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Calculus II                                                          Quiz #4                                             March 16, 2004

Name____________________                   R.  Hammack                                                 Score ______
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(1)  Consider the region contained between the graphs of y = (3 - x)^(1/2) ,  y = 0, x = -1,  and x = 3.
This region is revolved around the x-axis.  Find the volume of the resulting solid.


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The cross-section at x is a circle of radius A(x) = π((3 - x)^(1/2))^2 = π(3 - x).
Using the volume-by-slicing formula, the volume is
∫_ (-1)^3π(3 - x) dx = π∫_ (-1)^3 (3 - x) dx = π[3x - x^2/2] _ (-1)^3 =

π((3 (3) - 3^2/2) - (3 (-1) - (-1)^2/2)) = π(9 - 9/2 + 3 + 1/2) = 8π cubic units



(2)  Find the length of the curve y = 2/3x^(1/2)^3 + 1  between x = 0 and x = 1.
f(x) = 2/3x^(3/2) + 1

f ' (x) = x^(1/2) = x^(1/2)

L = ∫_0^1 (1 + (f ' (x))^2)^(1/2) dx = ∫_0^1 (1 + x^(1/2)^2)^(1/2) dx = ∫_0^1 (1 + x)^(1/2) dx =[2/3 ((1 + x)^(1/2))^3] _0^1
2/3[((1 + 1)^(1/2))^3 - ((1 + 0)^(1/2))^3] = 2/3[2^(1/2)^3 - 1] = 2/3[22^(1/2) - 1] = (42^(1/2) - 2)/3units