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

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Anonymous Anonymous · Answer 1 · 2016-10-13T10:20:31-04:00

hmm well, here an example y=3 ,

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

(9−18sinx+9sin2x) − (9−18cosx+9cos2x) (9−18sin⁡x+9sin2⁡x) − (9−18cos⁡x+9cos2⁡x)

= 18 (cosx−sinx) + 9 (sin2x−cos2x) = 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) − 92sin2x ]π/40π [ 18 (sin⁡x+cos⁡x) − 92sin⁡2x ]0π/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 ( 182–√ − 92 − 18 ) = π ( 182–√ − 452 ) or 9π2 ( 42–√ − 5 ) ,