Volume of cylinder with two different radius and one height

Question

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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.

Answer 1

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

Answer 2

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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:

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

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\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}

Answer 3

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