Transcript Volume: The Disc Method

Volume: The Disc
Method
Section 6.2
If a region in the plane is revolved
about a line, the resulting solid is a
solid of revolution, and the line is
called the axis of revolution. The
simplest such solid is a right
circular cylinder or disc, which is
formed by revolving a rectangle
about an axis .
Revolving a Function
• Consider a function f(x)
on the interval [a, b]
a
• Now consider revolving
b
that segment of curve
about the x axis
• What kind of functions generated these
solids of revolution?
f(x)
Disks
f(x)
• We seek ways of using
integrals to determine the
volume of these solids
• Consider a disk which is a
slice of the solid
dx
– What is the radius
– What is the thickness
– What then, is its volume?
Volume of slice =   f ( x)  dx
2
If a region of a plane is revolved about a line, the resulting solid is
the solid of revolution, and the line is the axis of revolution.
r
The simplest is a right circular cylinder or disk,
a rectangle revolved around the x-axis.
The volume is equal to the area of the
disk times the
width of the disk, V = πr2w.
w
The Disc Method
Volume of disc =
(area of disc)(width of disc)
V   r x
2
Disks
• To find the volume of the
whole solid we sum the
volumes of the disks
• Shown as a definite integral
b
V     f ( x) dx
2
a
f(x)
a
b
w
w
R
Axis of Revolution
R
Revolve this rectangle about the x-axis.
Revolve this function about the x-axis.
f x
A
B
Revolve this rectangle about the x-axis.
Forms a Disk.
Vdisk  Acircle  x
  r 2  x
x
x
Revolve this function about the x-axis.
f x
A
f x
B
A
f x
B
B
A
4
V     ri  x
2
i 1
b
V    f  x   dx
a
2
Revolving About y-Axis
• Also possible to revolve a function about
the y-axis
– Make a disk or a washer to be horizontal
• Consider revolving a parabola about the
y-axis
– How to represent the
radius?
– What is the thickness
of the disk?
Revolving About y-Axis
• Must consider curve as
x = f(y)
– Radius = f(y)
– Slice is dy thick
• Volume of the solid rotated
about y-axis
b
V     f ( y) dy
2
a
Horizontal Axis of Revolution
2
b
Volume = V =
  R( x) dx
a
Vertical Axis of Revolution
2
Volume = V =
d
  R( y) dy
c
Disk Method (to find the volume of a solid of revolution)
f x
Horizontal Axis of revolution:
B
A
B
f y 
Vertical Axis of revolution:
A
Disk Method (to find the volume of a solid of revolution)
f x
A
dx
Horizontal Axis of revolution:
b
B
V    f  x   dx
a
B
f y 
width
radius
Vertical Axis of revolution:
b
dy
2
A
V    f  y   dy
a
2
1.
Find the volume of the solid formed by
revolving f(x) over 0,  about the x-axis.
f  x   sin x
1.
Find the volume of the solid formed by
revolving f(x) over 0,  about the x-axis.
f  x   sin x

V 
2
-1
-2
sin x
 dx
0

1
0

2
dx
2

   sin x dx
0
    cos x 0

    cos   cos0
  1 1
 2
2.
Find the volume of the solid formed by revolving the
region formed by f(x) and g(x) about y = 1.
f  x   2  x 2 and g  x   1
2.
Find the volume of the solid formed by revolving the
region formed by f(x) and g(x) about y = 1.
f  x   2  x 2 and g  x   1
1
1
1
1
dx
2
1
2 x
2

 2x
 1 x 2
2
dx


1

2
  1
2
2
3
3
1
5
5

1
1
  1 32  51    1 32  51 
   2  34  52 
16

15
1
1
-2

2
 x x  x
-1
Length of Rectangle:
2
   1  2x 2  x 4 dx
1
-2

V    1 x
2
3.
Find the volume of the solid formed by
2
revolving the region formed by y  x over
y  0,4 about the y-axis.
3.
Find the volume of the solid formed by
2
revolving the region formed by y  x over
y  0,4 about the y-axis.
y x
4
4
4
V 
3
2
0
-1
0
4
dy
   y dy
1
-2
 y
2
0
2


1
2
 8
y
2

4
0
dy
The disc method can be extended
to cover solids of revolution with
holes by replacing the
representative disc with a
representative washer.
w
R
r
Washers
• Consider the area
between two functions
rotated about the axis
f(x)
g(x)
a
b
• Now we have a hollow solid
• We will sum the volumes of washers
• As an integral
b
V 
a
 f (x)   g(x) dx
2
2
Volume of washer = ( R  r )dx
2

2

V    R( x)  r ( x) d x
b
a
2
2
How do you find the volume of the figure formed by revolving
the shaded area about the x-axis?
g x
f x
A
B
How do you find the volume of the figure formed by revolving
the shaded area about the x-axis?
g x
f x
b
V    f  x   g  x  dx
2
2
a Outside
Inside
radius
radius
The volume we want is the difference between the two.
B
A
g x
f x
A
b
B
V    f  x   dx
2
a
Revolving this function creates a solid
whose volume is larger than we want.
B
A
b
V     g  x   dx
2
a
Revolving this function carves out the
part we don’t want.
Washer Method (for finding the volume of a solid of revolution)
g x
f x
A
B
Washer Method (for finding the volume of a solid of revolution)
Awasher    outside radius    inside radius 
2
2
g x
f x
b
Vfigure    f  x   g  x  dx
2
a Outside
radius
A
B
2
Inside
radius
1.
Find the volume of the solid generated by revolving
the area enclosed by the two functions about the xaxis.
f  x   x, g  x   x2
1.
Find the volume of the solid generated by revolving
the area enclosed by the two functions about the xaxis.
Find intersection points first.
f  x   x, g  x   x2
x  x2
x  x4
0  x4  x
0  x  x 3  1
x  0,1
g x f x
1
0.5
Washer Method:
1
1
V 
0
1
 x  x 
2
2

2
dx

   x  x 4 dx
0


1
2
x  x
2
1
5
5

1
0
   21  51    0  
3

10
2.
Find the volume of the solid formed by revolving
the area enclosed by the given functions about the yaxis.
y  x 2  1, y  0, x  0, x  1
2.
Find the volume of the solid formed by revolving
the area enclosed by the given functions about the yaxis.
y  x 2  1, y  0, x  0, x  1
This region is not always created by
the same two functions.
2
dy
1.5
The change occurs at y = 1.
We need to use two integrals to find the volume.
1
dy
0.5
1
For y in [0,1] we use disk method.
For y in [1,2] we use washer method.
Since we’re revolving about the y-axis, each
radius must be in terms of y.
y  x2  1
y  1  x2
y 1  x
1
2
(distance from the parabola to the x-axis)
(distance from the parabola to the y-axis)
V    1 dy    1   y  1 dy


0
2
1
2
2
1
2
V    1 dy    1   y  1 dy


2
0
1
2
2
1
2
   1dy

0
   y 0
1
  2  y  dy
2
1

2 2
 2y  y 1
1
2
  1  0    4  2    2  21  



2
3

2
Volume of Revolution - X
Find the volume of revolution about the x-axis of
f(x) = sin(x) + 2 from x = 0 to x = 2.
Volume of Revolution - X
Find the volume of revolution about the x-axis of
f(x) = sin(x) + 2
from x = 0 to x = 2
2
A    (sin( x)  2) dx
2
0
Use the TI to integrate this one!
Did you get 88.826 cubic units?
Volumes of Solids with Known
Cross Sections
1. For cross sections of area A(x)
taken perpendicular to the x-axis,
b
Volume =

a
A( x)dx
2. For cross sections of area A(y)
taken perpendicular to the y-axis,
d
Volume =
A( y )dy

c
Volume of Revolution - X
Let’s look at the cross
section or slice. What is
it?
It’s a circle.
What is the radius?
f(x).
What is the area?
A = r 2
A =  (f(x))2
Volume of Revolution - X
How wide (thick) is the disc?
dx
The volume of the disk is
V =  (f(x))2dx
How do we add up all the
disks from x = 1 to x = 4?
Example 1
Find the volume of the solid whose base is the region in the xy-plane
bounded by the given curves and whose cross-sections perpendicular
to the x-axis are (a) squares; (b) semicircles; and (c) equilateral
triangles.
y  x , x  0, y  0, x  2
2
Example 1
(a) Square Cross Sections
2
Here is the x curve!
My Advice???
1) Draw the 2-D
picture and
imagine the crosssectional shape
coming out in the
third dimension!
2) Try and think of
how to write the
area of these cross
sections using the
given function.
3) Set up the integral
and integrate.
Think: SKATE BOARD RAMP!!!
Example 1
(b) Semicircular Cross Sections
Here is the x2 curve!
Think: CORNICOPIA!!!
Example 1
(c) Equilateral Triangular
Cross Sections
Think: PYRAMID, kinda??
Here is the x2 curve!
Example 2
Find the volume of the solid whose base is the region in the xy-plane
bounded by the given curves and whose cross-sections perpendicular
to the x-axis are (a) squares; (b) semicircles; and (c) equilateral
triangles.
x
x
y  1  , y  1  , x  0
2
2
Example 2
(a) Square Cross Sections
Here is the 1 
x
curve !
2
Here is the  1 
Think: PYRAMID!!!
x
curve !
2
Example 2
(b) Semicircular Cross Sections
Here is the 1 
x
curve !
2
Here is the  1 
Think: CORNICOPIA AGAIN!!!
x
curve !
2
Example 2
(c) Equilateral Triangular
Cross Sections
Here is the 1 
x
curve !
2
Think: PYRAMID!!!
Here is the  1 
x
curve !
2
Example 2
(c) Equilateral Triangular
Cross Sections
Here is the 1 
x
curve !
2
Think: PYRAMID!!!
Here is the  1 
x
curve !
2
Theorem:
The volume of a solid with cross-section of area
A(x) that is perpendicular to the x-axis is given by
b
V   A  x  dx
a
a
b
Finding the volume of a solid with known cross-section is a 3-step process:
Theorem:
The volume of a solid with cross-section of area
A(x) that is perpendicular to the x-axis is given by
b
V   A  x  dx
a
***This circle is the base of a 3-D
figure coming out of the screen.
S
a
b
***This rectangle is a side of a
geometric figure (a cross-section of
the whole).
Finding the volume of a solid with known cross-section is a 3-step process:
Step #1: Find the length (S) of the rectangle used to create the base of the figure.
Step #2: Find the area A(x) of each cross-section (in terms of this rectangle).
Step #3: Integrate the area function from the lower to the upper bound.
Volume of a solid with cross-sections of area A(y) and perpendicular to the y-axis:
d
V   A  y  dy
d
S
c
c
Find the volume of the solid whose base is bounded by the
2
2
circle x  y  4 with cross-sections perpendicular to the x-axis.
These cross-sections are
a. squares
Step #1: Find S.
Step #2: Find A(x).
Step #3: Integrate the area function.
Find the volume of the solid whose base is bounded by the
2
2
circle x  y  4 with cross-sections perpendicular to the x-axis.
These cross-sections are
a. squares
S   4  x2   4  x2
Step #1: Find S.
x2  y 2  4
y 2  4  x2
S
-2
 2 4  x2
2
y   4  x2
Asquare  S2
A x   2 4  x2
Step #2: Find A(x).

S



2
 4 4  x 2  16  4x 2
Step #3: Integrate the area function.
2
V

2

2


128
 42.667
16  4x dx  2 16  4x dx 
3
0
2
2
Find the volume of the solid whose base is bounded by the
2
2
circle x  y  4 with cross-sections perpendicular to the x-axis.
These cross-sections are
b. equilateral triangles
Step #1: Find S.
Step #2: Find A(x).
Step #3: Integrate.
Find the volume of the solid whose base is bounded by the
2
2
circle x  y  4 with cross-sections perpendicular to the x-axis.
These cross-sections are
b. equilateral triangles
Step #1: Find S.
S  2 4  x2
S
-2
Step #2: Find A(x).
S
2
S2 3
3 2
Aequilateral triangle 

S
4
4
2
3
2
Ax 
2 4x
4
 3 4  x2




Step #3: Integrate.
2
V  3
2

2
32 3
 18.475
4  x dx  2 3   4  x  dx 
3
0
2

2
Solids with known cross sections
Find the volume of a solid where the base is an ellipse
with semi-major axis of 4 units and semi-minor axis of
3 units if all cross sections perpendicular to the major
axis are semicircles.
Volumes by Slicing
y
3
x2 y2

1
16 9
y
4
y
x
x
y 2
V 
2
2y
Now,
y2 
.x
9
(16  x 2 )
16
9
2
V 
16

x
.dx

32 4
4
9
2
4

16  x

16
.dx
0
4
9 
x3 

16x  
16 
3 0
 24 units3
y
The natural draft cooling tower
shown at left is about 500 feet
high and its shape can be
approximated by the graph of
this equation revolved about
the y-axis:
500 ft
x
x  .000574 y 2  .439 y  185
The volume can be calculated using the disk method with
a horizontal disk.

500
0
.000574 y
2
 .439 y  185 dy  24,700,000 ft 3
2
