Posted by **kchik** on Tuesday, December 11, 2012 at 12:54pm.

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.

- calculus -
**Steve**, Tuesday, December 11, 2012 at 2:34pm
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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