Volume of the frustrum is incorrect if you derive it!!!!

**Drigomaniac** · December 6th 2012, 01:53 PM

So most likely I am the one with a mistake...but hell I can't get it to work.
I am attaching my work. So I try to derive the volume of a cylinder from the area of the circle (A(circle) = pi*r^2)....from h=0 to h=H.
Same with the cone...but here we know that the radius depends on the h so A(circle_for_the_cone)= A(h) = (R/H)*h, you derive it and gold.

So I did the same with the cone but changing the integral limits....and doom! Cant get it to work. Could somebody help me out using this methodology???

The volume of the frustrum should be: V (frustrum)= pi*h/3*(R1^2+R1*R2+R2^2)

Check attached picture!

Thanks!

**skeeter** · December 6th 2012, 04:51 PM

see attached sketch for my variable labels ...

equation of line in quad I is $y = \frac{r}{h} \cdot x$

volume of small cone ...

$V = \pi \int_0^h \left(\frac{r}{h} \cdot x\right)^2 \, dx$

$V = \frac{r^2 \pi}{h^2} \int_0^h x^2 \, dx$

$V = \frac{r^2 \pi}{h^2} \left[\frac{x^3}{3}\right]_0^h$

$V = \frac{r^2 \pi}{h^2} \cdot \frac{h^3}{3}$

$V = \frac{\pi r^2 h}{3}$

frustrum volume = volume of large cone - volume of small cone

$V = \frac{\pi R^2(H+h)}{3} - \frac{\pi r^2 h}{3}$

$V = \frac{\pi}{3} \left[R^2(H+h) - r^2 h \right]$

$V = \frac{\pi}{3} \left(R^2H +R^2h - r^2 h \right)$

$V = \frac{\pi}{3} \left[R^2H + h(R^2 - r^2) \right]$

using similar triangles, $h = \frac{rH}{R-r}$

$V = \frac{\pi}{3} \left[R^2H + \frac{rH}{R-r} \cdot (R^2 - r^2) \right]$

$V = \frac{\pi}{3} \left[R^2H + rH(R+r) \right]$

$V = \frac{\pi H}{3} \left(R^2 + Rr + r^2 \right)$

**Drigomaniac** · December 6th 2012, 10:23 PM

Tried the same approach but how do you get h to be in terms of the other variables...this might be a stupid question...but i will appreciate if you can help me.

h= \frac{rH}{R-r}

Also instead of doing large cone - small cone, could we get the solution deriving the area of the frustrum as I have it in the attached document?

Thanks!

**MarkFL** · December 6th 2012, 10:49 PM

If I were going to derive the volume of the frustum of a cone using slicing, I would let the volume of an arbitrary slice be:

$dV=\pi r^2\,dx$

With $r_1$ as the radius of the top of the frustum and $r_2$ as the radius of the base, we may let the radius of the slice at $x$ be:

$r=\frac{r_2-r_1}{h}x+r_1$

Hence:

$dV=\pi \left(\frac{r_2-r_1}{h}x+r_1 \right)^2\,dx$

And so the volume is:

$V=\pi\int_0^h \left(\frac{r_2-r_1}{h}x+r_1 \right)^2\,dx=$

$\pi\int_0^h \left(\frac{r_2-r_1}{h}\right)^2x^2+\frac{2r_1(r_2-r_1)}{h}x+r_1^2\,dx=$

$\pi\left[\left(\frac{r_2-r_1}{h}\right)^2\cdot\frac{x^3}{3}+\frac{2r_1(r_2-r_1)}{h}\cdot\frac{x^2}{2}+r_1^2x \right]_0^h=$

$\pi\left(\left(\frac{r_2-r_1}{h}\right)^2\cdot\frac{h^3}{3}+\frac{2r_1(r_2-r_1)}{h}\cdot\frac{h^2}{2}+r_1^2h \right)=$

$\frac{\pi h}{3}\left((r_2-r_1)^2+3r_1(r_2-r_1)+3r_1^2 \right)=$

$\frac{\pi h}{3}\left(r_2^2-2r_1r_2+r_1^2+3r_1r_2-3r_1^2+3r_1^2 \right)=$

$\frac{\pi h}{3}\left(r_2^2+r_1r_2+r_1^2 \right)$

**skeeter** · December 7th 2012, 03:10 AM

Originally Posted by Drigomaniac

Tried the same approach but how do you get h to be in terms of the other variables...this might be a stupid question...but i will appreciate if you can help me.

$h= \frac{rH}{R-r}$

similar triangles ... sides are proportional

$\frac{h}{r} = \frac{h+H}{R}$

solve for $h$

**Drigomaniac** · December 7th 2012, 08:53 AM

Great! Got it....thanks again for all your help!

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