Calculus Application of Integration

356 views

Published on

Ppt on calculus Application of Integration

Published in: Education
  • Be the first to comment

Calculus Application of Integration

  1. 1. Shantilal Shah Engineering College Bhavnagar Gujarat Technological University Information Technology Department
  2. 2. 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
  3. 3. APPLICATION OF INTEGRATION
  4. 4.  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
  5. 5. Areas Between Curves To 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 a dxxgxfA )]()([
  6. 6. 2 1 2y x  2y x  2 2 1 2 x x dx    2 3 2 1 1 1 2 3 2 x x x    8 1 1 4 2 2 3 3 2                  8 1 1 6 2 3 3 2     36 16 12 2 3 6     27 6  9 2  Example:-find area of the region enclosed by parabola and the line . And . 2 1 2y x  2y x  solution:
  7. 7. 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 2 y x 2y x  2y x    2 2 0 2y y dy  2 2 3 0 1 1 2 2 3 y y y  8 2 4 3   10 3  Example: Find the area of the region by parabola and the line . dx y x 2y x  solution:
  8. 8. 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.
  9. 9. Volume by slicing
  10. 10. The Disk Method
  11. 11. 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 right circular cylinder or disk, which is formed by revolving a rectangle about an axis adjacent to one side of the rectangle, as shown in Figure 7.13. Figure 7.13 Axis of revolution Volume of a disk: R2w
  12. 12. The volume of such a disk is Volume 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
  13. 13. 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
  14. 14. 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
  15. 15. 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
  16. 16. 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
  17. 17. 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 representative rectangle in the upper graph in Figure 7.16, you can see that the radius of this solid is R(x) = f (x) Figure 7.16
  18. 18. Example 1 – Solution So, the volume of the solid of revolution is Apply disk method. Simplify. Integrate. [cont’d]
  19. 19. The Washer Method
  20. 20. • 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
  21. 21. 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 region Solid of revolution with hole
  22. 22. • 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
  23. 23. Example 3 – Using the Washer Method Find 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
  24. 24. 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.
  25. 25. Example 3 – Solution Simplify. Integrate. cont’d
  26. 26. 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
  27. 27. 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
  28. 28. 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
  29. 29. 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
  30. 30. 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
  31. 31. VOLUME BY CYLINDRICAL SHELL
  32. 32. Conclusion We Hope this ppt usefull for you to find area and volume with use of different methods likes disk , washer and cylindrical.

×