Tuesday, September 29, Page 453. Problem 5

Transcription

1 Tuesday, September 9, 15 Page 5 Problem 5 Problem. Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by y = x, y = x 5 about the x-axis. Solution. The inner radius is R 1 x) = x 5 and the outer radius is R x) = x. The extremities are x = and x = 1. So the volume is Problem 6 π x ) x 5 ) ) dx x x 1) dx 1 5 x x ) 11 = 6π 55. Problem. Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by y =, ] 1 y = x about the x-axis. Solution. The inner radius is R 1 x) = and the outer radius is R x) = x. The 1

2 extremities are x = and x =. So the volume is ) ) V x dx ) ) 16 x + x dx 16 ) 1 x + x dx 16 1x x x5 ] ) 5 ) Problem 11b) ) 8 = 168π x ) + 18 )] )5 )] about the y-axis. y = x, y =, x = Solution. Because we are revolving about the y-axis, we should write our functions as functions of y. So we have x = y. Now, the inner radius is R 1 y) = y and the

3 outer radius is R y) =. The extremities are y = and y =. So the volume is Problem 11d) π y ) ) dy 9 y ) dy 9y 15 ] y5 9 1 ) 5 9 = 6π 5. about the line x = 6. y = x, y =, x = Solution. The inner radius is from x = 6 to x =, so R 1 y) = 6 = and the outer radius is R y) = 6 y. The extremities are y = and y =. The volume is π 6 y ) ) dy 7 1y + y ) dy 7y y + 15 ] y5 7 ) + 15 ) ) 5 = 8π 5.

4 Problem 15 about the line y =. y = x, y =, x = Solution. We are revolving about a horizontal line, so we must integrate in the x direction and make vertical slices. The inner radius goes from y = to y =, so R 1 = 1. The outer radius goes from y = to y = x, so R = x. The extremities are from x = to x =. The volume is π x) 1 ) dx 15 8x + x ) dx 15x x + 1 x ] ) = 18π. Problem y = x, y =, y =, x = about the line x = 5.

5 Solution. We are revolving about a vertical line, so we must integrate in the y direction and make horizontal slices. The inner radius goes from x = 5 to x = y, so R 1 = 5 y) = y +. The outer radius goes from x = 5 to x =, so R = 5. The extremities are from y = to y =. The volume is Problem 1 π 5 y + ) ) dy 1 y y ) dy 1y y 1 ] y 8 8 ) = 9π. about the line x = 5. x = y, x = Solution. We are revolving about a vertical line, so we must integrate in the y direction and make horizontal slices. The inner radius goes from x = 5 to x =, so R 1 = 1. The outer radius goes from x = 5 to x = y, so R = 5 y. The extremities are from y = to y = =. The 5

6 volume is π 5 y ) 1 ) dy 1y + y ) dy y 1 y y ) 5 = 8π 15. ] Problem 9 y = x + 1, y = x + x + 5, x =, x = about the x-axis Solution. The graph is We are revolving about a horizontal line, so we must integrate in the x direction and make vertical slices. Furthermore, we see from the graph that the upper and lower function switch at x =, so we must set up two integrals. 6

7 In the first integral, the inner radius goes from y = to y = x +1, so R 1 = x +1. The outer radius goes from y = to y = x + x + 5, so R = x + x + 5. The extremities are from x = to x =. The first volume is V 1 = π x + x + 5) x + 1) ) dx x 8x + x + ) dx x 8 ] x + 1x + x 16 6 ) = 15π. In the second integral, the inner radius goes from y = to y = x + x + 5, so R 1 = x + x + 5. The outer radius goes from y = to y = x + 1, so R = x + 1. The extremities are from x = to x =. The second volume is V = Thus, the combined volume is π x + 1) x + x + 5) ) dx x + 8x x ) dx ] x + 8 x 1x x ) ) = 15π. )) 8 V 1 + V = 15π = 77π. + 15π 7

8 Problem y = x, y = 1 x +, x =, x = 8 about the x-axis Solution. The graph is We are revolving about a horizontal line, so we must integrate in the x direction and make vertical slices. Furthermore, we see from the graph that the upper and lower function switch at x =, so we must set up two integrals. In the first integral, the inner radius goes from y = to y = x, so R 1 = x. The outer radius goes from y = to y = 1x +, so R = 1 x +. The extremities 8

9 are from x = to x =. The first volume is V 1 1 x + ) ) x) dx ) 1 x 5x + 16 dx 1 1 x 5 ] x + 16x ) = 88π In the second integral, the inner radius goes from y = to y = x + x + 5, so R 1 = x + x + 5. The outer radius goes from y = to y = x + 1, so R = x + 1. The extremities are from x = to x = 8. The second volume is 8 V x) 1 ) x + ) dx 8 1 ) x + 5x 16 dx Thus, the combined volume is 1 1 x + 5 ] 8 x 16x 18 ) ) = 56π. 16 )) + 6 V 1 + V = 88π + 56π = 1π = 8π. 9

10 Problem 1 Problem. Find the volume of the solid generated by rotating region R 1 see diagram) about the line x =. Solution. The inner radius is R 1 = 1 x and the outer radius is R = 1. The extremities are x = and x = 1. The volume is π 1 1 x) ) dx ) x x dx x 1 ] 1 x 1 1 ) = π. Problem 5 Problem. Find the volume of the solid generated by rotating region R see diagram) about the line x =. Solution. The inner radius is R 1 = and the outer radius is R = x. The extremities are x = and x = 1. Because R 1 =, this is the disk method. The volume is 5. π x ) ) dx 1 5 x5 x dx ] 1 Problem 7 Problem. Find the volume of the solid generated by rotating region R see diagram) about the line x =. 1

11 Solution. Having worked problems 1 and 5, a quick way to get this volume is to subtract the other two volumes from the volume of all three regions rotated about the x-axis. That solid is a cylinder of radius 1 and height 1, so its volume is π. Then the volume of region R rotated must be V π π 5 = π 15. Let s use the washer method and see whether we get the same answer. The inner radius is R 1 = x and the outer radius is R = x. The extremities are x = and x = 1. The volume is Yep. = π 15. π x x ) ) dx x x ) dx 1 x 1 5 x5 ] 1 11