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Math 2260 Written HW #3 Solutions 1. Consider the following region: 9 8 7 6 5 4 3 2 1 y x2 y2x 1 2 3 (a) What is the volume of the solid generated by revolving this region around the x-axis? Answer: Revolving the region around the x-axis yields the solid shown between the two surfaces shown below: 0.0 0.5 1.0 1.5 2.0 4 2 0 -2 4 2 0 -2 -4 -4 Since we’re given both curves as graphs of functions of x, it’s going to be most convenient to integrate in terms of x. Therefore, it’s natural to use the cross-section method with vertical cross-sections. Such a cross section is a washer with outer radius 2x and inner radius x2 , so the cross-sectional area, as a function of x, is A(x) = π(2x)2 − π(x2 )2 = 4πx2 − πx4 = π(4x2 − x4 ). 1 Therefore, the volume of the solid is Z 2 Z A(x) dx = 0 2 π(4x2 − x4 ) dx 0 2 4 3 x5 =π x − 3 5 0 4 3 25 2 − − (0 − 0) =π 3 5 32 32 =π − 3 5 64π = . 15 (b) What is the volume of the solid generated by revolving this region around the y-axis? Answer: Rotating the region around the y-axis yields the solid which lies between the two surfaces shown below: 4 -2 2 1 3 0 -1 2 1 0 -2 0 -1 1 2 It’s still desirable to integrate with respect to x, so now we’ll use the cylindrical shell method. In this case, a cylindrical shell with radius x has height 2x − x2 , so its surface area is A(x) = 2πx(2x − x2 ) = 2π(2x2 − x3 ). Therefore, the volume of the solid is Z 2 Z A(x) dx = 0 2 2π(2x2 − x3 ) dx 0 2 2 3 x4 = 2π x − 3 4 0 2 3 24 = 2π 2 − − (0 − 0) 3 4 16 = 2π −4 3 8π = . 3 2 2. Suppose an ant travels along the floor on a path which can be closely approximated by the 3 1 curve x = y3 + 4y from y = 1 to y = 3: 3 2 1 2 4 6 8 The ant keeps track of the distance he walks in an exercise journal. Assuming all lengths are in centimeters, what distance should the ant put down for this journey? Answer: To find the distance traveled by the ant, we simply integrate the arc length element from y = 1 to y = 3: Z 3s Z 3p dx 2 2 dx + dy = 1+ dy. Distance = dy 1 1 Now, dx 1 = y2 − 2 , dy 4y 2 dx 1 1 = y4 − + dy 2 16y 4 so and dx 2 1 1 1 2 4 2 1+ =y + + = y + 2 . dy 2 16y 4 4y Therefore, the distance traveled by the ant is s s 2 2 Z 3 Z 3 dx 1 1+ dy = y 2 + 2 dy dy 4y 1 1 Z 3 1 = y 2 + 2 dy 4y 1 3 3 y 1 − = 3 4y 1 3 3 1 1 1 = − − − 3 12 3 4 107 1 = − 12 12 106 = 12 53 = . 6 3