Chapter 7. Applications of the Definite integral in Geometry, Science, and Engineering By Jiwoo Lee...
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Transcript of Chapter 7. Applications of the Definite integral in Geometry, Science, and Engineering By Jiwoo Lee...
![Page 1: Chapter 7. Applications of the Definite integral in Geometry, Science, and Engineering By Jiwoo Lee Edited by Wonhee Lee.](https://reader030.fdocuments.net/reader030/viewer/2022033020/56649ef25503460f94c042c6/html5/thumbnails/1.jpg)
Chapter 7. Applications of the Definite integral in Geometry, Science, and EngineeringByJiwoo Lee
Edited byWonhee Lee
![Page 2: Chapter 7. Applications of the Definite integral in Geometry, Science, and Engineering By Jiwoo Lee Edited by Wonhee Lee.](https://reader030.fdocuments.net/reader030/viewer/2022033020/56649ef25503460f94c042c6/html5/thumbnails/2.jpg)
Area Between Two Curves• If f and g are continuous functions o
n the interval [a,b] and if f(x) > g(x) for all x in [a,b] then the area of the region bounded by y=f(x), below by g(x), on the left by the line x=a, and on the right by the line x=b is
• ∫ba[f(x)-g(x)]dx
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7.1.2 Area Formula
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Step 1
Determine which function is on top
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Step 2Solve
Olive green area= ∫b
a[f(x)-g(x)]dx
Beige area= ∫b
a[g(x)]dx
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Tip1
g(x) f(x) • Sometimes it is easier to solve by integrating with respect to y rather than x
• ∫dc[f(x)-g(x)]dx
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Tip2
When finding the area enclosed by two functions, let the two functions equal each other and solve for the intersecting points to find a and b.
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Tip3
If the two functions switch top and bottom, then the regions must be subdivided at those points to find total area
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Solve
The area between the parabolasX=y2-5y and x=3y-y2
Solution: 1. Intersections at (0,0) and (-4,4)2. Determine upper function by either
plugging in points or graphing
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Upper function is 3y-y2 , therefore
∫403y-y2 –(y2-5y)dy
=64/3
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Solve
the area enclosed by the two functions y=x3-2x and y=(abs(x))1/2
Solution:1. Intersection at x= -1, 1.6662. Functions Switch top and bottom at
x=0 so the integral must be divided from -1 to 0 and 0 to 1.666
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∫0-1x3-2x-(-x)1/2dx+ ∫1.666
0(x)1/2-(x3-2x)dx =.083+2.283=2.367
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7.2 Volumes by Slicing; Disks and Washers
The volume of a solid can be obtained by integrating the cross-sectional area from one end of the solid to the other.
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Volume Formula
• Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x=a and x=b. If, for each x in [a, b], the cross-sectional area of S perpendicular to the x-axis is A(x), then the volume of the solid is
• ∫ba[A(x)]dx
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• ∫ba[A(x)]dx
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Example
The base of a solid is the region bounded by y=e-x ,the x-axis, the y-axis, and the line x=1. Each cross section perpendicular to the x-axis is a square. The volume of the Solid is...
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V= ∫10 (e-x)2dx = (1-1/e2)/2
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If each cross section is a circle...
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V= ∫ba ∏ (A(x))2dx
A special case, known as method of disks, often used to find areas of functions rotated around axis or lines.
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If there are two functions rotated, then subtract the lower region from the upper region, A method known
as method of washers
∫ba(∏(f(x))2dx- ∏(g(x))2dx]
f(x) g(x)
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Rotated around a line
A(X)
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height = A(x)-a Therefore, V= ∫b
a ∏ (A(x)-a)2dx
A(X) If “a” is below the x-axis, then “a” would be added to the height
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area between the two functions rotated around a
line G(x)
A(x)
∫ba(∏(G(x)-a)2dx- ∏(A(x)-a)2dx]
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Set up but do not solve for the area using washers method
1.y=3x-x2 and y=x rotated around the x-axis
2. y=x2 and y=4 rotated around the x-axis
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1
V= ∏ ∫20[(3x-x2)2-x2dx]
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2
V=2∏∫20[(4+1) 2-
(x2+1)2dx
=2∏∫20[24-x4-2x2]dx
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Volume by Cylindrical Shells
Another method to determine the volume of a solid
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• a b
• f(x)
•
• a b
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• a b
• f(x)
•
• a b
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• f(x)
•
• a b
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When the section is flattened...
width =dx height = f(x) length= 2∏x
Area of Cross section= 2∏xf(x)dx
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Volume of f(x) rotated around the y-axis = ∫b
a2∏xf(x)dx
• f(x)
• a b
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Reminder
Shells Method: ∫b
a[2∏x(fx)dx]
Washers Method: ∫b
a [∏(f(x))2dx- ∏(g(x))2dx]
When shells method is used and includes dx, then the function is rotated around the y axis
When washers method is used and uses dx, then the function is rotated around the x axis
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Solve
• the region bounded by y=3x-x2 and y=x rotated about the y-axis
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V= ∫ba2∏(2x2-x3)dx =
8∏/3
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Length of a Plane Curve
• If f(x) is a smooth curve on the interval [a,b] then the arc length L of this curve over [a,b] is defined as
• ∫ba√ 1 + [f’(x)]2 dx
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Length of a Plane Curve
• If no segment of the curve represented by the parametric equations is traced more than once as t increases from a to b, and if dx/dt and dy/dt are continuous functions for a<t<b, then the arc length is given by
• ∫ba√ (dx/dt) 2 +(dy/dt) 2 dx
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Area of a Surface of Revolution
• If f is smooth, nonnegative function on [a,b] then the surface area S of the surface of revolution that is generated by revolving the portion of the curve y = f(x) between x=a and x=b about the x-axis is defined as
• ∫ba2∏f(x) √1 + [f’(x)]2 dx
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Set up the integral for
The length the curve y2=x3 cut off by the line x=4
• Solution:• 2y dy/dx = 3 x2 , dy/dx = (3√x)/2• 2∫4
0√ 1 + 9x/4 dx
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Set up the integral for
The surface area of y=2x3 rotated around the x-axis from 2 to 7
• Solution: y’ =6x2
• ∫722∏2x3√1 + [6x2]2dx
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Work
• If a constant force of magnitude F is applied in the direction of motion of an object, and if that object moves a distance d, then we define the work W performed by the force on the object to be
• W=Fd
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Work
• Suppose that an object moves in the positive direction along a coordinate line over the interval [a,b] while subjected to a variable force F(x) that is applied in the direction of motion. Then we define work W performed by the force on the object to be
• W= ∫baF(x)dx
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Solve
• A square box with a side length of 7 feet is filled with an unknown chemical. How much work is required to pump the chemical to a connecting pipe on top of the box?
• Hint: The weight density of the chemical is found to be 90lb/ft3
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Square box with a side length of 7
Density of chemical found to be 90lb/ft3
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Solution
• Volume of each slice of chemical = 7*7*dy
• Increment of force= 90*49dy =4410dy• Distance lifted =7-y• Work=force * distance = 4410 ∫7
0(7-y)dy
=108045
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Thank you for your undivided attention