volumes by slicing. disk find the volume of revolution using the disk method washer find the volume...
DESCRIPTION
These are a few of the many industrial uses for volumes of revolutions.TRANSCRIPT
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Find the Volume of revolution using the diskdisk methodFind the volume of revolution using the washerwasher methodFind the volume of revolution using the shellshell methodFind the volume of a solid with known cross sectionsknown cross sections
Volume of REVOLUTIONVolume of REVOLUTION
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These are a few of the many industrial uses for volumes of revolutions.
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Volume of the washerR
r
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ExampExample:le:
Find the volume of the solid formed by revolving the region bounded by y = x and y = x² over the interval [0, 1] about the x – axis.
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The region between the curve , and the
y-axis is revolved about the y-axis. Find the volume.
1xy
1 4y
y x
1 12
3
4
1 .7072
1 .5773
12
We use a horizontal disk.
dy
The thickness is dy.
The radius is the x value of the function .1
y
24
1
1 V dyy
volume of disk
4
1
1 dyy
4
1ln y ln 4 ln1
02ln 2 2 ln 2
Example of rotating the region about y-axis
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Find the volume of the solid of revolution formed by rotating the finite region bounded by the graphs of about the x-axis.
Example: rotate it Example: rotate it around around xx = axis = axis
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The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis:
2.000574 .439 185x y y x
y
500 ft
500 22
0.000574 .439 185 y y dy
The volume can be calculated using the disk method with a horizontal disk.
324,700,000 ft
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Find the volume of the solid generated by revolving the regionsabout the x-axis.2y x x and y 0 bounded by
12
0
r dx 1
22
0
dx xx
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Find the volume of the solid generated by revolving the regionsabout the x-axis.2y 2sin2x, 0 x bounded by
/ 22
0
xr d
/ 2
2
0
dx2sin2x
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Find the volume of the solid generated by revolving the regionsabout the y-axis.1
2y x, x 0, y 2 bounded by
22
0
r dy 2
2
0
d2y y
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Find the volume of the solid generated by revolving the regionsabout the line y = -1.2y 3 x , y 1 bounded by
22
2
dxr
2 22
2
3 x 1 dx
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The region bounded by and is revolved about the y-axis.Find the volume.
2y x 2y x
The “disk” now has a hole in it, making it a “washer”.
If we use a horizontal slice:
The volume of the washer is: 2 2 thicknessR r
2 2R r dy
outerradius
innerradius
2y x
2y x
2y x
y x
2y x
2y x
2
24
0 2yV y dy
4 2
0
14
V y y dy
4 2
0
1 4
V y y dy 4
2 3
0
1 12 12y y
1683
83
Example of a washer
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2y x
If the same region is rotated about the line x=2:
2y x
The outer radius is:
22yR
R
The inner radius is:
2r y
r
2y x
2y x
2y x
y x
4 2 2
0V R r dy
2
24
02 2
2y y dy
24
04 2 4 4
4yy y y dy
24
04 2 4 4
4yy y y dy
14 2 20
13 4 4
y y y dy 43
2 3 2
0
3 1 82 12 3y y y
16 64243 3
83
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The volume of the solid generated by revolving the first quadrantregion bounded by the curve and the lines x = ln 3 andy = 1 about the x-axis is closest to
x / 2y e
a) 2.79 b) 2.82 c) 2.85 d) 2.88 e) 2.91
ln3
2 2x / 2
0
e 1 dx
CALCULATOR REQUIRED
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2
322
0
222
0
24 2
0
2 4
The volume of the solid generated by rotating about the x-axis
the region enclosed between the curve y 3x and the line y 6x is given by
A. 6x 3x dx
B. 6x 3x dx
C. 9x 36x dx
D. 36x 9x d
2
0
22
0
x
E. 6x 3x dx
CALCULATOR REQUIRED
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Let R be the region in the first quadrant above by the graph of f x 2Arc tanx and below by the graph of y = x. What is the volume
of the solid generated when R is rotated about the x-axis?A. 1.21
B. 2.28 C. 2.69 D. 6.66 E. 7.15
CALCULATOR REQUIRED
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Let R be the region in the first quadrant that is enclosed by thegraph of f x ln x 1 , the x-axis and the line x = e. What is
the volume of the solid generated when R is rotated about theline y = -1?A.
5.037 B. 6.545 C. 10.073 D. 20.146 E. 28.686
CALCULATOR REQUIRED
e
2 2
0
ln x 1 1 1 dx 20.14627352 D
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Volumes by Cylindrical Shells
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Shell Method
• Based on finding volume of cylindrical shells– Add these volumes to get the total volume
• Dimensions of the shell– Radius of the shell– Thickness of the shell– Height
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The Shell• Consider the shell as one of many of a solid
of revolution
• The volume of the solid made of the sum of the shells
f(x)
g(x)xf(x) – g(x)
dx
2 ( ) ( )b
a
V x f x g x dx
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Hints for Shell Method
• Sketch the graph over the limits of integration• Draw a typical shell parallel to the axis of
revolution• Determine radius, height, thickness of shell• Volume of typical shell
• Use integration formula 2 radius height thickness
2b
a
Volume radius height thickness
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• Consider the region bounded by x = 0, y = 0, and
28y x
2 22
0
2 8V x x dx
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Rotation About x-Axis• Rotate the region bounded by y = 4x and
y = x2 about the x-axis
• What are the dimensions needed?– radius– height– thickness
radius = y
height = 4yy
thickness = dy
16
0
24yV y y dy
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Rotation About Non-coordinate Axis
• Possible to rotate a region around any line
• Rely on the basic concept behind the shell method
x = a
f(x) g(x)
2sV radius height thickness
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Rotation About Non-coordinate Axis
• What is the radius?
• What is the height?
• What are the limits?
• The integral:
x = a
f(x) g(x)a – x
f(x) – g(x)x = c
r
c < x < a
( ) ( ) ( )a
c
V a x f x g x dx
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• Rotate the region bounded by 4 – x2 , x = 0 and, y = 0 about the line x = 2
• Determine radius, height, limits
4 – x2 r = 2 - x
2
0
22
0
2 (2 ) (4 )V x x dx
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u = x – 1 x = 1 u=0
x = u + 1 x = 2 u=1 du = dx
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Blobs in Space
Volume of a blob:
Cross sectional area at height h: A(h)
Volume =
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Volumes with cross-sections:
• We will be given a “boundary” for the base of the shape which will be used to find a length.
• We will use that length to find the area of a figure generated from the slice .
• The dy or dx will be used to represent the thickness.
• The volumes from the slices will be added together to get the total volume of the figure.
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Procedure: volume by slicing
o sketch the solid and a typical cross sketch the solid and a typical cross sectionsection
o find a formula for the area, find a formula for the area, A(x),A(x), of of the cross sectionthe cross section
o find limits of integrationfind limits of integration
o integrate integrate A(x)A(x) to get volume to get volume
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 y2 1
Bounds?
Top Function?
Bottom Function?
[-1,1]
y 1 x2
y 1 x2
Length? 1 x2 1 x2 2 1 x2
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x2 y2 1
We use this length to find the area of the square.
Length? 2 1 x2
Area? 2 1 x2 24 1 x2
4 1 x2 dx 1
1
Volume?
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Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles.
x2 y2 1
Length? 2 1 x2
Area?12
2 1 x2 2 1 x2 Volume? 2 1 x2 dx
0
1
Bounds? [0,1]
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Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all squares whose sides lie on the base of the circle.
First, find the length of a side of the squarethe distance from the curve to the x-axis is half the length of the side of the square … solve for y
2 2
2 2
2
4 4 4
x yy x
y x
22 4 x
length of a side is :
2
2 2
2
2 4 4 4 16 4Area x x
x
22
216 4 Volume x dx
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Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all squares whose sides lie on the base of the circle.
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The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections
perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.
y 2 sin x
Bounds:Top Function: Bottom Function:
[0,π]y 2 sin x y 0
Length: 2 sin x
Area of an equilateral triangle:
34
(2 sin x )2
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Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all equilateral triangles whose sides lie on the base of the circle.
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Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all semicircles whose sides lie on the base of the circle.
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Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all Isosceles right triangles whose sides lie on the base of the circle.
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Let R be the region marked in the first quadrant enclosed bythe y-axis and the graphs of as shown in the figure below
2y 4 x and y 1 2sinx
R
a) Setup but do not evaluate the integral representing the volume of the solid generated when R is revolved around the x-axis.
b) Setup, but do not evaluate the integral representing the volume of the solid whose base is R and whose cross sections perpendicular to the x-axis are squares.
1.102
2 22
0
4 x 1 2sinx dx
1.102 22
0
4 x 1 2sinx dx
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CALCULATOR REQUIRED
2 2
The region S is represented by the area between the graphs of
f x 0.5x 2x 4 and g x 2 4 4x x . Write, but do
not evaluate, a definite integral which represents:a. the volume of a solid with base S if eac
h cross section ofthe solid perpendicular to the x-axis is a semi-circle.
b. the volume generated by rotating region S around the line y = 5.
24
0
g x f xdx
2 2
4
2 2
0
5 f x 5 g x dx
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NO CALCULATORThe base of a solid is the region in the first quadrant bounded by
the curve y sinx for 0 x . If each cross section of the solid perpendicular to the x-axis is a square, the volume of thesolid, in cu
bic units, is:A. 0 B. 1 C. 2 D. 3 E. 4
2
00 0
sinx dx sinxdx cos x | 1 1 2 C
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NO CALCULATORThe base of a solid is a right triangle whose perpendicular sideshave lengths 6 and 4. Each plane section of the solid perpendicularto the side of length 6 is a semicircle whose diameter lies in theplane of the triangle. The volume, in cubic units, of the solid is:A. 2 B. 4 C. 8 D. 16 E. 24
2
6 62
0 0
3 60
2 x1 13 dx x dx2 2 18
x |544 B
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Let R be the region in the first quadrant under the graph of
3
8y for 1, 8x
a) Find the area of R.
b) The line x = k divides the region R into two regions. If the part of region R to the left of the line is 5/12 of the area of the whole region R, what is the value of k?
c) Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the x-axis are squares.
CALCULATOR REQUIRED
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Let R be the region in the first quadrant under the graph of
3
8y for 1, 8x
a) Find the area of R.
8
31
8 dx 36x
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Let R be the region in the first quadrant under the graph of
3
8y for 1, 8x
b) The line x = k divides the region R into two regions. If the part of region R to the left of the line is 5/12 of the area of the whole region R, what is the value of k?
k
31
8 dxx
A
5 3612
2/ 3 k112x | 15
2/32k 11 12 5
k 3.375
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Let R be the region in the first quadrant under the graph of
3
8y for 1, 8x
c) Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the x-axis are squares.
28
21
8 dx 192x
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Let R be the region in the first quadrant bounded above by thegraph of f(x) = 3 cos x and below by the graph of 2xg x e
a) Setup, but do not evaluate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis.
b) Let the base of a solid be the region R. If all cross sections perpendicular to the x-axis are equilateral triangles, setup, but do not evaluate, an integral expression of a single variable for the volume of the solid.
20.836 22 x
0
3cosx e dx
20.836 2
x
0
3 3cosx e dx4