How do you find the volume of $y = 11 sin x$ $y = 11sinx$ , $y = 0$ $y=0$ , $[0, π]$ $[ 0,pi ]$ revolving about the x axis?

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How do you find the volume of $y = 11 sin x$ $y = 11sinx$ , $y = 0$ $y=0$ , $[0, π]$ $[ 0,pi ]$ revolving about the x axis?

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Answer 1 · 2015-06-10T22:41:31

$V = \frac{121}{2} π^{2} \approx 597.111$ $V=121/2pi^2~~597.111$ .

Explanation:

This is the formula for calculating the volume of a solid obtained from a revolution around the x-axis:
$V = π \cdot \int_{a}^{b} f^{2} (x) d x$ $V=pi*int_a^b f^2(x) dx$
Our function $f (x) = 11 sin x and a = 0, b = π$ $f(x)=11sinx and a=0, b=pi$ .
So the volume is:
$V = π \cdot \int_{0}^{π} 121 {sin}^{2} (x) d x$ $V=pi*int_0^pi 121sin^2(x) dx$
$V = 121 π \cdot \int_{0}^{π} {sin}^{2} (x) d x$ $V=121pi*int_0^pi sin^2(x) dx$
$V = 121 π \cdot {[\frac{1}{2} (x - sin (x) cos (x))]}_{0}^{π} = \frac{121}{2} π {[x - sin (x) cos (x)]}_{0}^{π}$ $V=121pi*[1/2(x-sin(x)cos(x))]_0^pi=121/2pi[x-sin(x)cos(x)]_0^pi$
$V = \frac{121}{2} π (π - 0 - 0 + 0) = \frac{121}{2} π^{2} \approx 597.111$ $V=121/2pi(pi-0-0+0)=121/2pi^2~~597.111$ .
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How do you find the volume of $y = 11 sin x$ $y = 11sinx$ , $y = 0$ $y=0$ , $[0, π]$ $[ 0,pi ]$ revolving about the x axis?

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Explanation:

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How do you find the volume of y=11sinxy = 11sinx , y=0y=0, [0,π][ 0,pi ] revolving about the x axis?

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How do you find the volume of $y = 11 sin x$ $y = 11sinx$ , $y = 0$ $y=0$ , $[0, π]$ $[ 0,pi ]$ revolving about the x axis?