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Math 21B 1 (20 pts.) Brian Osserman Practice Exam 2 The region enclosed by the curves y = x1/2 , y = x1/4 , 0≤x≤1 is revolved about the y -axis. Find the volume of the resulting solid. Solution: We use cylindrical shells. For 0 ≤ x ≤ 1 we have x1/4 ≥ x1/2 , so the height of a shell is given by x1/4 − x1/2 . Since we revolve about the y -axis, the radius is x, so the volume is Z 1 Z 1 1/4 1/2 2π x(x − x )dx = 2π (x5/4 − x3/2 )dx 0 0 1 ! 4 9/4 2 5/2 = 2π x − x 9 5 0 4 2 = 2π − −0 9 5 4 = π. 45 2 (20 pts.) The parametric curve 2 x = t3/2 , 3 √ y = 2 t, 0≤t≤ √ 3 is revolved about the y -axis. Find the area of the resulting surface. Solution: The area is √ Z s 3 x 2π 0 dx 2 dy 2 + dt = 2π dt dt √ Z 0 √ 3 2 3/2 t 3 q (t1/2 )2 + (t−1/2 )2 dt 2 3/2 √ t t + t−1 dt 3 0 Z √3 √ 4 = π t4 + t2 dt 3 0 Z √3 √ 4 = π t t2 + 1dt. 3 0 Z = 2π 3 Using u = t2 + 1, du = 2tdt, this gives 2 π 3 Z 4 √ 1 2 π udu = 3 2 3/2 u 3 4 ! 1 2 2 2 = π ·8− ·1 3 3 3 14 2 π· = 3 3 28 = π. 9 3 (15 pts.) If a force of 90 N stretches a spring 1 m beyond its natural length, how much work does it take to stretch the spring 5 m beyond its natural length? Solution: If the force constant for the spring is k N/m, we have k · 1 = 90, so k = 90. Then the work to stretch the spring 5 m is (in Newtons): Z 5 90xdx = 45x2 5 0 0 = 45 · 25 − 0 = 1125. 4 (20 pts.) The shape of a water tank is determined by revolving the curve y = x2 , 0 ≤ x ≤ 4 about the y -axis. The container is lled with water. What is the work required to pump all the water out over the top edge of the tank? (the weight-density of water is 10, 000 N/m3 , and assume all distance units are in meters). Solution: If we take a horizontal slice of water at height y , with thickness dy , we get a disk of radius x = √ y . Since x goes from 0 to 4, we see that y goes from 0 to 16, so the tank has height 16, and we need to lift this slice a distance of 16 − y m. The weight of the slice in Newtons is √ 10, 000 · π( y)2 dy = 10, 000πydy. Thus, the work to pump out the slice is 10, 000πy(16 − y)dy N·m. To nd the total work to pump the water out, we have to integrate over all slices, 2 so we get (in N·m): 16 Z 1 8y 2 − y 3 3 10, 000πy(16 − y)dy = 10, 000π 0 16 ! 0 4096 = 10, 000π 2048 − −0 3 20, 480, 000 = π. 3 5 (15 pts.) A cubical tank with sides 3 m long is lled with water to a depth of 2 m. Find the total force of the water on one side of the tank (the weight-density of water is 10, 000 N/m3 ). Solution: At a depth of h meters, the pressure of the water on the side is 10, 000h N/m2 . The length of a side is 3 m, so (in Newtons) the total force is Z 2 Z 10, 000h · 3dh = 30, 000 0 2 hdh 0 = 30, 000 1 2 h 2 2 ! 0 = 30, 000(2 − 0) = 60, 000. 6 (10 pts.) Evaluate the following integral Z (csc x − sec x)(sin x + cos x)dx. Solution: Z (csc x − sec x)(sin x + cos x)dx Z (csc x sin x − sec x cos x + csc x cos x − sec x sin x)dx = Z = (1 − 1 + cot x − tan x)dx = ln | sin x| + ln | cos x| + C. 3