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Math 1432
Dr. Melahat Almus
[email protected]
http://www.math.uh.edu/~almus
Visit CASA regularly for announcements and course material!
If you email me, please mention the course (1432) in the subject line.
Please follow “classroom behavior” policies.
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Chapter 7- Applications of Integration
Section 7.4 - Volume
Now, we work on the problem of finding the volume of 3D objects using
integration. Integration is defined using Reimann sums and by taking their limit as
n goes to infinity. That is, we can interpret the integral to be a continuous sum.
Before we start, recall the formulas for finding volumes of basic geometric
shapes:
Volume of a sphere with radius r: 34
3V r
Volume of a box with dimensions l, h, w: ( )V l w h base area h
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Volume of a right cylinder with height h and radius r: 2( )V base area h r h
LATERAL surface area of such a cylinder: 2LA rh
Reason:
Volume of a right cone with height h, radius r: 21
3V r h
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We will use “disks” often: A disk is a 2D circular shape that includes the inside of
the circle. For example: 2 2 1x y describes a disk with center at (0,0) and radius
1.
The area of a disk with radius r is: 2A r
Note: Area of an equilateral triangle with one side s: 2 3
4
sA
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PART 1- Finding the Volume of a solid with known cross sections:
The figure below shows a plane region and a solid formed by translating
along a line perpendicular to the plane of . Such a solid is called a right cylinder
with cross-section .
If has area A and height h , then the volume is very simple: V A h
You have seen two elementary examples of this notion in Geometry:
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To calculate the volume of a more general solid, we introduce a coordinate axes,
and examine the cross-sections of that solid perpendicular to that coordinate axis.
Here is an example where we choose the x-axis as the coordinate axis:
By ( )A x , we mean the area of the cross-section at coordinate x .
If the cross-sectional area ( )A x continuously varies with x , we can find the
volume of the solid by integrating ( )A x from x a to x b .
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The idea can be generalized as: Volumes by Slicing
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In summary:
* If the cross section is perpendicular to the x-axis and its area is a function of x,
say A(x), then the volume of the solid from a to b is given by
b
aV A x dx
* If the cross section is perpendicular to the y-axis and its area is a function of y,
say A(y), then the volume of the solid from c to d is given by
d
cV A y dy
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Example: Let R be the region bounded by
2 6 5 ( 1)( 5)f x x x x x and the x-axis.
R is the base of a solid whose cross-sections perpendicular to the x-axis are
squares.
Click on this link to get the Geogebra file:
https://www.geogebra.org/m/DaRqTDa7
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Example: Let R be the region bounded by f 2x x 4x and the x-axis.
R is the base of a solid whose cross-sections perpendicular to the x-axis are
equilateral triangles.
Click on this link to get the Geogbera file. Move the slider to see different cross
sections.
https://www.geogebra.org/m/YGkVa7wy
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Example 1: Let R be the region bounded by ( ) 2f x x , 1x and the x-axis.
If R is the base of a solid whose cross-sections perpendicular to the x-axis are
squares, find the volume of this solid.
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Example 2: Let R be the region in the first quadrant bounded by 23y x and
3y . If R is the base of a solid whose cross-sections perpendicular to the y-axis
are semicircles, compute the volume of this solid.
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Example 3: Let R be the region bounded by 2( )f x x and 2( ) 8g x x .
If R is the base of a solid whose cross-sections perpendicular to the x-axis are
squares, set up the integrals that would give the volume of this solid.
Exercise: What if the cross-sections given in Example 3 were “perpendicular to
the y-axis”?
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Example 4: Let R be the region in the first quadrant bounded by 2( ) xf x e ,
0 1x . If R is the base of a solid whose cross-sections perpendicular to the x-
axis are squares, compute the volume of this solid.
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Exercise: Consider a solid whose base is the region inside the circle
x2 + y2 = 4. If cross sections taken perpendicular to the y-axis are squares,
compute the volume of this solid.
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Part 2 - Finding the volume of a solid of revolution:
When you revolve a plane region about an axis, the solid formed is called a solid of
revolution.
If there is no gap between the axis of rotation and the region, the cross-sections
(slices) are disks and the method is called “disk method”.
If there is a gap between the axis of rotation and the region, the cross-sections are
“flat rings” or “washers”; and the method is called the “washer method”.
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Volumes of solids of revolution by DISKS and WASHERS
Disk Method:
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Disk Method:
Revolving about the x-axis: 2
fb
aV x dx
Revolving about the y-axis: 2
gd
cV y dy
Example: Let R be the region bounded by the x-axis and the graphs of y x
and x = 4. Sketch and shade the region R. Label points on the x and y-axis.
a) Give the formula for the volume of the solid generated when the region R is
rotated about the x-axis.
b) Find the volume for the solid in (a).
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Example: Let R be the region bounded by the y-axis and the graphs of 2y x
and y = 2. Sketch and shade the region R. Label points on the x and y-axis.
a) Give the formula for the volume of the solid generated when the region R is
rotated about the y-axis.
b) Find the volume for the solid in (a).
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Example: Consider the region in the first quadrant enclosed by 24y x .
Set up the integral that gives the volume of the solid formed by revolving this
region about the x-axis.
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Example: Let R be the region in the first quadrant bounded by the y-axis and the
graphs of 2y x and y = 9. Sketch and shade the region R.
Give the formula for the volume of the solid generated when the region R is rotated
about the y-axis. Find the volume for the solid.
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WASHER METHOD
When we apply the same idea (parallel cross sections) to a region that is not
containing the axis of revolution, we might get “washers” instead of disks. That is,
if there is a gap between the region and the axis of rotation, the cross-sections
might look like washers.
The dimensions of a typical washer are:
Outer radius: ( )R x
Inner radius: ( )r x
The washer’s area is: 2 2 2 2( ) ( ) ( ) ( ) ( )A x R x r x R x r x
Consequently, the volume of the solid can be found by:
2 2( ) ( ) ( )
b b
a a
V A x dx R x r x dx
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Washer Method:
Revolving about the x-axis: 2 2b
aV f x g x dx
Revolving about the y-axis: 2 2d
cV F y G y dy
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Example: Let R be the region bounded by the graphs of f x x and
3( )g x x . Set up the formula that gives the volume of the solid generated by
rotating R about the x-axis.
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Example: Let R be the region bounded by the graphs of f x x and
( ) 2g x x . Set up the formula that gives the volume of the solid generated by
rotating R about the y-axis.
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What if we rotate around a different line?
Example: Consider the region enclosed by , 1, 4y x x x and 1y . Give the
formula for the volume of the solid formed by revolving this region around the line
1y .
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Example: Consider the region enclosed by 2y x , 0y , 3x . Set up the
integral that gives the volume of the solid formed by revolving this region around
the line 3x .
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Example: Consider the region enclosed by 2y x and the x-axis for 0 1x .
Set up the integral that gives the volume of the solid formed by revolving this
region around the line 1y .
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Example: Set up the integral(s) that give the volume of the region bounded by
, 0, 2y x x y being revolved about:
a. x – axis
b. x = 4
c. y = 2
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d. y = 3
e. 1y
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Remark:
If the region is below the axis of rotation, be careful:
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Example: Let R be the region in the first quadrant bounded by 2( ) 10f x x
and ( ) 3g x x . Set up the integrals that will give the volume of the solid formed
when
a) R is rotated about the x-axis.
b) R is rotated about the y-axis.
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EXERCISE: Let R be the region bounded by the graph of
1f x x , x=0 and y = 3.
Set up the formula that gives the volume of the solid generated by rotating R about
the line y=5.
Exercise: Let R be the region bounded by the graph of sinhf x x and
coshg x x for 0,ln10x . Setup the integral that gives the volume of the
solid generated by rotating R about the x-axis.
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SHELL METHOD
Sometimes, finding the volume of a solid using disk/washer method is very
difficult, or even impossible:
To find the volume of this solid, we would need a different method.
The idea in Shell Method is to use “nested cylinders” to find the volume of a solid
of revolution.
Shell: consider a solid cylinder with radius R and height h, and cut out a cylindrical
core from it that has radius r.
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When one shell is cut and open up, it looks like this:
Length of this “box” is: 2 (distance from the line to the axis of rotation)=2 r
Height is: ( )h f x ; the width is: dx
Surface Area is: ( ) 2A x rh ;
Volume of one shell: 2 rhdx
Hence, the volume of the solid is the continuous sum: 2
b
a
V rhdx
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In the Disk Method, the rectangle of revolution is perpendicular to the axis of
revolution.
In the Shell Method, the rectangle of revolution is parallel to the axis of
revolution.
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Example: Let R be the region in the first quadrant bounded by 24 , 0y x y . Find the volume of the solid formed by rotating R about the y –
axis using shell method.
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Example: Let R be the region in the first quadrant bounded by 24 , 0y x y . Find the volume of the solid formed by rotating R about the x –
axis using shell method.
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Example: Let R be the region bounded by the graph of 2 4f x x x and
2g x x . Set up the integral that gives the volume of the solid generated by
rotating R about
a) the y-axis.
b) the x-axis.
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Example: Let R be the region bounded by 2 4, 6, 0y x y x x .
Set up the integral that gives the volume of the solid formed by rotating R about
the y – axis.
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Exercise: Let R be the region bounded by 2, ,0 1y x y x x . Set up the
integral that gives the volume of the solid formed when R is revolved about the
line x=-2.
Exercise: Let R be the region bounded by the graph of 29f x x and
22g x x . Set up the integrals that give the volume of the solid generated by
rotating R about
a) the x-axis.
b) the y-axis.
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WORKSHEET --- VOLUME
Solve these exercises after class! These are pretty typical problems. Solutions
will not be posted; if you doubt your answer, you can ask those questions
during lab.
Exercise: Let be the region bounded by y x and 2y x , from x = 0 to x= 1.
a. Find the volume of the solid formed by rotating around the x-axis.
b. Find the volume of the solid formed by rotating around the y-axis.
Exercise: Let be the region bounded by 3y x and 8y x , from x = 0 to x= 2.
a. Find the volume of the solid formed by rotating around the x-axis.
b. Find the volume of the solid formed by rotating around the y-axis.
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Solve these exercises after class! These are pretty typical problems.
Exercises:
1. The region bounded by 3y x , x = 1 and the x-axis is rotated about the x-
axis. Find the volume of the solid formed.
2. The region bounded by cosy x , x = 0, / 2x and the x-axis is rotated
about the x-axis. Find the volume of the solid formed.
3. The region bounded by 3y x , y = 8 and the y-axis is rotated about the y-
axis. Find the volume of the solid formed.
4. The region bounded by 2y x , x = 1 and the x-axis is rotated about the x-
axis. Find the volume of the solid formed.
5. The region bounded by 2y x , y = 1 and the y-axis is rotated about the y-
axis. Find the volume of the solid formed.