Question

Find the volume of the solid generated by rotating the region above 𝑦 = 12 and below 𝑦 = sin 𝑥 for 0≤𝑥≤𝜋 about the 𝑦-axis for 1 complete revolution.

Answer 1

the area ABOVE? y = 12???
That is infinite as is the area below y = sin x. There is no "between " I can see here.

Answer 2

Sorry typo. It should be 1/2

Answer 3

the curves intersect at 𝜋/6 and 5𝜋/6, so the volume, using shells, is

v = ∫[π/6,5π/6] 2πrh dx
where r=x and h=sinx - 1/2
v = ∫[π/6,5π/6] 2πx(sinx-1/2) dx = π^2(√3 - π/3)

using discs, it's tougher:
v = 2∫[1/2,1] π(R^2-r^2) dy
where R=π-arcsin(y) and r=arcsin(y)
v = ∫[1/2,1] π((π-arcsin(y))^2-(arcsin(y))^2) dy
= π^2(√3 - π/3)