rotating about the x-axis ???
I will assume that.
intersection:
x^3 = 25x
x(x^2 - 5) = 0
x(x-5)(x+5) = 0
x = 0 or x = 5 , since x ≥ 0
Volume = π∫(625x^2 - x^6) dx
= π[ (625/3)x^3 - (1/7)x^7] from x =0 to 5
= π(78125/3 - 78125/7)
= π(312500/21)
Find the volume of the solid obtained by rotating the region enclosed by y=x^3, y=25x, x ≥ 0
1 answer