using discs of thickness dx,
v = ∫[0,1] πr^2 dx
where r=y=22(x-x^2)
v = ∫[0,1] 22π(x-x^2)^2 dx = 11π/15
Find the volume of the solid obtained by revolving the region bounded by
y=22*x−22*(x^2) and the x-axis
around the x-axis.
2 answers
I misplaced the 22 in the integral. It should be
∫[0,1] π*(22(x-x^2))^2 dx = 242π/15
check: using shells of thickness dy, we have
x = 1/2 ± 1/(2√11) √(11-2y)
The vertex is at (1/2,11/2)
using symmetry, we can just double the volume under one side of the axis of symmetry, so
v = ∫[0,11/2] 2πrh dy
where r=y and h=1/2-x = 1/2-(1/2 - 1/(2√11) √(11-2y)) = √(11-2y)/(2√11)
v = 2∫[0,11/2] 2πy√(11-2y)/(2√11) dy
= 2π/√11 ∫[0,11/2] y√(11-2y) dy = 242π/15
∫[0,1] π*(22(x-x^2))^2 dx = 242π/15
check: using shells of thickness dy, we have
x = 1/2 ± 1/(2√11) √(11-2y)
The vertex is at (1/2,11/2)
using symmetry, we can just double the volume under one side of the axis of symmetry, so
v = ∫[0,11/2] 2πrh dy
where r=y and h=1/2-x = 1/2-(1/2 - 1/(2√11) √(11-2y)) = √(11-2y)/(2√11)
v = 2∫[0,11/2] 2πy√(11-2y)/(2√11) dy
= 2π/√11 ∫[0,11/2] y√(11-2y) dy = 242π/15