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

Question

2111 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-11-12T20:17:18

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

enter image source here

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.

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