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

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

1 Answer

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Answer 1 · 2015-06-10T22:41:31

#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=pi*int_a^b f^2(x) dx#
Our function #f(x)=11sinx and a=0, b=pi#.
So the volume is:
#V=pi*int_0^pi 121sin^2(x) dx#
#V=121pi*int_0^pi sin^2(x) dx#
#V=121pi*[1/2(x-sin(x)cos(x))]_0^pi=121/2pi[x-sin(x)cos(x)]_0^pi#
#V=121/2pi(pi-0-0+0)=121/2pi^2~~597.111#.
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How do you find the volume of #y = 11sinx# , #y=0#, #[ 0,pi ]# revolving about the x axis?

1 Answer

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