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Calculus Application of Integration

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Shantilal Shah Engineering College Bhavnagar Gujarat Technological University Information Technology Department
Transcript
Page 1: Calculus Application of Integration

Shantilal Shah Engineering College Bhavnagar

Gujarat Technological University

Information Technology Department

Page 2: Calculus Application of Integration

Topic:- APPLICATION OF INTEGRATION

Name Enrollment No:-

Maniya sagar J 160430116056

Koriya vijay 160430116044

Bhuva yatin 160430116010

Chotliya bhadresh 160430116014

Made By M.S.J

Page 3: Calculus Application of Integration

APPLICATION OF INTEGRATION

Page 4: Calculus Application of Integration

Find the volume of a solid of revolution using the area between the curves method.

Find the volume of a solid of revolution using the volume slicing method.

Find the volume of a solid of revolution using the disk method.

Find the volume of a solid of revolution using the washer method.

Find the volume of a solid of revolution using the cylindrical shell method.

Objectives

Page 5: Calculus Application of Integration

Areas Between CurvesTo find the area:• divide the area into n strips of equal width• approximate the ith strip by a rectangle with base Δx and height f(xi) – g(xi).• the sum of the rectangle areas is a good approximation• the approximation is getting better as n→∞.

y = f(x)

y = g(x)

• The area A of the region bounded by the curves y=f(x), y=g(x), and the lines x=a, x=b, where f and g are continuous and f(x) ≥ g(x) for all x in [a,b], is

b

adxxgxfA )]()([

Page 6: Calculus Application of Integration

21 2y x

2y x

2 2

12 x x dx

23 2

1

1 123 2

x x x

8 1 14 2 23 3 2

8 1 16 23 3 2

36 16 12 2 36

276

92

Example:-find area of the region enclosed by parabola and the line . And .2

1 2y x 2y x

solution:

Page 7: Calculus Application of Integration

y x

2y x

y x

2y x

If we try vertical strips, we have to integrate in two parts:

dx

dx

2 4

0 2 2x dx x x dx

We can find the same area using a horizontal strip.

dy

Since the width of the strip is dy, we find the length of the strip by solving for x in terms of y.y x

2y x2y x

2y x

2 2

02 y y dy

22 3

0

1 122 3

y y y 82 43

103

Example: Find the area of the region by parabola and the line .

dx

y x 2y x

solution:

Page 8: Calculus Application of Integration

General Strategy for Area Between Curves:

1

Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.)

Sketch the curves.

2

3 Write an expression for the area of the strip.(If the width is dx, the length must be in terms of x. If the width is dy, the length must be in terms of y.

4 Find the limits of integration. (If using dx, the limits are x values; if using dy, the limits are y values.)

5 Integrate to find area.

Page 9: Calculus Application of Integration

Volume by slicing

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The Disk Method

Page 16: Calculus Application of Integration

The Disk Method

If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution.

The simplest such solid is a rightcircular cylinder or disk, which isformed by revolving a rectangleabout an axis adjacent to oneside of the rectangle,as shown in Figure 7.13.

Figure 7.13

Axis of revolution

Volume of a disk: R2w

Page 17: Calculus Application of Integration

The volume of such a disk isVolume of disk = (area of disk)(width of disk)

= πR2w

where R is the radius of the disk and w is the width.

The Disk Method

Page 18: Calculus Application of Integration

To see how to use the volume of a disk to find the volume of a general solid of revolution, consider a solid of revolution formed by revolving the plane region in Figure 7.14 about the indicated axis.

Figure 7.14

The Disk Method

Disk method

Page 19: Calculus Application of Integration

To determine the volume of this solid, consider a representative rectangle in the plane region. When this rectangle is revolved about the axis of revolution, it generates a representative disk whose volume is

Approximating the volume of the solid by n such disks of width Δx and radius R(xi) produces

Volume of solid ≈

The Disk Method

Page 20: Calculus Application of Integration

This approximation appears to become better and better as So, you can define the volume of the solid as

Volume of solid =

Schematically, the disk method looks like this.

The Disk Method

Page 21: Calculus Application of Integration

A similar formula can be derived if the axis of revolution is vertical.

Figure 7.15

The Disk Method

Horizontal axis of revolution Vertical axis of revolution

Page 22: Calculus Application of Integration

Example 1 – Using the Disk Method

Find the volume of the solid formed by revolving the region bounded by the graph of and the x-axis (0 ≤ x ≤ π) about the x-axis.

Solution: From the representativerectangle in the upper graphin Figure 7.16, you can see thatthe radius of this solid is

R(x) = f (x)Figure 7.16

Page 23: Calculus Application of Integration

Example 1 – Solution

So, the volume of the solid of revolution isApply disk method.

Simplify.

Integrate.

[cont’d]

Page 24: Calculus Application of Integration

The Washer Method

Page 25: Calculus Application of Integration

• The disk method can be extended to cover solids of revolution with holes by replacing the representative disk with a representative washer.

• The washer is formed by revolving a rectangle about an axis,as shown in Figure 7.18.

• If r and R are the inner and outer radii of the washer and w is the width of the washer, the volume is given by• Volume of washer = π(R2 – r2)w.

The Washer Method

Figure 7.18

Axis of revolution Solid of revolution

Page 26: Calculus Application of Integration

To see how this concept can be used to find the volume of a solid of revolution, consider a region bounded by an outer radius R(x) and an inner radius r(x), as shown in Figure 7.19.

Figure 7.19

The Washer Method

Plane regionSolid of revolutionwith hole

Page 27: Calculus Application of Integration

• If the region is revolved about its axis of revolution, the

volume of the resulting solid is given by

• Note that the integral involving the inner radius represents

• the volume of the hole and is subtracted from the integral involving the outer radius.

The Washer Method

Washer method

Page 28: Calculus Application of Integration

Example 3 – Using the Washer MethodFind the volume of the solid formed by revolving the region bounded by the graphs of about the x-axis, as shown in Figure 7.20.

Figure 7.20

Solid of revolution

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Example 3 – Solution

In Figure 7.20, you can see that the outer and inner radii are as follows.

Integrating between 0 and 1 produces

Outer radius

Inner radius

Apply washer method.

Page 30: Calculus Application of Integration

Example 3 – Solution

Simplify.

Integrate.

cont’d

Page 31: Calculus Application of Integration

So far, the axis of revolution has been horizontal and you have integrated with respect to x. In the Example 4, the axis of revolution is vertical and you integrate with respect to y. In this example, you need two separate integrals to compute the volume.

The Washer Method

Page 32: Calculus Application of Integration

Example 4 – Integrating with Respect to y, Two-Integral Case

Find the volume of the solid formed by revolving the region bounded by the graphs of y = x2 + 1, y = 0, x = 0, and x = 1 about y-axis, as shown in Figure 7.21.

Figure 7.21

Page 33: Calculus Application of Integration

Example 4 – Solution• For the region shown in Figure 7.21, the outer

radius is simply R = 1.

• There is, however, no convenient formula that represents the inner radius

• When 0 ≤ y ≤ 1, r = 0, but when 1 ≤ y ≤ 2, r is determined by the equation y = x2 + 1, which implies that

Page 34: Calculus Application of Integration

Example 4 – Solution

Using this definition of the inner radius, you can use two integrals to find the volume.

Apply washer method.

Simplify.

Integrate.

cont’d

Page 35: Calculus Application of Integration

Example 4 – Solution

• Note that the first integral represents the volume

• of a right circular cylinder of radius 1 and height 1.

• This portion of the volume could have been determined

• without using calculus.

cont’d

Page 36: Calculus Application of Integration

VOLUME BY CYLINDRICAL SHELL

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Conclusion

We Hope this ppt usefull for you to find area and volume with use of different methods likes disk , washer and cylindrical.


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