volume = π[integral]y^2 dx from x = 0 to x = 3
y=x(3-x)
y^2 = x^2(3-x)^2
= x^2(9 - 6x + x^2)
= 9x^2 - 6x^3 + x^4
the integral of that is
3x^3 - (3/2)x^4 + (1/5)x^5
so volume
= π(81 - 243/2 + 243/5 - 0)
= (17/2)π
check my arithmetic
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=0, y=x(3-x) about the axis x = 0
1 answer