Transcript Slide 1

6
APPLICATIONS OF INTEGRATION
Summary
In general, we calculate the volume of
a solid of revolution by using the basic
defining formula
b
V   A( x) dx
a
or
V   A  y  dy
d
c
APPLICATIONS OF INTEGRATION
6.2
Volumes
In this section, we will learn about:
Using integration to find out
the volume of a solid.
VOLUMES
We start with a simple type of solid
called a cylinder or, more precisely,
a right cylinder.
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.
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.
IRREGULAR SOLIDS
We start by intersecting S with a plane
and obtaining a plane region that is called
a 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.
IRREGULAR SOLIDS
The cross-sectional area A(x) will vary
as x increases from a to b.
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.
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 Px 1 and Px ) by a
i
i
cylinder with base area A(xi*) and ‘height’ ∆x.
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:
V (Si )  A( xi *)x
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.
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.
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
i 1
a
V  lim  A( xi *)x   A( x) dx
x 
VOLUMES
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.
VOLUMES
Notice that, for a cylinder, the cross-sectional
area is constant: A(x) = A for all x.
 So, our definition of volume gives:
V   Adx  A  b  a 
b
a
 This agrees with the formula V = Ah.
SPHERES
Example 1
Show that the volume of a sphere
of radius r is
V  r .
4
3
3
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,
is:
y r x
2
2
SPHERES
Example 1
So, the cross-sectional area is:
A( x)   y   (r  x )
2
2
2
SPHERES
Example 1
Using the definition of volume with a = -r and
b = r, we have:
V   A( x) dx     r  x  dx
r
r
r
r
r
 2  (r  x ) dx
2
2
2
2
(The integrand is even.)
0
r
 2
 3 r 
x 
 2  r x    2  r  
3 0
3


3
 r
4
3
3
3
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.
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 .
VOLUMES
Example 2
The area of the cross-section is:
A( x)   ( x )   x
2
The volume of the approximating cylinder
(a disk with thickness ∆x) is:
A( x)x   xx
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
1
x  
  
2 0 2
2
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.
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
VOLUMES
Example 3
So, the area of a cross-section through y is:
A( y)   x   ( y )   y
2
2
3
The volume of the approximating
cylinder is:
A( y)y   y y
2/3
2/3
VOLUMES
Example 3
Since the solid lies between y = 0 and
y = 8, its volume is:
8
V   A( y ) dy
0
8
   y dy
23
0
96

3
3
 5 y


 0
5
5
8
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.
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.
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.
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:
A( x)   x   ( x )
2
2 2
  (x  x )
2
4
VOLUMES
Example 4
1

   (x
Thus, we have: V 
0
A( x) dx
1
0
2
 x ) dx
4
1
x
x 
   
5 0
3
2

15
3
5
VOLUMES
Example 5
Find the volume of the solid obtained
by rotating the region in Example 4
about the line y = 2.
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.
VOLUMES
Example 5
The cross-sectional area is:
A( x)   (2  x )   (2  x)
2 2
2
VOLUMES
Example 5
So, the volume is:
1
V   A( x) dx
0
    2  x
0
1

2 2
 (2  x)  dx

2
    x  5 x  4 x  dx
1
2
4
0
1
x
x  8
x
  5  4  
2 0 5
3
5
5
3
2
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.
SOLIDS OF REVOLUTION
In general, we calculate the volume of
a solid of revolution by using the basic
defining formula
b
V   A( x) dx
a
or
V   A  y  dy
d
c
SOLIDS OF REVOLUTION
We find the cross-sectional area
A(x) or A(y) in one of the following
two ways.
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
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
SOLIDS OF REVOLUTION
Example 6
Find the volume of the solid obtained
by rotating the region in Example 4
about the line x = -1.
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 .
SOLIDS OF REVOLUTION
Example 6
So, the cross-sectional area is:
A( y)   (outer radius)   (inner radius)
2

  1 y

2
  1  y 
2
2
SOLIDS OF REVOLUTION
Example 6
The volume is:
1
V   A( y )dy
0


   1 y
0

1
1


2

 1  y  dy

2

   2 y  y  y dy
0
2
1
 4 y 2 y 2 y3  
 
   
2
3
2
 3
0
3