4 l 1 2 9x 3 4 4 9x 0 2 3 dx 4 2 9x 3

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Unformatted text preview: e the integral. 4 L= = 1+ 2 9x 3 + 4 4 9x 0 2 3 dx = 4 2 9x 3 + 4 0 9x 2 3 4 dx = 0 2 9x 3 + 4 3x 1 3 2 dx = 4 0 1 −1 2 x 3 9x 3 + 4 dx 3 1 We now substitute using u = 9x 3 + 4 and du = 6x− 3 dx This gives 1 6 2 9(4 3 )+4 4 2 11 13 3 u 2 du = u2 3 18 2 1 Evaluating this gives: 27 94 2 3 +4 9(4 3 )+4 4 3 2 −8 2. (5 points) Find the area of the surface generated by rotating the curve y = x3 between x = 0 and x = 2 about the x-axis. b Recall that the basic formula for finding surface area in this situation is: SA = 2πf (x) 1 + [f ′ (x)]2 dx a 2 Here, f (x) = 3x , so we get: ′ SA = 2π 2 0 x3 1 + [3x2 ]2 dx = 2π 4 0 √ x3 1 + 9x4 dx We continue by using the substitution: u = 1 + 9x4 and du = 36x3 dx This gives us the integral: 2π 145 1 π 11 u 2 du = 36 18 145 1 π 2 3 u2 u du = 18 3 1 2 145 1 3 π 145 2 − 1 = 27 Extra Credit: Derive a formula for the surface area of a sphere of radius r. [Give your work on the back of the quiz or attach additional work] √ Hint: Think of a sphere as being generated by revolving the semicircle given by f (x) = r2 − x2 about the x-axis and use a surface integral. The algebra is a little messy, but it works out....
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This document was uploaded on 03/11/2014 for the course MATH 262 at Minnesota State University Moorhead .

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