(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let R be the region enclosed by the curves y=6sin^{2}x and y=6 over the segment [ π/2,3π/2]. Find the volume of the solid that is obtained by rotating R about the line y=−1.1.

2. Relevant equations

[tex]

$\displaystyle \Large pi * \int _a^c x((r+yt)^2 - (r+yb)^2)dx$

[/tex]

3. The attempt at a solution

[tex] $\displaystyle \Large pi * \int _{pi/2}^{3pi/2} x((-1.1+6)^2 - (-1.1+6sin^2x)^2)dx$ [/tex]

==> [tex] $\displaystyle \Large pi * \int _{pi/2}^{3pi/2} 24.01x - (-1.1+6sin^2x)^2 xdx$ [/tex]

==> [tex] $\displaystyle \Large pi [* \int _{pi/2}^{3pi/2} 24.01x dx - \int _{pi/2}^{3pi/2} (-1.1+6sin^2x)^2 x dx ]$ [/tex]

==> [tex] $\displaystyle \Large pi [* (236.97 + C) - \int _{pi/2}^{3pi/2} (-1.1+6sin^2x)^2 x dx ]$ [/tex]

==> [tex] $\displaystyle \Large pi [* (236.97 + C) - \int _{pi/2}^{3pi/2} 36xsin^4x - 13.2xsin^2x + 1.21x dx ]$ [/tex]

==> [tex] $\displaystyle \Large pi [* (236.97 + C) - (36\int _{pi/2}^{3pi/2} xsin^4xdx - 13.2\int _{pi/2}^{3pi/2}xsin^2xdx + 1.21\int _{pi/2}^{3pi/2}xdx)]$ [/tex]

is this right so far?

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# Homework Help: Trigonometric Inegrals: Volume help

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