Let R be the region bounded by the x-axis and the graph of y=6x-x^2 Find the volume of the solid generated when R is revolved around the y-axis

Answer 1

using discs:

v = ∫[0,6]2πrh dx where
r = x
h = y = 6x-x^2

v = 2π∫[0,6]x(6x-x^2) dx
= 2π∫[0,6] 6x^2 - x^3 dx
= 2π(2x^3 - x^4/4)[0,6]
= 2π(432 - 324)
= 216π

Using discs:
v = ∫[0,9]π(R^2-r^2)dy
R = 3+√(9-y)
r = 3-√(9-y)
be careful with the algebra, and you again get 216pi, but it's a lot more work. Recall that

∫√(a^2-y^2) dy = 1/2 (x√(a^2-y^2) + a^2 arcsin(x/a))

Answer 2

oops. Top method is using shells, but as a clever calculus student, I'm sure you figured that out! ☻