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lisov135 [29]
3 years ago
7

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

Mathematics
1 answer:
Dominik [7]3 years ago
8 0
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=\dfrac{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}
=\dfrac{9\pi^2}2
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Answer:

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<em>hope it helps</em>

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