Okay, so I'm currently working on a strange proof for the volume of a cone and could use some guidance with a difficult integral. I've done many other easier proofs involving rotational solids and triple integrals so please don't just tell me to do it that way, I already have.

I believe the following integral should give the volume of a cone with radius $r$ and height $h$:



The first trouble I had was solving the density integral $\delta$. I simplified it to the following and tried using a typical trig sub with tangent.


After a while I managed to solve the integral using integration by parts:

$$\delta(y)=2h\sqrt{r^2-y^2}-\frac{2h}{r}y^2\left[\frac{\sec\theta\tan\theta+\color{red}{\ln|\sec x+\tan x|}}{2}\right]_0^{\arctan\left(\frac{\sqrt{r^2-y^2}}{y}\right)}$$


However, upon plugging into the initial volume integral, I got the following:

$$V(r,h)=2h\int_0^r\sqrt{r^2-y^2}\;dy-\color{red}{\frac{2h}{r}\int_0^ry^2\ln(r+\sqrt{r^2-y^2})\;dy}-\frac{2h}{r}\int_0^ry^2\ln y\;dy$$

The first integral can be solved fairly easily, and the third can be done with integration by parts and a little work. However, the second appears quite daunting and I don't know that it can be solved using elementary techniques. I've considered trying Laisant's inverse method, converting to polar, or using some trig identity a few steps back to alter this particular integrand but just don't know if I'd get anywhere.

The other thing I've considered is using a hyperbolic trig sub, but I'm less familiar with these and don't know how I'd reapply bounds to even solve the density integral, much less the remaining integral with respect to $y$. I've gotten a little past here but it seems to simply dead-end.


Any advice for proceeding (i.e. integrating that second log integral above or doing the hyperbolic one below)? I'd really appreciate it and may assign a bounty if it proves excessively daunting. I'd prefer the solution not be posted but if it necessitates non-elementary techniques feel free to do so as you've earned the right to post imo.

By the way if you do solve the problem and post it here, I am planning on putting this proof in a collection I am writing and would like to cite your name with your work if/when I submit it for jobs and other things.

Here's an outline of the final workings:

\noindent We consider a cone of height $h$ and radius $r$ formed by the following equation:

$$z(x,y)=h-\frac{h\sqrt{x^2+y^2}}{r}$$ \begin{center} \small{*Note the transformations placing the cone upright on the xy-plane} \end{center}

\noindent At some point $y=c$, we assign a density $\delta$ corresponding to the area of the hyperbola formed by the intersection of $z$, $y=c$, and the xy-plane.

$$\delta(y)=2h\int_0^{\sqrt{r^2-y^2}}\left[1-\frac{\sqrt{x^2+y^2}}{r}\right]\;dx$$ \* \noindent The volume of the cone is given by the following integral:

$$V(r,h)=2\int_0^r\delta(y)\;dy$$ \* $$V(r,h)=2\int_0^r2h\int_0^{\sqrt{r^2-y^2}}\left[1-\frac{\sqrt{x^2+y^2}}{r}\right]\;dx\;dy$$ \noindent First, let's integrate $\delta$ by parts:

$$\delta(y)=2h\sqrt{r^2-y^2}-\frac{2h}{r}\int_0^{\sqrt{r^2-y^2}}\sqrt{x^2+y^2}\;dx$$ \* \noindent Substituting $x=y\tan\theta$:

$$\delta(y)=2h\sqrt{r^2-y^2}-\frac{2h}{r}y^2\int_0^{\arctan\left(\frac{\sqrt{r^2-y^2}}{y}\right)}\sqrt{y^2(1+\tan^2\theta)}\cdot y\sec^2\theta\;d\theta$$ \* Simplifying:

$$\delta(y)=2h\sqrt{r^2-y^2}-\frac{2h}{r}y^2\int_0^{\arctan\left(\frac{\sqrt{r^2-y^2}}{y}\right)}\sec^3\theta\;d\theta$$ \* \noindent Integrating with respect to $\theta$: $$\delta(y)=2h\sqrt{r^2-y^2}-\frac{2h}{r}y^2\left[\frac{\sec\theta\tan\theta+\ln|\sec x+\tan x|}{2}\right]_0^{\arctan\left(\frac{\sqrt{r^2-y^2}}{y}\right)}$$ \* \noindent Applying bounds and simplifying: $$\delta(y)=2h\sqrt{r^2-y^2}-\frac{h}{r}y^2\left[\frac{r}{y}\cdot\frac{\sqrt{r^2-y^2}}{y}+\text{arcsech}\left(\frac{h}{r}\right)\right]$$ \* \noindent Simplifying: $$\delta(y)=2h\sqrt{r^2-y^2}-h\sqrt{r^2-y^2}-\frac{h}{r}y^2\text{arcsech}\left(\frac{h}{r}\right)$$

$$\delta(y)=h\sqrt{r^2-y^2}-\frac{h}{r}y^2\text{arcsech}\left(\frac{h}{r}\right)$$ \noindent Substituting $\delta$ into $V(r,h)$:


\noindent Breaking up:

$$V(r,h)=2h\int_0^r\sqrt{r^2-y^2}\;dy-\frac{2h}{r}\int_0^ry^2\text{arcsech}\left(\frac{h}{r}\right)\;dy$$ \* \noindent Substituting $y=r\sin\phi$ in the first integral: $$V(r,h)=2hr^2\int_0^{\pi/2}\cos^2\phi\;d\phi-\frac{2h}{r}\int_0^ry^2\text{arcsech}\left(\frac{h}{r}\right)\;dy$$

\noindent Using double angle formula:


\noindent Integrating with respect to $\phi$:

$$V(r,h)=hr^2\left[\phi-\frac{\sin 2\phi}{2}\right]_0^{\pi/2}-\frac{2h}{r}\int_0^ry^2\text{arcsech}\left(\frac{h}{r}\right)\;dy$$

\noindent Simplifying:

$$V(r,h)=\frac{\pi r^2 h}{2}-\frac{2h}{r}\int_0^ry^2\text{arcsech}\left(\frac{h}{r}\right)\;dy$$

\noindent Integrating by parts:

$$V(r,h)=\frac{\pi r^2 h}{2}-\frac{2h}{r}\left(\left[ \frac{y^3}{3}\text{arcsech}\left(\frac{y}{r}\right)\right]_0^r+\frac{2h}{3} \int_0^r \frac{y^2}{\sqrt{r^2-y^2}} \;dy\right)$$

\noindent Simplifying:

$$V(r,h)=\frac{\pi r^2h}{2}-\frac{2h}{3}\int_0^r\frac{y^2}{\sqrt{r^2-y^2}}\;dy$$

\noindent Substituting $y=r\sin\alpha$:

$$V(r,h)=\frac{\pi r^2h}{2}-\frac{2r^2h}{3}\int_0^{\pi/2}\sin^2\alpha\;d\alpha$$

\noindent Using double angle formula:

$$V(r,h)=\frac{\pi r^2h}{2}-\frac{r^2h}{3}\int_0^{\pi/2}1-\cos2\alpha\;d\alpha$$

\noindent Integrating with respect to $\alpha$:

$$V(r,h)=\frac{\pi r^2h}{2}-\frac{r^2h}{3}\left[\alpha-\frac{\sin 2\alpha}{2}\right]_0^{\pi/2}$$

\noindent Applying bounds:

$$V(r,h)=\frac{\pi r^2h}{2}-\frac{r^2h}{3}\cdot\frac{\pi}{2}$$

\noindent Simplifying:

$$V(r,h)=\frac{\pi r^2h}{2}-\frac{\pi r^2h}{6}$$

$$V(r,h)=\frac{1}{3}\pi r^2h$$

  • 1
    $\begingroup$ By parts, $$\int y^2 \log\left(r + \sqrt{r^2 - y^2}\right) \ dy = \frac{y^3}{3} \log\left(r + \sqrt{r^2 - y^2}\right) - \int \text{Something} \ dy$$ The $\text{Something}$ should be doable, since you can effectively get rid of the $\log$ post-differentiation. $\endgroup$ Jul 13 '17 at 18:38

HINT: I think there is a sign error in your post. I found that $$V(r,h)=2h\int_0^r\sqrt{r^2-y^2}\;dy-\frac{2h}{r} \left( \int_0^ry^2\ln(r+\sqrt{r^2-y^2})\;dy-\int_0^ry^2\ln y\;dy \right) $$ which differs in the sign of the last term when compared with your formula above.

Now by lumping the last two terms together I get

$$V(r,h)=2h\int_0^r\sqrt{r^2-y^2}\;dy-\frac{2h}{r} \int_0^r y^2 \, sech^{-1}\left(\frac{y}{r}\right)\;dy $$

which on integrating by parts seems to lead to a sensible result.

Edited to add some details of the integration by parts

Taking $u=sech^{-1}\left(\frac{y}{r}\right)$; $\frac{du}{dy}=-\frac{r}{y\sqrt{r^2-y^2}}$ and $\frac{dv}{dy}=y^2$ ; $v=\frac{y^3}{3}$ gives

$$\frac{2h}{r} \int_0^r y^2 \, sech^{-1}\left(\frac{y}{r}\right)\;dy =\frac{2h}{r} \left[ \frac{y^3}{3}sech^{-1}\left(\frac{y}{r}\right)\right]_0^r+\frac{2h}{3} \int_0^r \frac{y^2}{\sqrt{r^2-y^2}} \;dy$$

the first resultant term is zero so the integral of real interest is

$$\int_0^r \frac{y^2}{\sqrt{r^2-y^2}} \;dy = \left[-\frac{y}{2}\sqrt{r^2-y^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{y}{r}\right)\right]_0^r$$

the only other integral of interest being the first one

$$\int_0^r\sqrt{r^2-y^2}\;dy=\left[ \frac{y}{2}\sqrt{r^2-y^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{y}{r}\right)\right]_0^r$$

  • $\begingroup$ Thank you for the response, would you refer to my answer below please? $\endgroup$ Jul 15 '17 at 0:33
  • $\begingroup$ Responded here with an edit to my original post to clarify the main steps, as I cannot comment on your post. $\endgroup$ Jul 15 '17 at 1:49
  • $\begingroup$ This works out to the desired result, fantastic work. Could I cite you for your help with this integral? $\endgroup$ Jul 15 '17 at 4:40
  • $\begingroup$ Yes that's fine with me, assuming that is, you can't find this method or a similar one in your literature survey work. In the unlikely event your approach to the problem is original why not polish things up a bit and attempt to publish the calculation in a journal. $\endgroup$ Jul 15 '17 at 15:00
  • $\begingroup$ I'll see if any professors at my institute have seen such an approach. In the meantime, I updated the question with the work $\endgroup$ Jul 15 '17 at 15:56

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