2
$\begingroup$

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 $h$:

$$V(r,h)=2\int_0^r\delta(y)\;dy$$

$$\delta(y)=2h\int_0^{\sqrt{r^2-y^2}}\left[1-\frac{\sqrt{x^2+y^2}}{r}\right]\;dx$$

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.

$$\delta(y)=2h\sqrt{r^2-y^2}-\frac{2h}{r}y^2\int_0^{\arctan\left(\frac{\sqrt{r^2-y^2}}{y}\right)}\sec^3\theta\;d\theta$$

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

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

$$\delta(y)=h\sqrt{r^2-y^2}-\frac{h}{r}y^2\ln(r+\sqrt{r^2-y^2})-\frac{h}{r}y^2\ln(y)$$

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.

$$\delta(y)=2h\sqrt{r^2-y^2}-\color{red}{\frac{2h}{r}y^2\int_0^{\text{arcsinh}\left(\frac{\sqrt{r^2-y^2}}{y}\right)}\cosh^2\theta\;d\theta}$$

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 $V(r,h)$:

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

\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:

$$V(r,h)=hr^2\int_0^{\pi/2}1-\cos2\phi\;d\phi-\frac{2h}{r}\int_0^ry^2\text{arcsech}\left(\frac{h}{r}\right)\;dy$$

\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:

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

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

$\endgroup$
  • 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. $\endgroup$ – MathematicsStudent1122 Jul 13 '17 at 18:38
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Thank you for the response, would you refer to my answer below please? $\endgroup$ – Lanier Freeman 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. $\endgroup$ – James Arathoon 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? $\endgroup$ – Lanier Freeman 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. $\endgroup$ – James Arathoon 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 $\endgroup$ – Lanier Freeman Jul 15 '17 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.