A k = area of k th rectangle, f(c k ) – g(c k ) = height, x k = width. 6.1 Area between two...
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Transcript of A k = area of k th rectangle, f(c k ) – g(c k ) = height, x k = width. 6.1 Area between two...
Ak = area of k th rectangle,f(ck) – g(ck ) = height, xk = width.
6.1 Area between two curves
Figure 4.23: When the formula for a bounding curve changes, the area integral changes to match. (Example 5)Find the area of the region between the curves
( ) ( ) 1f x x and g x x
A
Section 6.2 Figure 5Approximating 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
Figure 5.6: The region (a) and solid (b) in Example 4.6. 2 Volumes – Solid of revolution
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. 2( )diskV R thickness
Figure 5.6: The region (a) and solid (b) in Example 4.Volumes – Solid of revolution
2( )diskV R thickness
y x is rotated about the x-axis on [0, 4]Find the volume of the solid generated.
24 4 4
2 2 2 3
00 0
1(4 0 ) 8
2 2diskV x dx xdx x units
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.
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 4.
2( )diskV R thickness2 44 4 1
2 3
1 1 1
2 14 4 4 1 3
1 4disky
V dy y dy unitsy
Figure 5.10: The cross sections of the solid of revolution generated here are washers, not disks, so the integral A(x) dx leads to a slightly different formula. b
a
WashersIf the region revolved does not border on or cross the axis of revolution, the solid has a hole in it. The cross sections perpendicular to the axis are washers.
V = Outside Volume – Inside Volume
. The 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 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 x-coordinates 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
2 2b b
a a
Vwasher R dx r dx
2 2
23 1b b
a a
Vwasher x dx x dx
1 1
2 4 2
2 2
( 6 9) 2 1x x dx x x dx
11 5 3
4 2 2 3
2 2
117( 6 8) 3 8
5 3 5
x xx x x dx x x units
Calculation of volume
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 solid. Find the volume of the solid.
2
yx y and x
Drawing indicates a dy integration so solve each
equation for x as a function of y
Set = to find y limits of integration2
2 24 4 02 4
y yy y y y y y
y = 0 and y = 4 are limits
y-axis rotation
The washer swept out by one slice perpendicular to the y-axis.
2 2d d
c c
Vwasher R dy r 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 solid. Find the volume of the solid.
2 2d d
c c
Vwasher R dy r dy
2 24 4
0 02
yy dy dy
44 2 2 3
3
0 0
8
4 2 12 3
y y yy dy units
calculation
Figure 5.17: Cutting the solid into thin cylindrical slices, working from the inside out. Each slice occurs at some xk between 0 and 3 and has thickness x. (Example 1)
6. 3 Cylindrical ShellsUsed to find volume of a solid of revolution by 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 = length height thickness.
volume of a shell
Vshell =2(radius)(height)(thickness)
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.
2b
shella
V rh dx
Vshell =2(radius)(height)(thickness)
3 3
2 2 3
0 0
2 (3 ) 2 3shellV x x x dx x x dx
23 0 0, 3x x x x
343 3
0
81 272 2 (27 ) 2 (0 0)
4 4 2
xx units
problem
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 y = /x, , the x –axis and the line x = 4 is revolved about the y-axis to generate a solid. Find the volume of the solid.
y x
problem
The region, shell dimensions, and interval of integration in
2b
shella
V rh dx
34 4 42
0 0 0
2 2 2rh dx x x dx x dx
45 5 5
32 2 2
0
2 4 1282 (4 0 )
5 5 5x units
The shell swept out by the rectangle in.
Summary-Volumes-which method is best
Axis of rotationx-axis
shellparallel
dx
dy
diskperpendicular dx
dy
y-axis
2d
c
V rh dy 2b
a
V rh dx
2b
a
V r dx 2d
c
V r dy
Lengths of Plane curves
2
1b
a
dyL dx
dx
Find the length of the arc formed by12( ) 4 5 [1,8]f x x on
122
dyx
dx
2
4dy
xdx
8
1
1 4L xdx u = 1 + 4xdu = 4dxdu/4 = dx 331 3 3 333
2 2 2 2
5 5
1 1 2 1(33 5 ) 29.73
4 4 3 6L u du u units
Follow the link to the slide.
Then click on the figure to play the animation.A
Figure 6.2.5
Figure 6.2.12
Figure 6.3.7
Section 6.3 Figures 3, 4Volumes by Cylindrical Shells
Section 1 / Figure 1
Computer-generated picture of the solid in Example 9
A