Revolve the region bounded by y=x and y=x^2 about the y axis. In cubic units, the resulting volume is?

Answer 1

You can use discs, integrating along y:

V = Int(pi (R^2 - r^2) dy)[0,1]
where R = y and r = sqrt(y)
= pi*Int(y - y^2)dy[0,1]
= pi(1/2 y^2 - 1/3 y^3)[0,1]
= pi(1/2 - 1/3)
= pi/6

Or, you can use shells, integrating along x:

V = Int(2pi*r*h dx)[0,1]
where r = x h = x-x^2
= 2pi*Int(x(x-x^2) dx)[0,1]
= 2pi(x^2 - x^3 dx)[0,1]
= 2pi(1/3 x^3 - 1/4 x^4)[0,1]
= 2pi(1/3 - 1/4)
= 2pi(1/12)
= pi/6

Answer 2

Revolve the region bounded by y = 4x and y = x2 about the y-axis. In cubic units, the resulting volume is