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

User NigelDcruz
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1 Answer

5 votes
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
User Thafer Shahin
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