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

Nov 12, 2014

Let us look at the polar curve $r = 3 \sin \theta$. The above is actually equivalent to the circle with radius $\frac{3}{2}$, centered at $\left(0 , \frac{3}{2}\right)$, whose equation is:

${x}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {\left(\frac{3}{2}\right)}^{2}$

by solving for $y$, we have

$y = \pm \sqrt{{\left(\frac{3}{2}\right)}^{2} - {x}^{2}} + \frac{3}{2}$

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

$V = \pi {\int}_{- \frac{3}{2}}^{\frac{3}{2}} \left[{\left(\sqrt{{\left(\frac{3}{2}\right)}^{2} - {x}^{2}} + \frac{3}{2}\right)}^{2} - {\left(- \sqrt{{\left(\frac{3}{2}\right)}^{2} - {x}^{2}} + \frac{3}{2}\right)}^{2}\right] \mathrm{dx}$

by simplifying the integrand,

$= 6 \pi {\int}_{- \frac{3}{2}}^{\frac{3}{2}} \sqrt{{\left(\frac{3}{2}\right)}^{2} - {x}^{2}} \mathrm{dx}$

since the integral can be interpreted as the area of semicircle with radus $\frac{3}{2}$,

$= 6 \pi \cdot \frac{\pi {\left(\frac{3}{2}\right)}^{2}}{2} = \frac{27 {\pi}^{2}}{4}$

I hope that this was helpful.