Determining the Volume of a Solid of Revolution
Key Questions

Let us look at the polar curve
#r=3sin theta# .The above is actually equivalent to the circle with radius
#3/2# , centered at#(0,3/2)# , whose equation is:#x^2+(y3/2)^2=(3/2)^2# by solving for
#y# , we have#y=pm sqrt{(3/2)^2x^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)^2x^2}+3/2)^2(sqrt{(3/2)^2x^2}+3/2)^2] dx# by simplifying the integrand,
#=6pi int_{3/2}^{3/2}sqrt{(3/2)^2x^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.
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