I assume you mean only one period of the curve. If so, then using discs of thickness dy,
v = ∫[0,9π] πr^2 dy
where r = x = sin(y/9)
v = ∫[0,9π] π(sin(y/9))^2 dy = 9π^2/2
Using shells of thickness dx, and the symmetry of the region, we have
v = 2∫[0,1] 2πrh dx
where r = x and h = 9π/2 - y = 9π/2 - 9arcsin(x)
v = 2∫[0,1] 2πx(9π/2 - 9arcsin(x)) dx = 9π^2/2
integrate arcsin(x) using integration by parts
Find the volume of the solid obtained when the region under the curve
y = 9 arcsin(x), x ≥ 0,
is rotated about the y-axis. (Use the Table of Integrals.)
1 answer