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Find the volume V obtained by rotating the region bounded by the curves about the given axis.
y = 3 sin2(x), y = 0, 0 ≤ x ≤ π; about the x−axis

Respuesta :

wizard123
Use Disc-method here:
[tex]V = \int A(x) dx = \int \pi r^2 dx [/tex]

The radius of each circle cross-section is equal to the y-value of function.
The limits are from 0 to pi.

[tex]V = \pi \int_0^\pi(3sin^2 x)^2 dx = 9\pi \int_0^\pi sin^4 x dx[/tex]

Integrate by using trig identities to transform sin^4.
[tex]sin^2 x = \frac{1 - cos(2x)}{2} \\ \\ sin^4 x = \frac{(1-cos(2x))^2}{4} = \frac{1 - 2cos(2x)+cos^2 (2x)}{4} \\ \\ :: cos^2 x = \frac{1+cos(2x)}{2} \\ \\ sin^4 x = \frac{3-4cos(2x)+cos(4x)}{8}[/tex]

Sub back into integral:
[tex]V = \frac{9\pi}{8} \int_0^\pi (3-4cos(2x)+cos(4x)) dx \\ \\ V = \frac{9\pi}{8} |_0^\pi 3x -2sin(2x)+\frac{1}{4}sin(4x) \\ \\ V = \frac{9\pi}{8}[3\pi - 0] \\ \\ V = \frac{27 \pi^2}{8}[/tex] 
okpalawalter8

We have that for the Question, it can be said that The volume V obtained by rotating the region bounded by the curves about the given axis is

[tex]V=\frac{27 \pi^2 }{8}[/tex]

From the question we are told

Find the volume V obtained by rotating the region bounded by the curves about the given axis.

y = 3 sin2(x), y = 0, 0 ≤ x ≤ π; about the x−axis

Generally the  volume of solid about x axis

[tex]V_x=\int _0^{\pi} y^2 dx\\\\\V_x=9 \pi \int -0^{\pi} sin^4x dx\\\\Where\\\\Cos2x=1-2sin^2x[/tex]

[tex]Therefore\\\\sin^2x=\frac[1-cos2x}{2}\\\\sin^4x=1/4 (1^2+cos^2x-2cos2x)\\\\sin^4x=1/8 (3+cos^4x-4cos2x)[/tex]

Hence

[tex]\int_0^{\pi} sin^4dx=\int_0^{\pi}1/8(3+cos^4x-4cos2x))dx\\\\\int_0^{\pi} sin^4dx=1/8(3 \pi +0-0-0)\\\\\int_0^{\pi} sin^4dx=\frac{3 \pi}{8}[/tex]

Therefore

[tex]Volume is given as\\\\V=9 \pi int_0^{\pi} sin^4x dx\\\\V=9 \pi *\frac{3 \pi}{8}\\\\[/tex]

Therefore

The volume V obtained by rotating the region bounded by the curves about the given axis is

[tex]V=\frac{27 \pi^2 }{8}[/tex]

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https://brainly.com/question/19007362

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