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Math 166 Quiz 2 Name: Directions: This quiz is worth a total of 10 points. To receive full credit, all work must be shown. 1. Recall that the length L of a curve given by x = g(y) for a ≤ y ≤ b is Z b L= p 1 + g 0 (y)2 dy. a Z 2 r 1+ Use this to find a curve through the point (0, 1) whose length is given by the integral 1 1 dy. y4 Solution. Note that the curve x = g(y) whose length is the integral given above must satisfy the equation 1 g 0 (y)2 = 4 . y Taking the square root of both sides gives g 0 (y) = ± 1 . y2 Therefore, 1 g(y) = ∓ + C y for some constant C. Note that since the curve must pass through (0, 1), it must be that g(1) = 0. That is, 0 = g(1) = ∓1 + C which implies C = ±1. Therefore, there are two possible curves, g(y) = 1 −1 y 1 or g(y) = − + 1, y both with y values in the range 1 ≤ y ≤ 2 which work. √ 2. Find the area of the surface generated by revolving the curve y = x + 1, 1 ≤ x ≤ 5, about the x-axis. Solution. We use here the formula for surface area Z 5 p A= 2πf (x) 1 + f 0 (x)2 dx, 1 where f (x) = √ x + 1. To apply the formula, we first note that f 0 (x) = 1 (x + 1)−1/2 , 2 which implies 1 1 + f 0 (x)2 = 1 + (x + 1)−1 . 4 Therefore p √ f (x) 1 + f 0 (x)2 = x + 1 r 1 1 + (x + 1)−1 = 4 r 1 x+1+ = 4 Thus Z = = = = 5 r 5 dx 4 1 3/2 4π 5 5 x+ 3 4 1 " 3/2 # 3/2 25 9 4π − 3 4 4 4π 125 27 − 3 8 8 49π . 3 A = 2π x+ r 5 x+ . 4