Determining the Volume of a Solid of Revolution

Key Questions

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

  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+(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`

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  I hope that this was helpful.

  Wataru · 1 · Nov 12 2014