# How can I find the volume of a cone using integration?

I'm not given any numbers, just $h$ and $r$. I have to prove this volume to eventually find a centroid. Using the $xy$ plane as the base, extending upwards into the $z$ plane, the cone has a height $h$ and radius $r$.

I know that I will integrate a value of $z$ from $0$ to $h$. So I know that it will be $\int\pi r^2 dz$. I keep getting stuck on how to find $r$ in terms of $z$. I know I will use similar triangles, but when I'm not given any numbers at all, how do I compare them? How do I get $r$ in terms of $z$?

Sorry, I'm not sure how to use any code on here yet. A noooooooob.

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I cannot draw a reasonable picture, so some visualization and drawing is left to you.

Assume that the cone is "upside down," that is, point down, with point at $(0,0,0)$.

Take a horizontal slice through the cone, at height $z$. The cross-section of the cone at height $z$ has a certain radius, which we can call $r(z)$. Then the volume of the cone is $$\int_0^h \pi(r(z))^2\,dz.$$ As you saw, we want to find a formula for $r(z)$.

Take a vertical slice through the apex of the cone. The cross-section is an isosceles triangle, of height $h$, with the top length equal to $2r$. If we look at the part of the triangle that goes up to height $z$, that part has top length equal to $2r(z)$. The small triangle is similar to the full triangle, and therefore $$\frac{2r(z)}{2r}=\frac{z}{h}.$$ We conclude that $r(z)=\dfrac{rz}{h}$.

Our volume is therefore $$\int_0^h \pi \frac{r^2}{h^2}z^2\,dz.$$

Remark: If you do not want to think of the cone as being point down, let $r(z)$ be the radius at distance $z$ from the bottom, that is, at distance $h-z$ from the top. Basically the same argument as the one above shows that $$\frac{2r(z)}{r}=\frac{h-z}{h},$$ and therefore our volume is $$\int_0^h \pi \frac{r^2}{h^2}(h-z)^2\,dz.$$ The integral is just a little bit harder, not much, if we make the substitution $u=h-z$.

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Awesome, that is exactly what I needed. Finding that the volume is pi*r^2*h/3... – Joel Sep 20 '12 at 3:44
Good, I am happy that you managed to puzzle through my "thousand words" where a simple picture would do the job. – André Nicolas Sep 20 '12 at 3:49
Now I need to find the centroid.... That's the zbar = int(z~ * dV)/int(dV) right? z~ in this would be h/2 correct? – Joel Sep 20 '12 at 3:52
Say point down? (Can solve that way, then adjust.) Is on central axis, only need $z$-value. Find the moment, divide by volume. Moment is $\int_0^h z\pi(r(z))^2\,dz$. Evaluate, divide by volume. I get $(3/4)h$. If your cone is flat face down, it is $h/4$. Can't be $h/2$, has to be closer to the fat end of the cone. – André Nicolas Sep 20 '12 at 4:05
Ah... I see where that is. I got 3h/4 as well. But that is for the upside down. Yes....okay. Concept is building. I tried doing that a couple times. h/4 is the answer I'm looking for (my prof gives us partials). Thanks! I now have to add it to the centroid of a cylinder to find the centroid of the composite. The cylinder shouldn't be so bad, since it too does not have x or y values, just z. And we know the zbar for that is is in the middle. Then its just the weighted average of the two. – Joel Sep 20 '12 at 4:14