Find the volume of the solid formed by revolving the region bounded by y = x and y = √x about the line y = -2.

Answer 1

sqrt(x) = x

x = x^2

0 = x^2 - x

0 = x(x-1)

x=0 or x=1

so the limits of integrations are x=[0,1]

the master function is g(x) = [sqrt(x) - -2]-[x- -2]

= [sqrt(x) + 2] - [x + 2]

= sqrt(x) + 2 - x - 2

= sqrt(x) - x

g(x)^2 = (sqrt(x) - x) ^2 = x - 2*x*sqrt(x) +x^2

= x - 2*x^(3/2) + x^2

pi * integral ( x - 2*x^(3/2) + x^2) = pi * [(1/2)x^2 - (4/5)x^(5/2) + (1/3)x^3 ]

limit x=0 makes it totally vanish.....

for x =1, pi * ( 1/2 - 4/5 + 1/3) = pi * (15 - 24 + 10)/30 = pi/30

Answer 2

The area bounded by the 2 curves in the first quadrant is 1/6.

The y centroid (c) within the bounded area is 1/2

The curves meet at x=0 & x=1.

Therefore, the volume of rotation about the line y=-2 has a revolving radius of 1/2 + 2 (below the x axis)

2*pi*(c+2)*Area = 2*pi*(.5+2)*(1/6) = .8333pi = 2.62