Chapter 7. Applications of the Definite integral in Geometry, Science, and EngineeringByJiwoo Lee

Edited byWonhee Lee

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

7.1.2 Area Formula

Step 1

Determine which function is on top

Step 2Solve

Olive green area= ∫b

a[f(x)-g(x)]dx

Beige area= ∫b

a[g(x)]dx

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

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.

Tip3

If the two functions switch top and bottom, then the regions must be subdivided at those points to find total area

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

Upper function is 3y-y2 , therefore

∫403y-y2 –(y2-5y)dy

=64/3

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

∫0-1x3-2x-(-x)1/2dx+ ∫1.666

0(x)1/2-(x3-2x)dx =.083+2.283=2.367

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.

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

• ∫ba[A(x)]dx

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...

V= ∫10 (e-x)2dx = (1-1/e2)/2

If each cross section is a circle...

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.

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)

Rotated around a line

A(X)

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

area between the two functions rotated around a

line G(x)

A(x)

∫ba(∏(G(x)-a)2dx- ∏(A(x)-a)2dx]

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

1

V= ∏ ∫20[(3x-x2)2-x2dx]

2

V=2∏∫20[(4+1) 2-

(x2+1)2dx

=2∏∫20[24-x4-2x2]dx

Volume by Cylindrical Shells

Another method to determine the volume of a solid

• a b

• f(x)

•

• a b

• a b

• f(x)

•

• a b

• f(x)

•

• a b

When the section is flattened...

width =dx height = f(x) length= 2∏x

Area of Cross section= 2∏xf(x)dx

Volume of f(x) rotated around the y-axis = ∫b

a2∏xf(x)dx

• f(x)

• a b

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

Solve

• the region bounded by y=3x-x2 and y=x rotated about the y-axis

V= ∫ba2∏(2x2-x3)dx =

8∏/3

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

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

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

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

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

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

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

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

Square box with a side length of 7

Density of chemical found to be 90lb/ft3

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

Thank you for your undivided attention