Calculus...Find the volume of the solid?

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y = 0, y = x(6 - x) about the axis x = 0

V = integral(low: a, up: b) A(x) dx

Answer 1

If you do the calculus, you will find that the region that we are interested in determining is given by the volume element:
dV = pi(9-y(x))^2 dx
How did I get the 9-y(x)?
y(x) = x(6-x)
y'(x) = 6-2x set= 0 so that x = 3, an extremum
y(3) = 9 the highest value y(x) reaches in the first quadrant
But the other area is the y-axis, so we need the area between the y-axis and y(x) to rotate to find the volume. So we need to consider the area 9 - y(x), the area between
What about the limits of the integral? Well, we go from 0, the origin, to x = 3 only because beyond that we are not calculating the region between the two curves y-axis and y(x).
What is the integral then?
V = integral|x=0 --> 3|dV = integral|0-->3|pi(9 - 6x + x^2)^2 dx

Answer 2

Obviously the region of integration is from x = 0 to x = 6, so we have
V = π∫y^2 dx
= π∫(0 to 6) (6x - x^2)^2 dx
= π∫(0 to 6) (36x^2 - 12x^3 + x^4) dx
= π [12x^3 - 3x^4 + x^5 / 5] [0 to 6]
= π (12.6^3 - 3.6^4 + 6^5 / 5 - 0)
= 1296π/5