1 / 42

10 .1 Finding Plane Areas by Integration

10. Application s of D efinite Integrals. Case Study. 10 .1 Finding Plane Areas by Integration. 10 .2 Volumes of Solids of Revolution. Chapter Summary. In some countries, there are very wide income gaps between the rich and the poor.

donat
Télécharger la présentation

10 .1 Finding Plane Areas by Integration

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 10 Applicationsof Definite Integrals Case Study 10.1Finding Plane Areas by Integration 10.2 Volumes of Solids of Revolution Chapter Summary

  2. In some countries, there are very wide income gaps between the rich and the poor. Yes, I know. How can we give a quantitative measure of the inequality of income distribution? Case Study To measure the inequality of income distribution in a society, we can make use of the Gini coefficient. The ‘Lorenz Curve’ is sketched such that any point (x, y) on the curve indicates that the poorest x% families share y% of the total income of all the families in the society. If the Gini coefficient is close to 1, then there is a very wide income gap between the rich and the poor; if it is close to 0, then the income is uniformly distributed among the families.

  3. If f (x) ³ 0 in the interval a£x£b, then area of the shaded region . In this case, gives the area of the region bounded by the curve yf (x), the x-axis, the lines xa and x b. 10.1 Finding Plane Areas by Integration A. Area of the Region Bounded by a Curve and the x-axis In Chapter 9, we learnt that the definite integral of f (x) from a to b is defined as follows: As shown in the figure, if f (x) ³ 0 for a£x£b, then f (zi)Dx stands for the area of a very thin rectangle drawn under the curve yf (x).

  4. If f (x) £ 0 in the interval a£x£b, then area of the shaded region . If f (x) ³ 0 in the interval a£x£c and f (x) £ 0 in the interval c£x£b, then area of the shaded region  A1 + A2 10.1 Finding Plane Areas by Integration A. Area of the Region Bounded by a Curve and the x-axis On the other hand, if f (x) £ 0 for a £x£b as shown in the figure, then f (zi)x is equal to the negative of the area of the thin rectangle. If f (x) takes both positive and negative values in the interval a£x£b, we can divide the shaded region into two parts: one part for f (x) ³ 0 and the other for f (x) £ 0.

  5. 10.1 Finding Plane Areas by Integration A. Area of the Region Bounded by a Curve and the x-axis Example 10.1T Find the area of the region bounded by the curve y  x2, the axes and the line x  3. Solution: The required area

  6. 10.1 Finding Plane Areas by Integration A. Area of the Region Bounded by a Curve and the x-axis Example 10.2T Find the area of the region bounded by the curve y = x2 – 6x + 8, and the x-axis in the interval 0 £x£4. Solution: When x2 – 6x + 8 = 0, The required area

  7. where f (x) ³g(x) in the interval a £x£ b. 10.1 Finding Plane Areas by Integration B. Area of the Region Bounded by y = f(x) and y = g(x) In the figure, y = f (x) and y = g(x) are two continuous functions such that f (x) ³g(x) in the interval a£ x£b. Once again, the area of the shaded region A can be approximated by a large number of thin, vertical rectangles of equal width ∆x. However, now the length of a rectangle is given by f (zi) – g(zi).

  8. 10.1 Finding Plane Areas by Integration B. Area of the Region Bounded by y = f(x) and y = g(x) Example 10.3T Find the area of the region bounded by the curves y = x2 and y = x3 . Solution: Consider the points of intersection of the two curves. (2) – (1): Hence the curves intersect when x = 0 and x = 1. The required area

  9. 10.1 Finding Plane Areas by Integration B. Area of the Region Bounded by y = f(x) and y = g(x) Example 10.4T Find the area of the region bounded by the curves y = x2, xy = 1 and the line y = 4x in the interval 0£ x £ 1. Solution: Solving y = x2 and xy = 1, the point of intersection is (1, 1). Solving xy = 1 and y = 4x, the point of intersection is . The required area

  10. 10.1 Finding Plane Areas by Integration B. Area of the Region Bounded by y = f(x) and y = g(x) In Example 10.3T, the function f (x) – g(x) does not change sign in the interval a £ x£b, so the area of the region bounded by the two curves can be simply calculated by using the formula However, if y = f (x) and y = g(x) intersect in the interval as shown in the figure, we need to consider the shaded region on each side of the point of intersection independently.

  11. 10.1 Finding Plane Areas by Integration B. Area of the Region Bounded by y = f(x) and y = g(x) Example 10.5T Find the area of the region bounded by the curves and y = sin x in the interval 0£x£ p. Solution: At the point of intersection, The coordinates are On the left hand side of the intersection, is positive. On the right hand side of the intersection, is negative.  The required area

  12. 10.1 Finding Plane Areas by Integration C. Area of the Region Bounded by a Curve and the y-axis Sometimes, the equation of a curve may be written in the form x = u(y), such as x = y2. Using similar arguments as we have studied in the previous section, we can find the area of the region bounded by the curve x = u(y), the y-axis and the lines y = c and y = d.

  13. 10.1 Finding Plane Areas by Integration C. Area of the Region Bounded by a Curve and the y-axis Example 10.6T Find the area of the region bounded by the curve y = x3 – 2, the y-axis and the line y = 6. Solution: Fory = x3– 2, The required area

  14. 10.1 Finding Plane Areas by Integration C. Area of the Region Bounded by a Curve and the y-axis Example 10.7T Find the area of the region bounded by the curve x = –y2 +2y + 3 and the y-axis in the interval –1 £y£ 6. Solution: The required area

  15. where u(y) ³v(y) in the interval c £y£ d. 10.1 Finding Plane Areas by Integration D. Area of the Region Bounded by x = u(x) and x = v(y) As shown in the figure, x = u(y) and x = v(y) are two continuous functions such that u(y) ³v(y) in the interval c£y £ d.

  16. 10.1 Finding Plane Areas by Integration D. Area of the Region Bounded by x = u(x) and x = v(y) Example 10.8T • In the figure, y  4x – 3 is a tangent to the curve y  x4. • Find the point of contact. • Find the area of the region bounded by the curve y  x4, the line y  4x – 3 and the x-axis. Solution: (a) Slope of y = 4x – 3 is 4. For y = x4, (b)The required area When 4x3 = 4,  The point of contact is (1, 1).

  17. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Consider the region bounded by the curve y+ 1, the x-axis, the y-axis and the line x = 4. If it is revolved about the x-axis, then it will generate a solid as shown in the figure. In this case, we call it a solid of revolution and the x-axis the axis of revolution. We will learn how to find the volume V of the solid of revolution using integration.

  18. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Consider the curve of a non-negative function y = f (x) defined in the interval a£x£b. The area is approximated by n thin rectangles of width Dx = and length f (zi), where i = 1, 2, ... , n. If the first rectangle from the left is revolved about the x-axis to form a disc, the volume of the disc is If n is getting larger indefinitely, the width Dx tends to zero. Then the sum of the volumes of the discs will approach the actual volume of the solid of revolution. This method of finding the volume of the solid of revolution by integration is called the disc method.

  19. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis As shown in the figure, the shaded region enclosed by the line y=x (where r, h > 0) and x=h is revolved about the x-axis. Since the equation of the function is y=x with a = 0 and b = h, the required volume = We can also find the general formula for calculating the volume of a cylinder and a sphere using the disc method.

  20. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Example 10.9T Find the volume of the solid generated by revolving the region enclosed by the curve y  , the x-axis and the line x= 6 about the x-axis. Solution: The required volume

  21. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Example 10.10T Find the volume of the solid generated by revolving the region enclosed by the curve y2  x2(x + 1) about the x-axis. Solution: When x2(x + 1) = 0, x = –1 or 0 The required volume

  22. The volume enclosed by the outer surface of this hollow solid is given by , while the volume of the empty space is equal to . 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Now, consider two continuous functions f (x) and g(x), where f (x) ³g(x) > 0 in the interval a£x£b. If the region bounded by the curves y = f (x) and y = g(x) is revolved about the x-axis, then a hollow solid of revolution will be formed.

  23. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Example 10.11T Find the volume of the solid generated when the region bounded by the curves y  x2 + 2 and y  –x2 + 4 is revolved about the x-axis. Solution: Substituting (1) into (2),  The curves intersect at (1, 3) and (1, 3). The required volume

  24. 10.2Volumes of Solids of Revolution A. Solids of Revolution Revolved about the x-axis Example 10.12T Find the volume of the solid generated when the region bounded by the curves y = ex and y = e–xin the interval 0 ≤x ≤ ln 2 is revolved about the x-axis. Solution: The required volume

  25. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Using the method, we can derive the formula for the volume of the solid generated by revolving about the y-axis. If u(y) ³ 0 in the interval c£y£d, then when the shaded region is revolved about the y-axis, the volume of the solid of revolution can be found by using the following formula:

  26. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Example 10.13T Find the volume of the solid generated when the region between the y-axis and the curve x = tan y, where is revolved about the y-axis. Solution: The required volume

  27. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Example 10.14T The region bounded by the curve y  , the line 3x – 4y – 8  0 and the y-axis with x > 0 is revolved about the y-axis, find the volume of the solid generated Solution:  The point of intersection is (4, 1). Substituting (1) into (2),

  28. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Example 10.14T The region bounded by the curve y  , the line 3x – 4y – 8  0 and the y-axis with x > 0 is revolved about the y-axis, find the volume of the solid generated Solution:  The required volume

  29. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Now let us consider two continuous functions u(y) and v(y), where u(y) ³v(y) > 0 in the interval c£y£d. If the region bounded by the curves x = u(y) and x = v(y) is revolved about the y-axis, then the volume V of the hollow solid formed is given by:

  30. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Example 10.15T A container is obtained by revolving the region bounded by the curves y  10 – x2, y  8 – x2 and the line y  –2 about the y-axis. Find the volume of the material needed to make the container. Solution: Rewrite the equations of the curvesy  10 – x2 and y  8 – x2 as x2  10 – y and x2  8 – y respectively. The required volume

  31. 10.2Volumes of Solids of Revolution B. Solids of Revolution Revolved about the y-axis Example 10.16T Find the volume of the solid generated when the regionbounded by the curve y  2– (x + 1)2 and the line y  1is revolved about the y-axis. Solution: Vertex  (1, 2), The required volume Since the x-coordinates of the point on the left side of vertex less than –1, the equation of it is , where 1 £y£2. The equation of the point on the right side of vertex is , where 1 £y£2.

  32. 10.2Volumes of Solids of Revolution C. Solids of Revolution Revolved about Lines Parallel to the Coordinate Axes Sometimes, the solid of revolution may be formed by revolving about a line parallel to the x-axis or the y-axis. In this case, we need to modify the formulas obtained in Sections 10.2 A and B in order to find the volume of the solid of revolution. For example, as shown in the figure, the region bounded by the curve y = f (x) and the line y = h (where h > 0) in the interval a£x£ b is revolved about the line y = h to form the solid of revolution. If we translate the curve y = f (x) and the line y = h downwards by h units, then we can see that it is just the same as revolving the curvey = f (x) –h about the x-axis ! Thus, we can conclude that, the volume V of the solid of revolution is given by

  33. 10.2Volumes of Solids of Revolution C. Solids of Revolution Revolved about Lines Parallel to the Coordinate Axes Example 10.17T Find the volume of the solid generated when the region bounded by the curve y  , the lines y = 2 and x  is revolved about the line y  2. Solution: when y  2, The required volume

  34. 10.2Volumes of Solids of Revolution C. Solids of Revolution Revolved about Lines Parallel to the Coordinate Axes Example 10.18T Find the volume of the solid generated when the region bounded by the curve x  cos y andthe lines x  y + 1 and x  2 in the interval is revolved about the line x  2. Solution: For xy + 1, when x 2, y 1 The required volume

  35. 10.2Volumes of Solids of Revolution D. Shell Method In the disc method mentioned previously, the volumes of solids of revolution are found by approximating with circular discs. But in some situations, it may be easier to use the shell method, in which thin cylindrical shells are used to approximate the volume of the solid of revolution. For example, consider the region bounded by the curve y = f (x), the x-axis, the lines x = a and x = b as shown in the figure, where f (x) ³ 0 and 0 £a£b. A hollow solid is obtained when the region is revolved about the y-axis.

  36. 10.2Volumes of Solids of Revolution D. Shell Method Now, if we approximate the region by a number of vertical rectangles and all these rectangles are revolved about the y-axis, each of them will generate a thin cylindrical shell of thickness Dx, height f (x) and radius x as shown in the figure. If Dx is very small, the volume of such a shell is approximately equal to the volume of a thin rectangular sheet of thicknessDx, height f (x) and length equal to the base circumference of the shell, that is, 2px. Hence the volume of a shell is approximately equal to 2pxf (x)Dx and thus the total volume of the solid is

  37. 10.2Volumes of Solids of Revolution D. Shell Method Similarly, if the region bounded by the function x = g(y), the lines y = c and y = d is revolved about the x-axis, then Note: 1. In the above discussion, f (x) and g(y) are assumed to be non-negative within the interval of integration. If not, the formulas should be modified as • In the shell method, when the region is revolved about the y-axis, the volume is integrated along x; when it is revolved about the • x-axis, the volume is integrated along y. It is opposite to the formulas we learnt in the disc method. Students should pay attention to this.

  38. 10.2Volumes of Solids of Revolution D. Shell Method Example 10.19T Find the volume of the solid generated when the region bounded by the curve y  , the x-axis and the line x  9 is revolved about y-axis. Solution: The required volume

  39. 10.2Volumes of Solids of Revolution D. Shell Method Example 10.20T Find the volume of the solid generated when the region bounded by the curve y = x3, the x-axis and the line x = 1 is revolved about the line x = 2. Solution: The required volume

  40. Chapter Summary 10.1 Finding Plane Areas by Integration 1. For a£b £c, 2. For a£b £c, 3. For d£e £f, 4. For d£e £f,

  41. Chapter Summary 10.2 Volumes of Solids of Revolution • Hollow Solid of revolution about the x-axis • In the figure, by the disc method, 2. Hollow Solid of revolution about the y-axis In the figure, by the disc method, When the region bounded by the curve y = f (x), the linesx = a, x = b and y = h is revolved about y = h, the volume of solid of revolution is given by

  42. Chapter Summary 10.2 Volumes of Solids of Revolution In the figure, by the shell method, 1. 2. Similarly, when the region bounded by the curve x = g(y), the lines y = c and y = d is revolved about the x-axis,

More Related