Download - A k = area of k th rectangle, f(c k ) – g(c k ) = height, x k = width. 6.1 Area between two curves
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Ak = area of k th rectangle,f(ck) – g(ck ) = height, xk = width.
6.1 Area between two curves
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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
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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
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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
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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
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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
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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
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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
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. 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.
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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
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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
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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
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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
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The washer swept out by one slice perpendicular to the y-axis.
2 2d d
c c
Vwasher R dy r dy
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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
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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.
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)
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)
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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
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The shell swept out by the kth rectangle.Notice this axis or revolution is parallel to the red
rectangle drawn.
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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
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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
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The shell swept out by the rectangle in.
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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
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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
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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
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Section 6.3 Figures 3, 4Volumes by Cylindrical Shells
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Section 1 / Figure 1
Computer-generated picture of the solid in Example 9
A