1. Chapter 6 Applications of Integration 机动 目录 上页 下页 返回 结束 6.1 Area Between Curves 6.2 Volume 6.3 Volume by Cylindrical Shell 6.5 Average Value of a Function

2. 6.1 Area Between Curves So far we have defined and calculated areas of the regions that lie under the graphs of functions. In this section we use integrals to find areas of more general regions. Type 1 of Regions S: Bounded by two curves y=f(x) and g(x) and between two vertical line y=a and y=b. Integral Formula: o ab x y y=f(x)y=f(x) y=g(x)y=g(x) S Figure 1

3. S: Lies between two curves x=f(y) and g(y) and between two line x=c and y=d. Integral Formula: Type 2 of Region o c d x y y=dy=d y=cy=c S x=g(y)x=g(y)x=f(y)x=f(y) Example 1 Find the area of the region bounded by the parabola y=x 2 and y=2x-x 2. Figure 2 x o (1,1) Figure 3 y

4. Example 2 Find the area of region bounded by the line y=x-1 and the parabola y 2 =2x+6. (1) y=x-1 -3 (-1,-2) (5,4) A1A1 A2A2 x y (2) x=y+1 -3 (-1,-2) (5,4) x y x=y 2 /2-3. Figure 4 Figure 5

5. If we are asked to find the area between the curves y=f(x) and y=g(x) where f(x)>g(x) for some values of x but g(x)>f(x) for other values of x, then we split the given region into several regions S 1, S 2, …with areas A 1, A 2, …., as shown in Figure 6. Since We have the following expression for A: S1S1 S2S2 S3S3 xba y Figure 6

6. Example 3 Find the area of the region bounded by the curve y=sinx, y=cosx, x=0 and x= y=cosx y=sinx A1A1 A2A2 x y o Figure 7 /section 6.1 end

7. 6.2 Volume 1. Volume of a Cylinder A cylinder is bounded by a plane region B 1, called the base, and a congruent region B 2 in a parallel plane(see Figure 1(a)). If the area of the base is A and the height of the cylinder is h, then the volume V of the cylinder is defines by V=Ah B1B1 B2B2 h (a) Cylinder V=Ah Figure 1 h r (b) Circular Cylinder

8. 2. Volume of a Solid Let S be any solid. The intersection of S with a plane is a plane region that is called a cross-section of S. Suppose that the area of the cross-section of S in a plane P x is A(x), where a<x<b.(see figure 2) x y A(a)A(a) A(b)A(b) A(x)A(x) PxPx x ab Figure 2

9. Let us consider a partition P of the interval [a, b] by point x i such that a=x 0 < x 1 <x 2 <…< x n =b. The plane P x will slice S into smaller “slabs”. If we choose in [x i-1, x i ], we can approximate the ith slab by a cylinder with base area A( ) and height (see Figure 3) x y a b x i-1 x i { Figure 3

10. So an approximation to the volume of the ith slab S i is Adding the volumes of these slabs, we get an approximation to the total volume: As ||P|| 0 we recognize the limit of this Riemann sum as an a definite integral and so we have the following definition: Definition of Volume Let S be a solid that lies between the planes. If the cross- section area of S is A(x), where A(x) is an integrable function, then the volume of S is (1)

11. Example 1 Show that the volume of a sphere of radius r is x y r x

12. 3. Solid of Revolution (1)Let S be a solid obtained by revolving the plane region R bounded by y=f(x), y=0, x=a and x=b about the x-axis. x y y=f(x)y=f(x) a b x y y=f(x)y=f(x) a S Figure 6 R

13. The area of the cross-section through x perpendicular to the x-axis is The use of this Formula is often called the disk method. Example 2 Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to1.(see figure 7) (2) Thus, using the basic volume formula (1), we have the following formula for a volume of revolution:

14. o 1 x y o 1 x y Figure 7 rotate

15. (2) Formula 2 applies only when the axis of rotation is the x-axis. If the region bounded by the curves x=g(y), x=0, y=c, and y=d is rotated about the y-axis, then the corresponding volume of revolution is (3) y d c x=g(y)x=g(y) xo d c x=g(y)x=g(y) xo y rotate Figure 8

16. Example 3 Find the volume of the solid obtained by the region bounded by y=x 3, y=8, and x=0 around the y-axis. o 8 x y o 8 x y Figure 9

17. (3) If the region bounded by the curves y=f(x), y=g(x), x=a, and x=b [where f(x)>g(x)] is rotated about the x-axis, then the volume of revolution is (4) This method is often called the washer method. Example 4 The region R bounded by the curves y=x and y=x 2 is rotated by the x-axis. Find the volume of the solid. y=x2y=x2 o x y y=xy=x R o x y Figure 9

18. Example 5 Find the volume of the solid obtained by rotating the region in Example 4 about the axis y=2. 4. Some Other Examples We conclude this section by finding the volumes of the solids that are not solids of revolution. Example 6 A solid has a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid. x y 1 A y x A B B x

19. Example 7 A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle 30 o along a diameter of the cylinder.Find the volume of the wedge. x y 30 o y AB C /section 6.2 end

20. 6.3 Volumes by Cylindrical Shells 1. Cylindrical Shell r2r2 r1r1 r Figure 1 Figure 1 shows a cylinder shell with inner radius r 1, outer radius r 2, and height h. Its volume is If we let (the thickness of the shell) and (the average of the shell), then this formula becomes (1)

21. 2. Method of Cylindrical Shells (1) Region: (Type 1) a bx y y=f(x)>0y=f(x)>0 o Axis for rotating: y-axis Solid of revolution: Figure 3 y=f(x)y=f(x) a bx y o Figure 2 Figure 3 Let P be a partition of the interval [a, b] by point x i such that a=x 0 < x 1 <x 2 <…< x n =b and let be the midpoint of [x i-1, x i ].

22. If the rectangle with base [x i-1,x i ] and height f( ) is rotated about the y-axis, then the result is a cylindrical shell with average radius, height f( ), and thickness (see Figure 4), so by formula 1 its volume is Therefore an approximation to the volume V of S is given by the sum of the volumes of these shells: Taking limit as ||P|| 0, we obtain the following volume formula for the solid in figure 3: (2) x Figure 4 y x i-1 xixi

23. Example 1 Find the volume of the solid obtained by rotating about the y-axis the region bounded by y=x(x-1) 2 and y=0. 1x y Example 2 Find the volume of the solid obtained by rotating about the y-axis the region bounded by y=x 2 and y=x. 0 x y 1 Figure 5 Figure 6

24. (2)Region: (Type 2) Axis for rotating: x-axis The volume of the solid of revolution ( Figure 8) is o c d x y y=dy=d y=cy=c R x=f(y)x=f(y) Figure 7 o d x y R x=f(y)x=f(y) Figure 8

25. Example 3 Find the volume of the solid obtained by rotating the region bounded by y=x-x 2 and y=0 about the line x=2. 12 34 Figure 9 /section 5.3 end

26. 6.4 Average Value of a Function This section aims to compute the average value of a function y=f(x), a<x<b. We start by dividing the interval [a, b] into n equal subintervals, each with length. Then we choose points in successive subintervals and calculate the average of the numbers : The limiting value as n approaches infinity is We define the average value of a function f on [a, b] as (1)

27. The question arises: Is there a number c at which the value of f is exactly equal to the average value of the function, that is, f(c) =f ave ? The following theorem says that this true for continuous functions: Mean Value Theorem for Integrals If f is a continuous function on [a, b], then there exists a number c in [a, b] such that (2) Example 1: Find the average value of the function f(x)=1+x 2 over the interval [-1, 2] and then find c such that f(c) =f ave.

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