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Calculus I Problem Set 11 Answers p 1. A solid is formed over the region in the first quadrant bounded by the curve y = 10 x so that the section by any plane perpendicular to the x-axis is a semicircle. What is the volume of this solid? Answer. We sweep out along the x-=axis. The section at x is a semicircle of radius y=2, so has area A(x) = (π =2)(y=2)2 = (π =8)(10 x). Thus V = 3.5 3 2.5 2 1.5 1 0.5 0 p = π 8 Z 10 (10 x)dx = 0 π 10x 8 x2 10 25π = 2 0 4 y= : p 10 x dx 2 0 4 6 8 10 p 2. A solid is formed over the region in the first quadrant bounded by the curve y = 4 by any plane perpendicular to the x-axis is a square. What is the volume of this solid? Answer. The section at x has area y 2 = 4 x so that the section x, so Z V 4 = (4 x)dx = 8 : 0 y= 2 1.5 = p = p 4 x dx 1 = 0.5 0 p 0 0.5 1 1.5 2 2.5 3 3.5 p 4 3. A solid is formed over the region in the first quadrant bounded by the curve y = 2x x 2 so that the section by any plane perpendicular to the x-axis is a semicircle. What is the volume of this solid? Answer. As in problem 1, π y 2 π π 2 2 2 ( ) = (2x x ) dx = (4x 4x3 + x4 )dx : 2 2 8 8 Integrating dV from 0 to 2, we get π 32 32 2π V= ( 16 + ) = : 8 3 5 15 dV = y = 2x 1 p = = 0.8 0.6 p = 0.4 0.2 = 0 0 0.5 1 1.5 2 x2 p = p dx 4. The region in the first quadrant bounded by y = Find the volume of the resulting solid. p2 x 1; y = 0; x = 1; x = 4 is revolved around the x-axis. Answer. Here we find that at a typical x between 1 and 4, dV V = 18π . = = π (x 2 y = 2x p p = π r 2 dx = 1)dx. Integrating, we get x2 4 3 = 2 = 1 0 = 0 0.5 1 1.5 2 2.5 3 3.5 4 p = = p dx 5. Find the volume of the solid obtained by rotating about the y-axis the region bounded by y = x 2 ; x = 2 and the x-axis. Answer. Here we will use the washer method, sweeping out along the y-axis, with y ranging from 0 to 4. At p a typical y, dV = (π R 2 π r2 )dy, and R = 2; r = y. Thus the volume is V = =π Z 4 y)dy = 8π (4 0 : p = p 7 6 5 = = = = 4 3 2 1 0 0 p = dy p y = x2 = 0.5 1 1.5 2 = 2.5 6. The region in the first quadrant under the curve y = 2x the resulting solid. x 2 is rotated about the y-axis. Find the volume of Answer. Here we sweep out along the x-axis from x = 0 to x = 2, using the shell method. At a typical x, dV = 2π xydx = 2π (2x 2 x3 )dx, and V = = 2π Z 2 (2x x3 )dx = 2π ( 2 0 2x3 3 x4 2 8π ) = 4 0 3 : p = p = = = p 1 = 0.8 = 0.6 = p dx p 2x = 0.4 0.2 0 p = 0 0.5 1 1.5 2 x2 7. The region in the first quadrant bounded by y = x 4 and x = 1 is revolved around the y-axis. Find the volume of the resulting solid. p Answer. Here let’s use the shell method, sweeping out along the x-axis (compare this with problem 5). = dV = 2π xydx = 2π x 5 dx. Integrating from 0 to 1, we get Vp= π =53. = p = p = = = = 2.5 p = = 2 y = x4 = 1.5 p 1 = = x=1 0.5 0 0 0.2 0.4 0.6 0.8 dx 1 x 2 and y = x 8. The region in the first quadrant bounded by y = x the volume of the resulting solid. Answer. Using the washer method, at a typical x between 0 and 1, dV p x2=)2 ]dx. After some algebra, we obtain = p Z V 1 = (2x 0 = 3 p = = π( 3x4 + x6 )dx = p = (π R 2 π r2 )dx = π [(x 3 1 + ) = :0428π 5 7 x 3 )2 : = = = p 0.3 y=x = 0.4 = = = = p = 1 2 x3 is revolved around the x-axis. Find y=x 0.2 x2 = = 0.1 0 0 0.2 0.4 0.6 0.8 1 dx 9. The average value of a function y = f (x) defined over an interval [a; b] is defined to be yave = Z 1 b a b f (x)dx : a x3 (x Find the average of y = sin x over the interval [0; π ]. yave = 1 π Z π sin xdx 0 = 1 π ( cos x)j0 π = 2 : π 10. Let g(x) = x2 + x3 for x in the interval 0 x 10. Find the average, or mean, value of g on the interval. Find the average slope of the graph of y = g(x) on the interval. Answer. The average value of the function is 1 10 Z 10 2 3 (x + x )dx = 0 1 x3 10 3 + x4 10 100 1000 = + 4 0 3 4 = 283:33 Since the slope is y0 = g0 (x)m the average slope is 1 10 Z 10 g0 (x)dx = 0 1 (g(10) 10 g(0)) = 110 : :