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
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
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.
)
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)
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
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