Section 7.2 – Volume: The
Disk Method
Find the volume of the following cylinder:
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6 ft
12 ft
V    3 12  108  339.292 ft
2
3
Calculate the volume V of the solid obtained by
rotating the region between y = 5 and the x-axis about
the x-axis for 1≤x≤7.
2
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V    radius   height
V    5  6
2
5
6
V  150
V  471.239
Volumes of Solids of Revolution with
Riemann Sums
The Riemann Sum is set up by considering this cross
sections of the solid (circles) each with thickness dx:
xk
Volume 
Radius
a
b

lim
max xk 0
n
   radius
k 1
b
    radius  dx
2
a
b
   radius  dx
a
2
2
 xk
Volumes of Solids of Revolution:
Disk Method
• Sketch the bounded region and the line of
revolution. (Make sure an edge of the region is
on the line of revolution.)
• If the line of revolution is horizontal, the
equations must be in y= form. If vertical, the
equations must be in x= form.
• Sketch a generic disk (a typical cross section).
• Find the length of the radius and height of the
generic disk.
Disk Method =
• Integrate with the following formula: No hole in the
b
V    radius  height
a
2
solid.
Example 1
Calculate the volume of the solid obtained by rotating the region
bounded by y = x2 and y=0 about the x-axis for 0≤x≤2.
Sketch a Graph
Find the Boundaries/Intersections
x  0, 2
Line of Rotation
Radius =
x2
Integrate the Volume of Each
Generic Disk
  x
2
0
Height = dx
Make Generic Disk(s)

2 2
dx
32
 
5
Example 2
Calculate the volume of the solid obtained by rotating the
region bounded by y = x2 and y=4 about the line y = 4.
Line of Rotation
Sketch a Graph
Find the Boundaries/Intersections
x2  4
x  2, 2
Radius =
4 - x2
Integrate the Volume of Each
Generic Disk

2
2
Height = dx
Make Generic Disk(s)
4  x 
2 2
512


15
dx
NOTE:
Because of the
square, the
order of
subtraction
does not
matter.
Example 3
Calculate the volume of the solid obtained by rotating the
region bounded by y = x2, x=0, and y=4 about the y-axis.
Sketch a Graph
Since the Line of Revolution is
Vertical, Solve for x
y  x2
x
x y
Radius =
√y
We only
need 0≤x≤2
Find the Boundaries/Intersections
Remember:
0≤x≤2
Make Generic
Disk(s)
Height = dy
Line of Rotation
y  02
y0
y4
Integrate the Volume of Each
Generic Disk

4
0
 y
2
dy
 8
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White Board Challenge
Find the volume of the following three-dimensional
shape:
6 ft
12 ft
2 ft
V    3 12   1 12
2
2
    3  1  12
2
2
 96  301.593 ft
3
Calculate the volume V of the solid obtained by
rotating the region between y = 5 and the y = 2 about
the x-axis for 1≤x≤7.
2
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V    router   h    rinner   h
2
2

V    router    rinner    h


5
2
6
2
2
2

V   5 2  6
V  126
V  395.841
Area of a Washer
The region between two concentric circles is
called an annulus, or more informally, a
washer:
2
2
Area   R
Rinner
Router
outer
Area    R
 R
inner
R
inner
2
outer
2

Volumes of Solids of Revolution:
Washer Method
Always a difference
of squares.
of
• Sketch the bounded region and the line
revolution.
• If the line of revolution is horizontal, the equations
must be in y= form. If vertical, the equations must be
in x= form.
• Sketch a generic washer (a typical cross section).
• Find the length of the outer radius (furthest curve
from the rotation line), the length of the inner radius
(closest curve to the rotation line), and height of the
generic washer.
Washer Method
• Integrate with the following formula:
b
V     r
a
2
outer
r
2
inner
= Hole in the
solid.
  height

Example 1
Calculate the volume V of the solid obtained by rotating the
region bounded by y = x2 and y=0 about the line y = -2 for 0≤x≤2.
Sketch a Graph
Find the Boundaries/Intersections
Rinner =
Line of Rotation
0 - -2 = 2
x  0, 2
Router =
x2 - -2 = x2 + 2
Integrate the Volume of Each
Generic Washer
   x  2   22  dx
2
0
Height = dx
Make Generic Washer(s)

2
2
256


15

Example 2
Calculate the volume V of the solid obtained by rotating the
region bounded by y = ex and y=√(x +2) about the line y = 2.
Line of Rotation
Sketch a Graph
Find the Boundaries/Intersections
e  x2
x  1.981, 0.448
Router =
x
x
2-e
Integrate the Volume of Each
Generic Washer


2
x 2


2  e   2  x  2  dx


1.981 

Rinner =
0.448
2- √(x +2)
Height = dx
Make Generic Washer(s)
 8.536
CAN DO NOW:
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“Warm-up”: 1985 Section I
Volume of a Right Solid
A right solid is a geometric solid whose
sides are perpendicular to the base. The
volume of a right solid is the area of the
base times the height.
HSolid
Volume  ABase  H Solid
ABase
Volumes of Solids:
Slicing Method
• Sketch the bounded region.
• If the cross section is perpendicular to the x-axis,
the equations must be in y= form. If the y-axis,
the equations must be in x= form.
• Sketch a generic slice (a typical cross section).
• Find the area of the base and the height of the
generic slice.
Must Answer #1:
What does the
length across the
• Integrate with the following formula:
Must Answer #2: How
does the length
across the bounded
region help find the
area of the base of
the generic slice?
b
V   ABase  height
a
bounded region
represent in your
generic slice?
Example 1
Find the volume of the solid created on a region who base is
bounded by y = √x and the x-axis for 0≤x≤9. Let each cross
section be perpendicular to the x-axis and be a square.
Sketch a Graph
Find the Boundaries/Intersections
x  0,9
Height = dx
ABase
= ASquare
Side Length
= side2
= (√x)2
Make Generic Slice(s)
Integrate the Volume of Each
Generic Slice
  x
9
0
2
dx
81

2
Example 2
Find the volume of the solid created on a region who base is
bounded by x2 + y2 = 1. Let each cross section be perpendicular
to the x-axis and be a squares with diagonals in the xy-plane.
Sketch a Graph
Since the Cross Sections are Per.
to the x-axis, solve for y
ABase
Height
2
2
=
A
x

y
1
Square
= dx
If diameter is
known, a side
length is…
d
2
Diagonal
= side2




d
Make Generic Slice(s)
1 x   1 x
2
2
2 1 x
2
2

2
y   1  x2
Find the
 Boundaries/Intersections
2
x  1,1
2
Integrate the Volume of
Each Generic Slice

1

2 1 x
2
1
2

2
8
dx 
3
A solid has base given by the triangle with vertices
(-4,0), (0,8), and (4,0). Cross sections perpendicular to
the y-axis are semi-circles with diameter in the plane.
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White Board Challenge
What is the volume of the solid?
x   12 y  4
Height
= dy
x  12 y  4
Diameter
ABase
= ½πr2
Radius =
-½y+4

8
1
0 2
   y  4   dy
1
2
64
 
3
2