Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by: y = sin(x), y = 0, 0 ≤ x ≤ 𝜋; about y = −3

Question

I know it is integrating from 0 to pi, but the line that it is rotating around is confusing me a bit. The answer I came up with was integrating from 0 to pi of pi times (sin^2x+6sinx+9), which is wrong.

Answer 1

Answer 2

r=3+sinx

using disks,

dV=π(3+sinx)²dx, for volume as a whole

need to subtract cylinder dV_c=π3²= 9πdx

dV=π(6sinx+sin²x)dx for whole volume minus cylinder

Answer 3

You are close, but not exactly right. All you need to do is subtract the 9 from your parenthesis that is multiplied by pi.

The volume is best calculated by the washer method. The strip between y = 0 and y = sin x is revolved around the line y = - 3, creating an annulus (washer) of outside diameter 3 + sin x and inside diameter 3. Hence, its cross sectional area is (3 + sin x)² - 3² = ((sin x)² + 6 sin x) * pi.