Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.

Question

y = 8 sin x, y = 8 cos x, 0 ≤ x ≤ π/4;
about y = −1

Answer 1

using discs (washers)

v = ∫[0,π/4] π (R^2-r^2) dx
where R=1+8cosx and r=1+8sinx

v = π∫[0,π/4] (1+8cosx)^2 - (1+8sinx)^2 dx
= 16π (sinx+2sin2x+cosx) [0,π/4]
= 16π(1+√2)

Answer 2

A cross-section is a washer with an inner radius of 8sin(x) - (-1) and an outer radius of 8cos(x) - -(1), so its area would be:
A(x) = π[(8cos(x) + 1)^2 − (8sin(x) + 1)^2]
= π[64cos^2(x) + 16cos(x) + 1 - 64sin^2(x) − 16sin(x) − 1]
= π[64cos(2x) + 16cos(x) - 16sin(x)]
=> V(x) = ∫[0,π/4] π[64cos(2x) + 16cos(x) - 16sin(x)] dx
= π[32sin(2x) + 16sin(x) + 16cos(x)] |[0,π/4]
= π[32sin(π/2) + 16√2/2 + 16√2/2 - 16]
= π(32 - 16 + 16√2) = π(16 + 16√2)
The volume of the region is π(16 + 16√2).

Answer 3

lol please send help gave the correct answer 6 and 1/2 years after the question was asked