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UT M 408C - Areas and Volumes

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M408C: Areas and VolumeNovember 4, 2008The area between the curves y = f(x) and y = g(x) and b etween x = a and x = b isAn analogous formula holds for the area between the curves y = c and y = d and x = f (y) andx = g(y):For a solid S lying b e tween x = a and x = b with cross-sectional area function A(x) (A continuous),the volume of S isAgain, an analogous formula holds for the volume of a solid S lying between y = c and y = d withcross-sectional area function A(y) (A continuous). In this case,1.2.13. (6.1.50) Find the number a such that the line x = a bisects the area under the curve y = 1/x2for 1 ≤ x ≤ 4. Also, find the number b such that the line y = b bisects the same area.Solution: We want our line x = a to split our areas. That is, we want an a so thatRa11x2dx =R4a1x2dx. Taking the integrals, we find 1 −1a=1a−14. Solving for a, we find that54=2a, i.e. a =85.For the second part, we want to split the areas using a horizontal line y = b. A horizontal linewill give us a wedge above the line and a region below that includes a rectangle sitting below theline y = 1/16. The area of the lower region is 3 ·116+Rb1/161√y− 1dy. The first term comesfrom the rectangle of length 3 between 1 and 4 and height 1/16. The second terms comes fromdetermining the area between the curve x = 1/√y and x = 1. T he upper region has area given byR1b1√y− 1dy. We set these two equal to each other and integrate to get316+h2y12− yib1/16=h2y12− yi1b=⇒316+ 2√b − b − 2 ·14+116= 2 − 1 − 2√b + b.We s implify this to get 2b −4√b+54= 0. To solve for b, we set c2= b and consider 2c2−4c +54= 0.We solve for c using the quadratic formula to get c = 1 ±√64. Now√b < 1 since b < 1. Thusc =√b < 1, thus c = 1 −√64. Squaring this we get b = c2= 1 −√62+616≈ 0.150.4. (6.2.14) Find the volume obtained by rotating the region bounded by y = x, y = 0, x = 2 andx = 4 about the line x = 1.Solution: The solid looks like the space between a funnel and a cylinder. We split this solidinto two different regions: the lower one consists of a cylinder with a hole and the upper regionis a triangle rotated around the line x = 1. The lower region has cross-sectional area AL(y) =π(92− 12) = 8π and the upper region has cross-sectional area AU(y) = π(9 − (y − 1)2). Then wefindV =Z20AL(y) dy +Z42AU(y) dy =Z208π dy +Z42π(9 − (y − 1)2) dy=8πy20+ πZ42(8 + 2y − y2) dy = 16π + π8y + y2−13y342= 16π + π28 −563=763π5. Suppose 2 ≤ f(x) ≤ 4 for all x. Which of the following is false:(A)Z42f(x) dx ≥ 0. (B) The antiderivative of f is always positive. (C) 10 ≤Z50f(x) dx ≤ 20.(D) The antiderivative of f is always negative. (E)Z10f(x) dx ≥ 2Solution: (A) is true since f(x) ≥ 2 for all x. (C) is true since 2 ≤ f(x) ≤ 4 implies that2(5 −0) ≤R50f(x) dx ≤ 4(5 −0). (E) is true sinceR10f(x) dx ≥R102 dx = 2. For (B) and (D), notethat an antiderivative of f is g(x) =Rxaf(x) dx by the Fundamental Theorem of Calculus. Sincef(x) > 0, g(x) > 0 for any x. Thus (B) is true while (D) is false. Hence the answer is


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