#### Transcript Chapter 6

```6.1
Area between two curves
Ak = area of k th rectangle,
f(ck) – g(ck ) = height,
xk = width.
Figure 4.23: When the formula for a bounding curve
Find the
of the changes
region to
between
the curves
changes,
the area
area integral
match. (Example
5)
f ( x)  x and g ( x)  x  1
Section 6.2  Figure 5
Approximating the volume of a sphere with radius 1
(a) Using 5 disks,
V 4.2726
(b) Using 10 disks,
V 4.2097
(c) Using 20 disks,
V 4.1940
A
Figure 5.6: 6.The
region (a) and
solidof
(b)revolution
in Example 4.
2 Volumes
– Solid
y = f(x) is rotated about x-axis on [a,b].
Find the volume of the solid generated.
A cross-sectional slice is a circle and a
slice is a disk. Vdisk   R2 (thickness)
Figure 5.6: Volumes
The region–(a)
and of
solid
(b) in Example 4.
Solid
revolution
y  x is rotated about the x-axis on [0, 4]
Find the volume of the solid generated.
Vdisk   R (thickness)
2
4
Vdisk   
0
2
4
4
1 2
 2 2
x dx    xdx   x  (4  0 )  8 units 3
2 0 2
0
 
Volumes by disk-y axis rotation
Find the volume of the solid generated by revolving a region
between the y-axis and the curve x = 2/y from y = 1 to y = 4.
Find the volume of the solid generated by revolving a region
between the y-axis and the curve x = 2/y from y = 1 to 4.
Vdisk   R2 (thickness)
4
Vdisk
2
4
1 4
2
y
2
     dy    4 y dy  4
y
1
1 
1
1
1 
 4   1  3 units 3
4 
Figure 5.10: The cross sections of the solid of revolution
Washers
generated
here are
washers, does
not disks,
so
the integral
If
the
region
revolved
not
border
on or
b
slightly different
 a A(x)the
cross
axistoofa revolution,
theformula.
solid has a hole
in it. The cross sections perpendicular to the
axis are washers.
V = Outside Volume – Inside Volume
region bounded by the curve y = x2 +1 and
the line y = -x + 3 is revolved about the x-axis to
generate a solid. Find the volume of the solid of
revolution.
. The
The inner and outer radii of the washer swept out by one slice.
Outer radius R = - x + 3 and the inner radius r = x2 +1
The inner and outer radii of the washer swept out by one slice.
Outer radius R = - x + 3 and the inner radius r = x2 +1
Find the limits of integration by finding the xcoordinates of the points of intersection.
x2 + 1= - x + 3
x2 + x –2=0
( x+ 2 )(x – 1) = 0
x = -2
x=1
Outer radius R = - x + 3 and the inner radius r = x2 +1
Calculation of volume
b
2
b
2
Vwasher     R  dx     r  dx
a
b
a
2
b


2
Vwasher      x  3 dx    x 2  1 dx
a
1
  ( x 2  6 x  9)dx  
2
a
1

2

x 4  2 x 2  1 dx
1
  x5 x3

117
4
2
2
  ( x  x  6 x  8)dx   
  3x  8 x  
units 3
3
5
 5
 2
2
1
The region bounded by the parabola y = x2 and the line y = 2x
in the first quadrant is revolved about the y-axis to generate a
y-axis rotation
solid. Find the volume of the solid.
Drawing indicates a dy
integration so solve each
equation for x as a function of y
y
x  y and x 
2
Set = to find y limits
of integration
y
y2
y  y
 4 y  y2  y2  4 y  0
2
4
y = 0 and y = 4 are limits
The washer swept out by one slice perpendicular to the y-axis.
d
2
d
2
Vwasher     R  dy     r  dy
c
c
The region bounded by the parabola y = x2 and the line y = 2x
in the first quadrant is revolved about the y-axis to generate a
calculation
solid. Find the volume of the solid.
2
d
2
d
Vwasher     R  dy     r  dy
4

0
c
2
c
2
4
 
 y
y dy      dy
2
0 
3 4

y
y 
y
8
   y  dy        units 3
4 
 2 12  0 3
0
4
2
2
Figure 5.17: Cutting the solid into thin cylindrical slices,
6.
3 Cylindrical
Shells
working
from the inside
out. Each slice occurs at some xk
Used 0toand
find
volume
of a solid
of revolution
by
between
3 and
has thickness
 x. (Example
1)
summing volumes of thin cylindrical shells or
sleeves or tree rings.
Imagine cutting and unrolling a cylindrical shell to get a
(nearly) flat rectangular solid. Its volume is approximately V =
volume of a shell
length  height  thickness.
)
problem
The region enclosed by the x-axis and the parabola y =
f(x) = 3x – x2 is revolved about the y – axis. Find the
volume of the solid of revolution.
3x  x 2  0  x  0, x  3
b
Vshell  2   rh dx
a
3


3


Vshell  2  x(3x  x 2 ) dx  2  3x 2  x3 dx
0
3
0
 3 x4 
81
27
2  x    2 (27  )  2 (0  0)   units 3
4 
4
2

0
The shell swept out by the kth rectangle.
Notice this axis or revolution is parallel to the red
rectangle drawn.
The region bounded by the curve yy =
 /x,x , the x –axis and the
line x = 4 is revolved about the y-axis to generate a solid. Find
problem
the volume
of the solid.
The region, shell dimensions, and interval of integration in
b
Vshell  2   rh dx
a
4
4

2   rh dx  2  x x
0
5 4
2 2
x 

 2
 5

0
0

5
4
 (4 2
5
4 3
dx  2  x 2


0
5
128
2
0 )
5

dx


units 3
The shell swept out by the rectangle in.
Summary-Volumes-which method is best
Axis of rotation
x-axis
y-axis
dy
disk
perpendicular
shell
dx
b
d
 
 
V    r 2 dx
V    r 2 dy
a
c
dy
parallel
d
V  2   rh dy
c
dx
b
V  2   rh dx
a
b
Lengths of Plane curves L  
a
2
 dy 
1    dx
 dx 
Find the length of the arc formed by
f
1
( x)  4 x 2
dy
dx
33 1
1
u 2 du
4
5
on [1,8]
2
1
 2x 2
 dy 
   4x
 dx 
u = 1 + 4x
du = 4dx
du/4 = dx
L
5

33
3
1 2 2 
 u 
4 3


5

3
1
(33 2
6
8
L   1  4 xdx
1
3
 5 2 )  29.73 units
A