Posted by kchik on Tuesday, December 11, 2012 at 12:54pm.
the curves intersect at (1,2) and (3,0)
A = ∫[1,3] (3-x) - (x^2-5x+6) dx
= -x^3/3 + 2x^2 - 3x [1,3]
= 4/3
the parabola dips below the x-axis on the interval (2,3), so we want to rotate between the curves on [1,2] and under the line on [2,3]:
V = ∫[1,2] pi (3-x)^2 - (x^2-5x+6)^2 dx
+ ∫[2,3] pi (3-x)^2
= 12/10 pi + 1/3 pi = 23/15 pi
for the pool, multiply each strip of area by its depth:
V = ∫[1,3] (x+2)((3-x) - (x^2-5x+6)) dx
= -x^4/4 + 2/3 x^3 - 5/2 x^2 - 6x [1,3]
= 16/3
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