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## The Area Between Two Curves

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### The Area Between Two Curves

### Volumes – The Disk Method

### Volume: The Shell Method

### Arc Length and Surfaces of Revolution

### Work

### Moments, Center of Mass, Centroids

### Fluid Pressure and Fluid Force

Lesson 6.1

When f(x) < 0

- Consider taking the definite integral for the function shown below.
- The integral gives a ___________ area
- We need to think of this in a different way

a

b

f(x)

Another Problem

- What about the area between the curve and the x-axis for y = x3
- What do you get forthe integral?
- Since this makes no sense – we need another way to look at it

Solution

- We can use one of the properties of integrals
- We will integrate separately for _________ and __________

We take the absolute value for the interval which would give us a negative area.

General Solution

- When determining the area between a function and the x-axis
- Graph the function first
- Note the ___________of the function
- Split the function into portions where f(x) > 0 and f(x) < 0
- Where f(x) < 0, take ______________ of the definite integral

Try This!

- Find the area between the function h(x)=x2 – x – 6 and the x-axis
- Note that we are not given the limits of integration
- We must determine ________to find limits
- Also must take absolutevalue of the integral sincespecified interval has f(x) < 0

Area Between Two Curves

- Consider the region betweenf(x) = x2 – 4 and g(x) = 8 – 2x2
- Must graph to determine limits
- Now consider function insideintegral
- Height of a slice is _____________
- So the integral is

The Area of a Shark Fin

- Consider the region enclosed by
- Again, we must split the region into two parts
- _________________ and ______________

Slicing the Shark the Other Way

- We could make these graphs as ________________
- Now each slice is_______ by (k(y) – j(y))

Practice

- Determine the region bounded between the given curves
- Find the area of the region

Horizontal Slices

- Given these two equations, determine the area of the region bounded by the two curves
- Note they are x in terms of y

Assignments A

- Lesson 7.1A
- Page 452
- Exercises 1 – 45 EOO

Integration as an Accumulation Process

- Consider the area under the curve y = sin x
- Think of integrating as an accumulation of the areas of the rectangles from 0 to b

b

Integration as an Accumulation Process

- We can think of this as a function of b
- This gives us the accumulated area under the curve on the interval [0, b]

Try It Out

- Find the accumulation function for
- Evaluate
- F(0)
- F(4)
- F(6)

Applications

- The surface of a machine part is the region between the graphs of y1 = |x| and y2 = 0.08x2 +k
- Determine the value for k if the two functions are tangent to one another
- Find the area of the surface of the machine part

Assignments B

- Lesson 7.1B
- Page 453
- Exercises 57 – 65 odd, 85, 88

Lesson 7.2

Revolving a Function

- Consider a function f(x) on the interval [a, b]
- Now consider revolvingthat segment of curve about the x axis
- What kind of functions generated these solids of revolution?

f(x)

a

b

Disks

- To find the volume of the whole solid we sum thevolumes of the disks
- Shown as a definite integral

f(x)

a

b

Try It Out!

- Try the function y = x3 on the interval 0 < x < 2 rotated about x-axis

Revolve About Line Not a Coordinate Axis

- Consider the function y = 2x2 and the boundary lines y = 0, x = 2
- Revolve this region about the line x = 2
- We need an expression forthe radius_______________

Washers

- Consider the area between two functions rotated about the axis
- Now we have a hollow solid
- We will sum the volumes of washers
- As an integral

f(x)

g(x)

a

b

Application

- Given two functions y = x2, and y = x3
- Revolve region between about x-axis

What will be the limits of integration?

Revolving About y-Axis

- Also possible to revolve a function about the y-axis
- Make a disk or a washer to be ______________
- Consider revolving a parabola about the y-axis
- How to represent the radius?
- What is the thicknessof the disk?

Revolving About y-Axis

- Must consider curve asx = f(y)
- Radius ____________
- Slice is dy thick
- Volume of the solid rotatedabout y-axis

Flat Washer

- Determine the volume of the solid generated by the region between y = x2 and y = 4x, revolved about the y-axis
- Radius of inner circle?
- f(y) = _____
- Radius of outer circle?
- Limits?
- 0 < y < 16

Cross Sections

- Consider a square at x = c with side equal to side s = f(c)
- Now let this be a thinslab with thickness Δx
- What is the volume of the slab?
- Now sum the volumes of all such slabs

f(x)

c

a

b

Cross Sections

- We could do similar summations (integrals) for other shapes
- Triangles
- Semi-circles
- Trapezoids

f(x)

c

a

b

Try It Out

- Consider the region bounded
- above by y = cos x
- below by y = sin x
- on the left by the y-axis
- Now let there be slices of equilateral triangles erected on each cross section perpendicular to the x-axis
- Find the volume

Assignment

- Lesson 7.2A
- Page 463
- Exercises 1 – 29 odd
- Lesson 7.2B
- Page 464
- Exercises 31 - 39 odd, 49, 53, 57

Lesson 7.3

Find the volume generated when this shape is revolved about the y axis.

We can’t solve for x, so we can’t use a horizontal slice directly.

Shell Method

- Based on finding volume of cylindrical shells
- Add these volumes to get the total volume
- Dimensions of the shell
- _________of the shell
- _________of the shell
- ________________

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

dx

f(x)

f(x) – g(x)

x

g(x)

Try It Out!

- Consider the region bounded by x = 0, y = 0, and

Hints for Shell Method

- Sketch the __________over the limits of integration
- Draw a typical __________parallel to the axis of revolution
- Determine radius, height, thickness of shell
- Volume of typical shell
- Use integration formula

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

thickness = _____

_______________ = y

Rotation About Non-coordinate Axis

- Possible to rotate a region around any line
- Rely on the basic concept behind the shell method

g(x)

f(x)

x = a

Rotation About Non-coordinate Axis

- What is the radius?
- What is the height?
- What are the limits?
- The integral:

r

g(x)

f(x)

a – x

x = c

x = a

f(x) – g(x)

c < x < a

Try It Out

- Rotate the region bounded by 4 – x2 , x = 0 and, y = 0 about the line x = 2
- Determine radius, height, limits

Try It Out

- Integral for the volume is

Assignment

- Lesson 7.3
- Page 472
- Exercises 1 – 25 odd
- Lesson 7.3B
- Page 472
- Exercises 27, 29, 35, 37, 41, 43, 55

Lesson 7.4

Arc Length

- We seek the distance along the curve fromf(a) to f(b)
- That is from P0 to Pn
- The distance formula for each pair of points

P1

Pi

Pn

•

P0

•

•

•

•

•

b

a

What is another way of representing this?

Arc Length

- We sum the individual lengths
- When we take a limit of the above, we get the integral

Arc Length

- Find the length of the arc of the function for 1 < x < 2

Surface Area

Δx

- Suppose we rotate thef(x) from slide 2 aroundthe x-axis
- A surface is formed
- A slice gives a __________

P1

Pi

Pn

•

P0

•

•

•

•

•

•

xi

b

a

Δs

Surface Area

- We add the cone frustum areas of all the slices
- From a to b
- Over entire length of the curve

Surface Area

- Consider the surface generated by the curve y2 = 4x for 0 < x < 8 about the x-axis

Surface Area

- Surface area =

Limitations

- We are limited by what functions we can integrate
- Integration of the above expression is not _________________________
- We will come back to applications of arc length and surface area as new integration techniques are learned

Assignment

- Lesson 7.4
- Page 383
- Exercises 1 – 29 odd also 37 and 55,

Lesson 7.5

b

x

Hooke's Law- Consider the work done to stretch a spring
- Force required is proportional to _________
- When k is constant of proportionality
- Force to move dist x =
- Force required to move through i th interval, x
- W = F(xi) x

Hooke's Law

- We sum those values using the definite integral
- The work done by a ____________force F(x)
- Directed along the x-axis
- From x = a to x = b

Hooke's Law

- A spring is stretched 15 cm by a force of 4.5 N
- How much work is needed to stretch the spring 50 cm?
- What is F(x) the force function?
- Work done?

Winding Cable

- Consider a cable being wound up by a winch
- Cable is 50 ft long
- 2 lb/ft
- How much work to wind in 20 ft?
- Think about winding in y amt
- y units from the top 50 – y ft hanging
- dist = y
- force required (weight) =2(50 – y)

Pumping Liquids

- Consider the work needed to pump a liquid into or out of a tank
- Basic concept: Work = weight x _____________
- For each V of liquid
- Determine __________
- Determine dist moved
- Take summation (__________________)

b

a

Pumping Liquids – Guidelines- Draw a picture with thecoordinate system
- Determine _______of thinhorizontal slab of liquid
- Find expression for work needed to lift this slab to its destination
- Integrate expression from bottom of liquid to the top

Pumping Liquids

4

- Suppose tank has
- r = 4
- height = 8
- filled with petroleum (54.8 lb/ft3)
- What is work done to pump oil over top
- Disk weight?
- Distance moved?
- Integral?

8

___________

Work Done by Expanding Gas

- Consider a piston of radius r in a cylindrical casing as shown here
- Let p = pressure in lbs/ft2
- Let V = volume of gas in ft3
- Then the work incrementinvolved in moving the pistonΔx feet is

Work Done by Expanding Gas

- So the total work done is the summation of all those increments as the gas expands from V0 to V1
- Pressure is inversely proportionalto volume so p _________ and

Work Done by Expanding Gas

- A quantity of gas with initial volume of1 cubic foot and a pressure of 2500 lbs/ft2 expands to a volume of 3 cubit feet.
- How much work was done?

Assignment A

- Lesson 7.5
- Page 405
- Exercises 1 – 41 EOO

Lesson 7.6

Mass

- Definition: mass is a measure of a body's ____________to changes in motion
- It is ___________ a particular gravitational system
- However, mass is sometimes equated with __________ (which is not technically correct)
- Weight is a type of ___________… dependent on gravity

Mass

- The relationship is
- Contrast of measures of mass and force

Centroid

- Center of mass for a system
- The point where all the mass seems to be concentrated
- If the mass is of constant density this point is called the __________________

4kg

10kg

6kg

•

Centroid

- Each mass in the system has a "moment"
- The product of ____________________________ from the origin
- "First moment" is the __________of all the moments
- The centroid is

4kg

10kg

6kg

Also notated Mx,moment about x-axis

Centroid- Centroid for multiple points
- Centroid about x-axis

First moment of the system

Also notated My, moment about y-axis

Centroid

- The location of the centroid is the ordered pair
- Consider a system with 10g at (2,-1), 7g at (4, 3), and 12g at (-5,2)
- What is the center of mass?

Centroid of Area Under a Curve

- First moment with respectto the y-axis
- First moment with respectto the x-axis
- Mass of the region

Try It Out!

- Find the centroid of the plane region bounded by y = x2 + 16 and the x-axis over the interval 0 < x < 4
- Mx = ?
- My = ?
- m = ?

Theorem of Pappus

- Given a region, R, in the plane and L a line in the same plane and not intersecting R.
- Let c be the centroid and r be the distance from L to the centroid

R

L

r

c

Theorem of Pappus

- Now revolve the region about the line L
- Theorem states that the volume of the solid of revolution iswhere A is the area of R

R

L

r

c

Assignment

- Lesson 7.6
- Page 504
- Exercises 1 – 41 EOO also 49

Lesson 7.7

Fluid Pressure

- Definition: The pressure on an object at depth h is
- Where w is the weight-density of the liquid per unit of volume
- Some example densitieswater 62.4 lbs/ft3mercury 849 lbs/ft3

Fluid Pressure

- Pascal's Principle: pressure exerted by a fluid at depth h is transmitted _______in all __________________
- Fluid pressure given in terms of force per unit area

Fluid Force on Submerged Object

- Consider a rectangular metal sheet measuring 2 x 4 feet that is submerged in 7 feet of water
- Rememberso P = 62.4 x 7 = 436.8
- And F = P x Aso F = 436.8 x 2 x 4 = 3494.4 lbs

Fluid Pressure

- Consider the force of fluidagainst the side surface of the container
- Pressure at a point
- Density x g x depth
- Force for a horizontal slice
- Density x g x depth x Area
- Total force

Fluid Pressure

- The tank has cross sectionof a trapazoid
- Filled to 2.5 ft with water
- Water is 62.4 lbs/ft3
- Function of edge
- Length of strip
- Depth of strip
- Integral

(-4,2.5)

(4,2.5)

(2,0)

(-2,0)

Assignment A

- Lesson 7.7
- Page 511
- Exercises 1-25 odd

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