posted by kchik on .
The region A is bounded by the curve y=x^2-5x+6 and the line y = -x + 3.
(a) Sketch the line and the curve on the same set of axes.
(b) Find the area of A.
(c) The part of A above the x-axis is rotated through 360degree about the x-axis. Find the volume of the solid generated.
(d) The surface of a swimming pool takes the shape of A where the depth of the water at any point (x,y) in A is given by (x+2). Find the volume of the water in the swimming pool.
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]
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]