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- 1. Active Learning Assignment CALCULAS(2110014) Branch : Chemical Engineering Sem : 1st Academic Year : 2014(odd)
- 2. APPLICATION OF INTEGRATION
- 3. Objectives 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.
- 4. 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 [ f (x) g(x)]dx a
- 5. 2 y1 2 x 2y x 2 2 1 2 x x dx 2 1 1 3 2 1 2 x x x 3 2 8 1 1 4 2 2 3 3 2 8 1 1 6 2 3 3 2 27 36 16 12 2 3 6 6 9 2 Example:-find area of the region enclosed by parabola 2 and the line . And . 1 2yx 2yx solution:
- 6. Example: Find the area of the region by parabola and the line . y x y x 2 yx y x 2 If we try vertical strips, we have to integrate in two parts: dx dx 2 4 0 2 x dx x x 2 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 y x terms of y. 2 y x y x 2 y 2 x 2 2 0 y 2 y dy 2 1 1 2 3 0 y 2 y y 2 3 8 2 4 3 10 3 dx yx 2yx solution:
- 7. General Strategy for Area Between Curves: 1 Sketch the curves. 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.) 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.
- 8. Volume by slicing
- 9. The Disk Method
- 10. 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. Axis of revolution Volume of a disk: R2w Figure 7.13
- 11. The Disk Method 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.
- 12. The Disk Method 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. Disk method Figure 7.14
- 13. The Disk Method 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 ≈
- 14. The Disk Method 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.
- 15. The Disk Method A similar formula can be derived if the axis of revolution is vertical. Horizontal axis of revolution Vertical axis of revolution Figure 7.15
- 16. 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
- 17. Example 1 – Solution So, the volume of the solid of revolution is Apply disk method. Simplify. Integrate. [cont’d]
- 18. The Washer Method
- 19. The Washer Method • 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. Axis of revolution Solid of revolution Figure 7.18
- 20. The Washer Method 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 Plane region Solid of revolution with hole
- 21. The Washer Method • If the region is revolved about its axis of revolution, the volume of the resulting solid is given by Washer method • Note that the integral involving the inner radius represents • the volume of the hole and is subtracted from the integral involving the outer radius.
- 22. 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
- 23. Example 3 – Solution In Figure 7.20, you can see that the outer and inner radii are as follows. Outer radius Inner radius Integrating between 0 and 1 produces Apply washer method.
- 24. Example 3 – Solution Simplify. Integrate. cont’d
- 25. The Washer Method 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.
- 26. 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
- 27. 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
- 28. 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
- 29. 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
- 30. VOLUME BY CYLINDRICAL SHELL
- 31. Conclusion We Hope this ppt usefull for you to find area and volume with use of different methods likes disk , washer and cylindrical.
- 32. THANKS………. Reference: google

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