Transcript Calculus 7.3 Day 2

7.3 day 2
Disks, Washers and Shells
Limerick Nuclear Generating Station, Pottstown, Pennsylvania
A little review of what we have done in the past….
2
Suppose I start with this curve.
y x
My boss at the ACME Rocket
Company has assigned me to
build a nose cone in this shape.
1
0
1
2
3
4
So I put a piece of wood in a
lathe and turn it to a shape to
match the curve.

2
How could we find the volume
of the cone?
y x
One way would be to cut it into a
series of thin slices (flat cylinders)
and add their volumes.
1
0
1
2
3
4
The volume of each flat
cylinder (disk) is:
 r 2  the thickness

 x
2
dx
In this case:
r= the y value of the function
thickness = a small change
in x = dx

2
The volume of each flat
cylinder (disk) is:
y x
 r 2  the thickness
1

0
1
2
3
 x
2
dx
4
If we add the volumes, we get:
   x
2
4
0
dx
4
   x dx
0


2
4
 8
x2
0

This application of the method of slicing is called the
disk method. The shape of the slice is a disk, so we
use the formula for the area of a circle to find the
volume of the disk.
If the shape is rotated about the x-axis, then the formula is:
b
V   f (x )2dx
a
b
A shape rotated about the y-axis would be: V   f ( y )2 dy
a

1
The region between the curve x 
, 1  y  4 and the
y
y-axis is revolved about the y-axis. Find the volume.
y
x
1
1
2
3
4
1
 .707
2
1
 .577
3
1
2
We use a horizontal disk.
The thickness is dy.
4
3
2
The radius is the x value of the
1
function 
.
dy
y
1
2
 1 
V  
dy
 y 
1


4
0
1
4
 
1
1
dy
y
volume of disk
  ln y 1
4
0
   ln 4  ln1
  ln 22  2 ln 2

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

4
3
y  2x
2
y  x2
The region bounded by
y  x2 and y  2 x is
revolved about the y-axis.
Find the volume.
1
If we use a horizontal slice:
yx
y  2x
y
x
2
2
yx
0
1
2
The volume of the washer is:

V  
0


4
 y
2
 y
 
2
2

 dy

1 2

V     y  y  dy
0
4 

4
V 
4
0
1 2
y  y dy
4
The “disk” now has a hole in
it, making it a “washer”.
 R
 R
2
  r 2   thickness
2
 r 2  dy
outer
radius
4
1 
1
   y 2  y3 
12  0
2
inner
radius
 16 
  8  
3

8

3

This application of the method of slicing is called the
washer method. The shape of the slice is a circle
with a hole in it, so we subtract the area of the inner
circle from the area of the outer circle.
The washer method formula is:
b
V    R2  r 2 dx
a

y  x2
4
3
y  2x
2
1
0
1
r
y  2x
y
x
2
y  x2
yx
2
r  2 y
y2
   4  2 y   4  4 y  y dy
0
4
1
4
1 2
   3 y  y  4 y 2 dy
0
4
4
V    R 2  r 2 dy
0
2

y

  2   2 y
0
2

 dy
2


4
 3
1
8 
    y 2  y3  y 
12
3 0
 2
16 64 
8

    24    
3
3 3

3
2

y2 
    4  2 y    4  4 y  y dy
0
4 

4
The outer radius is:
y
R  2
2
The inner radius is:
R
4
4
If the same region is
rotated about the line x=2:

5
Find the volume
the
1 dy
 4region

1 4   y of
2
bounded by y  x  1 , x  2 ,
5
and y  0
the y  revolved
5  y dy  4about

1
axis.
5
4
3
y  x2  1
2
5
1 

 5 y  y 2   4
2 1

1
0
2
1
We can use the washer method ifwe split
25  itinto1 two
 parts:
  25     5    4
y 1  x2
5
 2 
2
1
outer
radius


x  y 1

2
y  1 dy    2 1
inner
radius
2
cylinder
thickness
of slice
2 
2 
 25 9 
     4
 2 2

16
 4
2
8  4
 12

5
4
Here is another
way we could
approach this
problem:
3
y  x2  1
2
1
0
1
2
cross section
If we take a vertical slice and revolve it about the y-axis
we get a cylinder.
If we add all of the cylinders together, we can reconstruct
the original object.

5
4
3
y  x2  1
2
1
0
2
1
cross section
The volume of a thin, hollow cylinder is given by:
Lateral surface area of cylinder  thickness
 circumference  height  thickness
=2 r  h  thickness


=2 x x 2  1 dx
r
h
circumference thickness
r is the x value of the function.
h is the y value of the function.
thickness is dx.

5
4
This is called the
shell method
because we use
cylindrical shells.
3
y  x2  1
2
1
0
2
1
cross section
If we add all the cylinders from the
smallest to the largest:

2
0
=2 r  h  thickness


=2 x x 2  1 dx
r
h
circumference thickness


2 x x 2  1 dx
2 4  2
2
2  x3  x dx
0
2
1 4 1 2 
2  x  x 
2 0
4
12

Find the volume generated
when this shape is revolved
about the y axis.
4
3
2
1
0
1
2
3
y
4

5
6
4 2
x  10 x  16
9
7
8

For this problem, the
washer method can’t be
used.
Why not?
We can’t solve for x,
so we can’t use a
horizontal slice
directly.

If we take a
vertical slice
and revolve it
about the y-axis
we get a cylinder.
4
3
2
1
0
Shell method:
1
2
3
y
4

5
6
4 2
x  10 x  16
9
7
8

Lateral surface area of cylinder
=circumference  height
=2 r  h
Volume of thin cylinder  2 r  h  dx

4
3
2
1
0
1
Volume of thin cylinder  2 r  h  dx
 4 2

2 2 x  9 x 10x  16  dx

8

r
circumference
h
thickness
2
3
y
4

5
6
4 2
x  10 x  16
9
7
8

 160
 502.655 cm3
Note: When entering this into the calculator, be sure to enter
the multiplication symbol before the parenthesis.

When the strip is parallel to the axis of rotation, use the
shell method.
When the strip is perpendicular to the axis of rotation,
use the washer method.
