Asked by kchik
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.
(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.
Answers
Answered by
Steve
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
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