Volume created by the revolution of a polar function around the initial line

[progressive]

Hello,

I was wondering if anyone could help me with deriving the volume created by the rotation of a polar equation around the initial line.

So, I thought about adding the surface area of cones (multiplied by $$d\theta$$) if each cone the triangle created with s-length of $$f(\theta)$$ and r-length of $$f(\theta)sin(\theta)$$.

Basically, this was entirely premised that cone shells could work exactly like cylindrical shells when trying to find volume - but I don't think it works.

The formula for the surface area of a cone (not including the circle at the bottom) is $$\pi*r*s$$.

http://math.about.com/od/formulas/ss/surfaceareavol_2.htm" is a site that has info on r and s.

So if I wanted to rotate the circle r=1 around the initial line but only from $$0<\theta<{\pi}$$, then the integral summing the cone surface areas could be expressed by:

$$\int_0^\pi \pi*r*s*d\theta =<br /> \int_0^\pi \pi f(\theta)*f(\theta)*sin(\theta) d\theta =<br /> \int_0^\pi \pi f(\theta)^2 sin (\theta) d\theta$$

I spent a lot of time on this problem, including looking at spherical sectors - circular sectors are needed to calculate the *area* of a polar graph.If anyone could help me with this derivation, it would help me significantly. Thanks so much for your help and time!

 

[progressive]

Hello again,

I forget to say that the integral I mention above does not calculate the correct volume e.g. for a circle r=1. It gives a volume of 2pi, when it should be 4/3 pi.



Thanks again for any help you can offer!