APPLICATIONS OF INTEGRATION

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

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

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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.

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VOLUMES

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

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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.

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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

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CYLINDERS

• In particular, if the base is a circle with
• radius r, then the cylinder is a circular
• cylinder with volume V pr2h.

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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.

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IRREGULAR SOLIDS

• For a solid S that isnt 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.

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IRREGULAR SOLIDS

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

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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.

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IRREGULAR SOLIDS

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

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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.

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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.

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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

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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 ? 8.
• Think of the slices as becoming thinner and
  thinner.

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IRREGULAR SOLIDS

• Therefore, we define the volume as the limit
• of these sums as n ? 8).
• However, we recognize the limit of Riemann
• sums as a definite integral and so we have
• the following definition.

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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

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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.

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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.

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SPHERES
Example 1

• Show that the volume of a sphere
• of radius r is

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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

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SPHERES
Example 1

• So, the cross-sectional area is

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SPHERES
Example 1

• Using the definition of volume with a -r and
• b r, we have
• 
  (The integrand is even.)

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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
• The slabs are circular cylinders, or disks.

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SPHERES
The three parts show the geometric interpretations
of the Riemann sums when n 5, 10, and
20 if we choose the sample points xi to be the
midpoints .
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SPHERES

• Notice that as we increase the number
• of approximating cylinders, the corresponding
• Riemann sums become closer to the true
• volume.

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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.

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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 .

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VOLUMES
Example 2

• The area of the cross-section is
• The volume of the approximating cylinder
• (a disk with thickness ?x) is

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VOLUMES
Example 2

• The solid lies between x 0 and x 1.
• So, its volume is

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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.

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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

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VOLUMES
Example 3

• So, the area of a cross-section through y is
• The volume of the approximating
• cylinder is

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VOLUMES
Example 3

• Since the solid lies between y 0 and
• y 8, its volume is

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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.

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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.

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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.

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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

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VOLUMES
Example 4

• Thus, we have

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VOLUMES
Example 5

• Find the volume of the solid obtained
• by rotating the region in Example 4
• about the line y 2.

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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.

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VOLUMES
Example 5

• The cross-sectional area is

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VOLUMES
Example 5

• So, the volume is

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SOLIDS OF REVOLUTION

• The solids in Examples 15 are all
• called solids of revolution because
• they are obtained by revolving a region
• about a line.

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SOLIDS OF REVOLUTION

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

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SOLIDS OF REVOLUTION

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

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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 p(radius)2

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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
  p(outer radius)2 p(outer radius)2

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SOLIDS OF REVOLUTION
Example 6

• Find the volume of the solid obtained
• by rotating the region in Example 4
• about the line x -1.

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SOLIDS OF REVOLUTION
Example 6

• The figure shows the horizontal cross-section.
• It is a washer with inner radius 1 y and
• outer radius

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SOLIDS OF REVOLUTION
Example 6

• So, the cross-sectional area is

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SOLIDS OF REVOLUTION
Example 6

• The volume is

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VOLUMES

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

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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.

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VOLUMES
Example 7

• Lets 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.

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VOLUMES
Example 7

• As B lies on the circle, we have
• So, the base of the triangle ABC is
• AB

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VOLUMES
Example 7

• Since the triangle is equilateral, we see
• that its height is

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VOLUMES
Example 7

• Thus, the cross-sectional area is

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VOLUMES
Example 7

• The volume of the solid is

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VOLUMES
Example 8

• Find the volume of a pyramid
• whose base is a square with side L
• and whose height is h.

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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.

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VOLUMES
Example 8

• We can express s in terms of x by observing
• from the similar triangles that
• Therefore, s Lx/h
• Another method is to observe that the line OP
  has slope L/(2h)
• So, its equation is y Lx/(2h)

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VOLUMES
Example 8

• Thus, the cross-sectional area is

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VOLUMES
Example 8

• The pyramid lies between x 0 and x h.
• So, its volume is

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NOTE

• In the example, we didnt need to place
• the vertex of the pyramid at the origin.
• We did so merely to make the equations simple.

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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

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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.

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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
• -4 x 4

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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 and whose height
• is BC y tan 30

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VOLUMES
Example 9

• Thus, the cross-sectional area is

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VOLUMES
Example 9

• The volume is