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

y = 6 sin x, y = 6 cos x, 0 ≤ x ≤ π/4; about y = - 1

Hmm well, here an example y=3,

y=3, rather than the x - x - axis.) Your integrand looks fine and reduces to

(9-18sinx+9 sin2 x) - (9-18cosx+9 cos2 x) (9-18sin⁡x+9 sin2 ⁡x) - (9-18cos⁡x+9 cos2 ⁡x)

= 18 (cosx-sinx) + 9 (sin2 x - cos2 x) = 18 (cosx-sinx) - 9 cos2x. = 18 (cos⁡x-sin⁡x) + 9 (sin2 ⁡x - cos2 ⁡x) = 18 (cos⁡x-sin⁡x) - 9 cos⁡2x.

The evaluation of the volume is then

π [ 18 (sinx+cosx) - 92 sin2x ] π/4 0 π [ 18 (sin⁡x+cos⁡x) - 92 sin⁡2x ] 0 π/4

= π ([ 18 (2-√ 2 + 2-√ 2) - 92 ⋅1 ] - [ 18 (0+1) - 92 ⋅0 ]) = π ([ 18 (22 + 22) - 92 ⋅1 ] - [ 18 (0+1) - 92 ⋅0 ])

= π (18 2-√ - 92 - 18) = π (18 2-√ - 452) or 9π 2 (4 2-√ - 5),