#### Transcript Chapter 6

6.1 Area between two curves Ak = area of k th rectangle, f(ck) – g(ck ) = height, xk = width. Figure 4.23: When the formula for a bounding curve Find the of the changes region to between the curves changes, the area area integral match. (Example 5) f ( x) x and g ( x) x 1 Section 6.2 Figure 5 Approximating the volume of a sphere with radius 1 (a) Using 5 disks, V 4.2726 (b) Using 10 disks, V 4.2097 (c) Using 20 disks, V 4.1940 A Figure 5.6: 6.The region (a) and solidof (b)revolution in Example 4. 2 Volumes – Solid y = f(x) is rotated about x-axis on [a,b]. Find the volume of the solid generated. A cross-sectional slice is a circle and a slice is a disk. Vdisk R2 (thickness) Figure 5.6: Volumes The region–(a) and of solid (b) in Example 4. Solid revolution y x is rotated about the x-axis on [0, 4] Find the volume of the solid generated. Vdisk R (thickness) 2 4 Vdisk 0 2 4 4 1 2 2 2 x dx xdx x (4 0 ) 8 units 3 2 0 2 0 Volumes by disk-y axis rotation Find the volume of the solid generated by revolving a region between the y-axis and the curve x = 2/y from y = 1 to y = 4. Find the volume of the solid generated by revolving a region between the y-axis and the curve x = 2/y from y = 1 to 4. Vdisk R2 (thickness) 4 Vdisk 2 4 1 4 2 y 2 dy 4 y dy 4 y 1 1 1 1 1 4 1 3 units 3 4 Figure 5.10: The cross sections of the solid of revolution Washers generated here are washers, does not disks, so the integral If the region revolved not border on or b dx leads slightly different a A(x)the cross axistoofa revolution, theformula. solid has a hole in it. The cross sections perpendicular to the axis are washers. V = Outside Volume – Inside Volume region bounded by the curve y = x2 +1 and the line y = -x + 3 is revolved about the x-axis to generate a solid. Find the volume of the solid of revolution. . The The inner and outer radii of the washer swept out by one slice. Outer radius R = - x + 3 and the inner radius r = x2 +1 The inner and outer radii of the washer swept out by one slice. Outer radius R = - x + 3 and the inner radius r = x2 +1 Find the limits of integration by finding the xcoordinates of the points of intersection. x2 + 1= - x + 3 x2 + x –2=0 ( x+ 2 )(x – 1) = 0 x = -2 x=1 Outer radius R = - x + 3 and the inner radius r = x2 +1 Calculation of volume b 2 b 2 Vwasher R dx r dx a b a 2 b 2 Vwasher x 3 dx x 2 1 dx a 1 ( x 2 6 x 9)dx 2 a 1 2 x 4 2 x 2 1 dx 1 x5 x3 117 4 2 2 ( x x 6 x 8)dx 3x 8 x units 3 3 5 5 2 2 1 The region bounded by the parabola y = x2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a y-axis rotation solid. Find the volume of the solid. Drawing indicates a dy integration so solve each equation for x as a function of y y x y and x 2 Set = to find y limits of integration y y2 y y 4 y y2 y2 4 y 0 2 4 y = 0 and y = 4 are limits The washer swept out by one slice perpendicular to the y-axis. d 2 d 2 Vwasher R dy r dy c c The region bounded by the parabola y = x2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a calculation solid. Find the volume of the solid. 2 d 2 d Vwasher R dy r dy 4 0 c 2 c 2 4 y y dy dy 2 0 3 4 y y y 8 y dy units 3 4 2 12 0 3 0 4 2 2 Figure 5.17: Cutting the solid into thin cylindrical slices, 6. 3 Cylindrical Shells working from the inside out. Each slice occurs at some xk Used 0toand find volume of a solid of revolution by between 3 and has thickness x. (Example 1) summing volumes of thin cylindrical shells or sleeves or tree rings. Imagine cutting and unrolling a cylindrical shell to get a (nearly) flat rectangular solid. Its volume is approximately V = volume of a shell length height thickness. ) Vshell =2(radius)(height)(thickness) problem The region enclosed by the x-axis and the parabola y = f(x) = 3x – x2 is revolved about the y – axis. Find the volume of the solid of revolution. Vshell =2(radius)(height)(thickness) 3x x 2 0 x 0, x 3 b Vshell 2 rh dx a 3 3 Vshell 2 x(3x x 2 ) dx 2 3x 2 x3 dx 0 3 0 3 x4 81 27 2 x 2 (27 ) 2 (0 0) units 3 4 4 2 0 The shell swept out by the kth rectangle. Notice this axis or revolution is parallel to the red rectangle drawn. The region bounded by the curve yy = /x,x , the x –axis and the line x = 4 is revolved about the y-axis to generate a solid. Find problem the volume of the solid. The region, shell dimensions, and interval of integration in b Vshell 2 rh dx a 4 4 2 rh dx 2 x x 0 5 4 2 2 x 2 5 0 0 5 4 (4 2 5 4 3 dx 2 x 2 0 5 128 2 0 ) 5 dx units 3 The shell swept out by the rectangle in. Summary-Volumes-which method is best Axis of rotation x-axis y-axis dy disk perpendicular shell dx b d V r 2 dx V r 2 dy a c dy parallel d V 2 rh dy c dx b V 2 rh dx a b Lengths of Plane curves L a 2 dy 1 dx dx Find the length of the arc formed by f 1 ( x) 4 x 2 dy dx 33 1 1 u 2 du 4 5 on [1,8] 2 1 2x 2 dy 4x dx u = 1 + 4x du = 4dx du/4 = dx L 5 33 3 1 2 2 u 4 3 5 3 1 (33 2 6 8 L 1 4 xdx 1 3 5 2 ) 29.73 units A Follow the link to the slide. Then click on the figure to play the animation. Figure 6.2.5 Figure 6.3.7 Figure 6.2.12 Section 6.3 Figures 3, 4 Volumes by Cylindrical Shells Computer-generated picture of the solid in Example 9 Section 1 / Figure 1 A