Question

Y = 3 sin2x, y = 0, 0 ≤ x ≤ π; about the x−axis

Answer 1

I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.

Via the disk method,


`\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx`

Recall the half-angle identity for sine:


`\sin^2t=\frac{1-\cos2t}2`

`\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx`

`=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_(x=0)^(x=\pi)`

`=\frac{9\pi^2}2`