of
Summary
In general, we calculate the volume of
a solid of revolution by using the basic
defining formula

3. 6.2 Volumes In this section, we will learn about:
APPLICATIONS OF INTEGRATION
6.2
Volumes
In this section, we will learn about:
Using integration to find out
the volume of a solid.

4. We start with a simple type of solid
VOLUMES
We start with a simple type of solid
called a cylinder or, more precisely,
a right cylinder.

5. 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.

6. For a solid S that isn’t a cylinder, we first
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.

7. We start by intersecting S with a plane
IRREGULAR SOLIDS
We start by intersecting S with a plane
and obtaining a plane region that is called
a cross-section of S.

8. Let A(x) be the area of the cross-section of S
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.

9. The cross-sectional area A(x) will vary as x increases from a to b.
IRREGULAR SOLIDS
The cross-sectional area A(x) will vary
as x increases from a to b.

10. We divide S into n ‘slabs’ of equal width ∆x
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.

11. If we choose sample points xi* in [xi - 1, xi], we
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 and ) by a
cylinder with base area A(xi*) and ‘height’ ∆x.

12. The volume of this cylinder is A(xi*).
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
is:

13. Adding the volumes of these slabs, we get an
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):
This approximation appears to become better and better as n → ∞.
Think of the slices as becoming thinner and thinner.

14. Therefore, we define the volume as the limit of these sums as n → ∞).
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.

15. Let S be a solid that lies between x = a and x = b.
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:

16. When we use the volume formula , it is important to remember
VOLUMES
When we use the volume formula
, it is important to remember
that A(x) is the area of a moving
cross-section obtained by slicing through
x perpendicular to the x-axis.

17. Notice that, for a cylinder, the cross-sectional
VOLUMES
Notice that, for a cylinder, the cross-sectional
area is constant: A(x) = A for all x.
So, our definition of volume gives:
This agrees with the formula V = Ah.

18. Show that the volume of a sphere of radius r is
SPHERES
Example 1
Show that the volume of a sphere
of radius r is

19. If we place the sphere so that its center is
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,
is:

20. So, the cross-sectional area is:
SPHERES
Example 1
So, the cross-sectional area is:

21. Using the definition of volume with a = -r and b = r, we have:
SPHERES
Example 1
Using the definition of volume with a = -r and
b = r, we have:
(The integrand is even.)

22. Find the volume of the solid obtained by
VOLUMES
Example 2
Find the volume of the solid obtained by
rotating about the x-axis the region under
the curve from 0 to 1.
Illustrate the definition of volume by sketching
a typical approximating cylinder.

23. The region is shown in the first figure.
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

24. The area of the cross-section is:
VOLUMES
Example 2
The area of the cross-section is:
The volume of the approximating cylinder
(a disk with thickness ∆x) is:

25. The solid lies between x = 0 and x = 1. So, its volume is:
VOLUMES
Example 2
The solid lies between x = 0 and x = 1.
So, its volume is:

26. Find the volume of the solid obtained
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.

27. As the region is rotated about the y-axis, it
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

28. So, the area of a cross-section through y is:
VOLUMES
Example 3
So, the area of a cross-section through y is:
The volume of the approximating
cylinder is:

29. Since the solid lies between y = 0 and y = 8, its volume is:
VOLUMES
Example 3
Since the solid lies between y = 0 and
y = 8, its volume is:

30. The region R enclosed by the curves y = x
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.

31. The curves y = x and y = x2 intersect at the points (0, 0) and (1, 1).
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.

32. A cross-section in the plane Px has the shape
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.

33. Thus, we find the cross-sectional area by
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:

34. VOLUMES
Example 4
Thus, we have:

35. Find the volume of the solid obtained
VOLUMES
Example 5
Find the volume of the solid obtained
by rotating the region in Example 4
about the line y = 2.

36. Again, the cross-section is a washer.
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.

37. The cross-sectional area is:
VOLUMES
Example 5
The cross-sectional area is:

38. VOLUMES
Example 5
So, the volume is:

39. The solids in Examples 1–5 are all called solids of revolution because
they are obtained by revolving a region
about a line.

40. In general, we calculate the volume of
SOLIDS OF REVOLUTION
In general, we calculate the volume of
a solid of revolution by using the basic
defining formula

41. We find the cross-sectional area A(x) or A(y) in one of the following
SOLIDS OF REVOLUTION
We find the cross-sectional area
A(x) or A(y) in one of the following
two ways.

42. If the cross-section is a disk, we find
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

43. If the cross-section is a washer, we first find
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

44. Find the volume of the solid obtained
SOLIDS OF REVOLUTION
Example 6
Find the volume of the solid obtained
by rotating the region in Example 4
about the line x = -1.

45. The figure shows the horizontal cross-section.
SOLIDS OF REVOLUTION
Example 6
The figure shows the horizontal cross-section.
It is a washer with inner radius 1 + y and
outer radius

46. So, the cross-sectional area is:
SOLIDS OF REVOLUTION
Example 6
So, the cross-sectional area is:

47. SOLIDS OF REVOLUTION
Example 6
The volume is: