Chapter 6

Report
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. V d isk   R 2 ( th ickn ess )
Figure 5.6: Volumes
The region–(a)
and of
solid
(b) in Example 4.
Solid
revolution
y
is rotated about the x-axis on [0, 4]
Find the volume of the solid generated.
x
V d isk   R ( th ickn ess )
2
2
4
V disk  

0
x

4
dx  
 xdx  
0
1
2
4
x
2

0

2
(4  0 )  8 units
2
2
3
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.
V d isk   R ( th ickn ess )
2
4
V disk
2
4
1
2
y
2
     dy    4 y dy  4
y
1
1
1
4
1
1

3
  4   1   3 units
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
dx leads
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
V w asher  
R
2
dx  
a
b
V w asher  
2
b
   x  3
r 
a
2
b
dx  
a
1


2
dx


2
x  1 dx
a
1
( x  6 x  9) dx  
2
2

4
2

x  2 x  1 dx
2
1
3
  x5

x
117 
4
2
2
3
  (  x  x  6 x  8) dx   

 3x  8x 
units
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
x
y and x 
y
2
Set = to find y limits
of integration
y 
y
2
 y
y
2
2
2
 4y  y  y  4y  0
4
y = 0 and y = 4 are limits
The washer swept out by one slice perpendicular to the y-axis.
d
V w a sh er  
R
c
2
d
dy  
 r 
c
2
dy
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
R
V w a sh er  
dy  
c


y

0

y 
  y 
dy  

4 
0
4
2
 r 
dy
c
2
4
2
d
2
4
 y
dy      dy
2
0
4
y
y 
8
3



units


12 
3
 2
0
2
3
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.
)
Vshell =2(radius)(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.
Vshell =2(radius)(height)(thickness)
2
3 x  x  0  x  0, x  3
b
V shell  2
  rh dx
a
3
V shell  2

0
3

x (3 x  x ) dx  2
2
3

2
3x  x
3
dx
0
 3 x 
81
27
3
2  x 

2

(27

)

2

(0

0)

 units

4 
4
2

0
4
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
V shell  2
  rh dx
a

2   rh dx  2  x x dx  2   x 2 dx


0
0
0


4
5
5
5

2 2
4
128
3
2
2
 2 x    (4  0 ) 
units
5
5
5



0
4
4


4
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
dx
V 
b

r
2
d

V 
dx
a

r
2
dy
c
dy
shell
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
1
f ( x)  4 x 2  5
dy
1
 2x2
dx
on [1, 8]
 dy 

  4x
 dx 
u = 1 + 4x
du = 4dx
du/4 = dx
L
1
4
33

5
8
2
33
L

1  4 x dx
1
3
3


1 2 2
1
2
u du   u   (33 2  5 2 )  29.73 units
4 3
6


5
1
3
A
Follow the link to the slide.
Then click on the figure to play the animation.
Figure 6.2.5
Figure 6.3.7
Figure 6.2.12
Section 6.3  Figures 3, 4
Volumes by Cylindrical Shells
Computer-generated picture of the solid in Example 9
Section 1 / Figure 1
A

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