# Volume of a Cone — Stuck On My Approach

I'd like to calculate the volume of a right circular cone via my way.

If I have a right-triangle with base $D$ and height $H$ then its area is $\frac{1}{2}HD$. Now if we imagine rotating this shape through space about the $z$ axis then we know it will trace out the cone I desire. Appropriate integration will give me this volume: $$\int \frac{1}{2}HD$$

The hard part, obviously, is filling in the details. So I suppose the best place to start is looking at $dx dy dz$, $$\int \int \int \frac{1}{2}HD dx dy dz$$ I'm imagining rotation about the $z$ axis such that $dz =0$ so we can simplify our integration a little: $$\int \int \frac{1}{2}HD dx dy$$ Now how about finding $dx$ and $dy$... If we imagine the following scenario:

then we can obviously approximate $dx = -y d\theta$ and $dy=xd\theta$ but in our case $y=H$ and $D=x$ such that $$\int_0^{2\pi}\int_0^{2\pi} \frac{1}{2}HD (-H d\theta) ( D d\theta)$$ $$\int_0^{2\pi}\int_0^{2\pi} -\frac{1}{2}H^2 D^2 d\theta d\theta$$ but this doesn't seem to work. Can you tell me where I went wrong and how I can fix this so that it will work how I intend?

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Revolving around $z$ doesn't mean that $dz=0$. You still have to add up vertical pieces. If you revolve the triangle, the problem is that various parts are at different distances from the axis. This is why people do shells or disks-the shells are all at the same distance and we know the answer for disks. You can certainly do a 3d integral, but it won't have anything to do with rotation. – Ross Millikan Dec 13 '12 at 20:55
@RossMillikan, Thank you for your comment. I'm a little confused by your comment that a 3d integral won't have to do with rotation. In the case of a finding a rectangle with dimensions $A,B,C$, don't we take $\int_0^C AB dC = ABC$ for the volume? Instead of adding the individual areas along $dc$ can't we just add the individual areas of a triangle around an angle $d\theta$ to get the cone volume? That's what I imagine when I say "rotation" – sciencenewbie Dec 14 '12 at 2:06
If you do a 3d integral the volume of interest is encoded in the limits. So if the radius is $r$ and the height is $h$ the integral is $\int_0^h \int_{-r}^r \int_{\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\; dy \; dx \; dz$ – Ross Millikan Dec 14 '12 at 3:39
In your integral the units aren't right-you have four lengths multiplied together. In polar coordinates you have $r \;d\theta\; dr$ which gives an area. – Ross Millikan Dec 14 '12 at 3:42
The problem in your logic here is your assumption that ${dz=0}$. If you rotate around the z axis ${z=2\pi r}$ so ${dz=2\pi dr}$ therefore, your change in the angle around the z axis is ${2\pi}$ NOT ${0}$ – Chris Jan 16 '14 at 14:22

Unless you are pursuing this multiple integral approach for a reason, why not just use the method of disks to compute the volume of the cone? (Video #28 here is a great resource.)

The triangle passes through the point $(0,H)$ and $(D,0)$; the line between these points is $y=-\frac{H}{D}x+H$, or equivalently, $x=-\frac{D}{H}y+D$ (this is the form we will need in a moment).

Then, using the method of disks, the volume of the cone is given by $$V=\int_0^H \pi\cdot(\text{radius})^2\,dy=\int_0^H \pi \left(-\frac{D}{H}y+D\right)^2\,dy=\frac{1}{3}\pi D^2H,$$ which is the well-known formula for the volume of a right circular cone of base radius $D$ and height $H$.

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Thank you for your comment. I've certainly seen this approach when I took calculus many years ago, but I was hoping to attack it this way as a personal learning experience. – sciencenewbie Dec 14 '12 at 2:09

Start with the observation that axes are arbitrary. Next we need to find the equation of the line in a cylinder with the height on the Z axis. $${Z=mX +H}$$ Where H is the Z intercept (the apex of the cone)

Change the X axis to ${\rho}$ then the slope of the line is $${(H/R)}$$ Where H is the apex of the cone and R is the radius of the base. (the limit of the base on ${\rho}$)

THen the equation of the slant is $${Z=(H/R)\rho + H}$$ IF we assume the slope is negative then the equation becomes $${Z=H-(H/R)\rho}$$

Now we can integrate $${\int_0^R\int_0^{2\pi}\int^H_{(H/R)p}(\rho )d\rho d\theta dz}$$ (don't forget the jacobian)

Integrating over z will give you the equation of the slant but the ${\rho}$ keeps you from dragging it out of the integral with respect to ${\rho}$ So now we have $${\int_0^R \int_0^{2\pi} \rho[H-(H/R)\rho]d\rho d\theta}$$ ... $${\int_0^R\int_0^{2\pi}H\rho-(H\rho^2)/R d\rho d\theta}$$.. $${\int_0^{2\pi} (HR^2)/2-(HR^3)/3R d\theta}$$... $${\int_0^{2\pi}[(3HR^2)/6-(2HR^2)/6 = \int_0^{2\pi}( HR^2/6) d\theta}$$... finally $${(2\pi HR^2)/6 = ( \pi R^2 H)/3}$$

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It may be easier to see ( at least it was for me) if you inscribe the cone in a unit- sphere since the height is equal to the radius, the triangle integrates over ${2\pi}$ much nicer.

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