Volume of cylinder with two different radius and one height image
If I have top radius $R_1$,bottom radius $R_2$ (where $R_1>R_2$), total height $h$ and another height $p$.
Then how can I calculate the volume of lower part with height of $p$? 
I am confused whether it is cylinder? Image is attached above.
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Add and subtract a 'small cone' of height $\ds{h'}$. Then, you evaluate the difference of volume of the 'big' cone and the 'small' one:

$$
{h' \over R_{2}} =
{h \over R_{1} - R_{2}} \implies h' = {R_{2} \over R_{1} - R_{2}}\,h 
$$

\begin{align}
V & =
{1 \over 3}\,\pi R_{1}^{2}\pars{h + h'} -
{1 \over 3}\,\pi R_{2}^{2}h' =
{1 \over 3}\,\pi R_{1}^{2}\,{R_{1} \over R_{1} - R_{2}}h -
{1 \over 3}\,\pi R_{2}^{2}\,{R_{2} \over R_{1} - R_{2}}\,h
\\[5mm] & =\
\bbox[8px,border:1px groove navy]{%
{1 \over 3}\,\pi\pars{R_{1}^{2} + R_{1}R_{2} + R_{2}^{2}}h}
\end{align}
A: Hint. Consider the the volume of the solid as the "sum" of the volumes $\pi R^2(t)\,dt$ of the thin cylinders of radius $R(t)$ and height $dt$:
$$\mbox{Volume of lower part with height of $p$}=V_p=\int_{t=0}^p \pi R^2(t)\,dt$$
Here the function $R(t)$ is linear and goes from $R_2=R(0)$ to $R_1=R(h)$:
$$R(t)=R_2+\frac{(R_1-R_2)t}{h}.$$
It does not matter whether $R_1\geq R_2$ or $R_2\geq R_1$.
P.S. Finally you will find
$$V_p=\frac{\pi p(R_2^2+R(p) R_2+ R(p)^2)}{3}$$
where $\displaystyle R(p)=R_2+\frac{(R_1-R_2)p}{h}$.
For more details see: http://mathworld.wolfram.com/ConicalFrustum.html
A: I had the same problem its easier to do:
Volume of cylinder=h(pi)r^2
Volume of unequal = (h1(pi)r1^2 + h2(pi)r2^2/2
  Just take the average of both cylinders
