How do I determine the volume of the solid obtained by revolving the curve #r=3sin(theta)# around the polar axis?

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Answer 1 · 2014-11-12T20:17:18

Let us look at the polar curve #r=3sin theta#.

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The above is actually equivalent to the circle with radius #3/2#, centered at #(0,3/2)#, whose equation is:

#x^2+(y-3/2)^2=(3/2)^2#

by solving for #y#, we have

#y=pm sqrt{(3/2)^2-x^2}+3/2#

By Washer Method, the volume of the solid of revolution can be found by

#V=pi int_{-3/2}^{3/2}[(sqrt{(3/2)^2-x^2}+3/2)^2-(-sqrt{(3/2)^2-x^2}+3/2)^2] dx#

by simplifying the integrand,

#=6pi int_{-3/2}^{3/2}sqrt{(3/2)^2-x^2} dx#

since the integral can be interpreted as the area of semicircle with radus #3/2#,

#=6pi cdot {pi(3/2)^2}/2={27pi^2}/4#

I hope that this was helpful.