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					                6
APPLICATIONS OF INTEGRATION
APPLICATIONS OF INTEGRATION




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

In trying to find the volume of a solid,
we face the same type of problem as
in finding areas.
VOLUMES

We have an intuitive idea of what volume
means.

However, we must make this idea precise
by using calculus to give an exact definition
of volume.
VOLUMES

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

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.
CYLINDERS

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
as:
          V = Ah
CYLINDERS

In particular, if the base is a circle with
radius r, then the cylinder is a circular
cylinder with volume V = πr2h.
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 Pxi 1 and Pxi ) by a
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
              V  lim  A( xi *)x   A( x) dx
                                      b

                  x                 a
                         i 1
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   A dx  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
                                     2     2
           r                 r
                  r
         2  (r  x ) dx
                      2   2
                                   (The integrand is even.)
                  0
                               r
              2    x    3
                             3 r              3
         2  r x    2  r  
                   3 0        3
         r
          4
          3
                  3
SPHERES

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 4   4.18879
                   3

    The slabs are circular cylinders, or disks.
SPHERES

The three parts show the geometric
interpretations of the Riemann sums
  n             n

 A( xi )x    (1  xi )x when n = 5, 10,
i 1          i 1
                    2    2


and 20 if we choose the sample points xi*
to be the midpoints xi .
SPHERES

Notice that as we increase the number
of approximating cylinders, the corresponding
Riemann sums become closer to the true
volume.
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
                       
                       2 0 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     3         2      2/3




The volume of the approximating
cylinder is:
            A( y )y   y y2/3
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

                  3 y 3   96
                        5   8
                5
                 
                        0
                             5
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
Thus, we have: V    
                     0
                             A( x ) dx
                         1
                   ( x  x ) dx 2         4
                     0
                                                 1
                   x x       3         5
                   
                   3 5 0
                  2
                
                  15
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              (2  x)  dx
                   1
                               2 2                2
                0
                                                
                                                 
               x  5 x  4 x  dx
                   1
                           4       2
                   0
                                              1
              x  x   x  8
                       5       3          2
             5  4  
              5  3   2 0 5
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

                          V   A  y  dy
       b                        d
 V   A( x) dx     or
       a                        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                 2



                           1  y 
                       2
        1 y
                                          2
SOLIDS OF REVOLUTION                  Example 6

The volume is:
                  1
             V   A( y )dy
                  0

                    1 y
                                             1  y   dy
                      1                   2
               
                                                            2
                  0
                                                        
                                                         
                                                   
                      1
                  2 y  y  y dy             2
                      0
                                                        1
                   4 y 2 y 2 y3  
                                  3

                         
                   3
                  
                          2   3
                                 0
                                    2
VOLUMES

In the following examples, we find
the volumes of three solids that are
not solids of revolution.
VOLUMES                         Example 7

The figure shows a solid with a circular base
of radius 1. Parallel cross-sections
perpendicular to the base are equilateral
triangles.

Find the volume of the solid.
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.
VOLUMES                     Example 7

As B lies on the circle, we have y  1  x   2




So, the base of the triangle ABC is
|AB| = 2 1  x2
VOLUMES                    Example 7

Since the triangle is equilateral, we see
that its height is 3 y  3 1  x2
VOLUMES                      Example 7

Thus, the cross-sectional area is :


     A( x)   2 1  x  3 1  x
             1
             2
                         2               2


           3(1  x )2
VOLUMES                              Example 7

The volume of the solid is:
      1
V   A( x) dx
      1
        1                        1
          3(1  x ) dx  2 
                   2
                                       3(1  x ) dx
                                                 2
      1                     0
                       1
            x    3
                  4 3
    2 3 x   
            3 0  3
VOLUMES                 Example 8

Find the volume of a pyramid
whose base is a square with side L
and whose height is h.
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.
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)
VOLUMES                 Example 8

Thus, the cross-sectional area is:
                        2
                     L 2
          A( x)  s  2 x
                  2

                     h
VOLUMES                                Example 8

The pyramid lies between x = 0 and x = h.

                             h
So, its volume is: V 
                         
                         0
                                 A( x) dx
                                  2
                             h   L 2
                                 2
                                     x dx
                         0       h
                                       h
                     L x    2
                            Lh     3        2
                     2  
                     h 3 0 3
NOTE

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.
NOTE

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
       hL
V   2 (h  y ) dy
                2
     0 h

      2
    Lh
  
      3
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.
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
y  16  x2 , -4 ≤ x ≤ 4
VOLUMES                       Example 9

A cross-section perpendicular to the x-axis at
a distance x from the origin is a triangle ABC,
whose base is   y  16  x2 and whose height
is |BC| = y tan 30° = 16  x 2 3.
VOLUMES                   Example 9

Thus, the cross-sectional area is:

                      1
     A( x)  16  x 
            1
            2
                     2
                         16  x 2

                       3
             16  x 2
           
              2 3
VOLUMES                                  Example 9

The volume is:
           4
    V   A( x) dx
          4

               16  x   2

                                        16  x  dx
           4                         1    4
                          dx                      2
          4                              0
                2 3                  3
                                 4
        1        x    128  3
         16 x  3  
         3         0 3 3

				
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