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July 1

Deriving volume of a cone formula with double integration

I am trying to derive $V=\frac{1}{3}\pi h{r}^2$ using the following double integral:

$\underset{-r}{\overset{r}{\int }}\underset{-\sqrt{r^2-x^2}}{\overset{\sqrt{r^2-x^2}}{\int }}h-\frac{h}{r}\sqrt{x^2+y^2}dydx$

because I observed that a cone can be defined by that equation. The problem is, it gets messy after the first integration:

$\underset{-r}{\overset{r}{\int }}\underset{-\sqrt{r^2-x^2}}{\overset{\sqrt{r^2-x^2}}{\int }}h-\frac{h}{r}\sqrt{x^2+y^2}dydx=\underset{-r}{\overset{r}{\int }}(hy-\frac{h}{r}(\frac{x^2\ln |\sqrt{x^2+y^2}+y|}{2}+\frac{y\sqrt{x^2+y^2}}{2})){|}_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}dx=\underset{-r}{\overset{r}{\int }}2h\sqrt{r^2-x^2}-\frac{h}{2r}(x^2\ln \left|\frac{r+\sqrt{r^2-x^2}}{r-\sqrt{r^2-x^2}}\right|+2r\sqrt{r^2-x^2})dx$

The first term easily integrates to $\pi h{r}^2$ because I recognize it as the area of a circle times the height, while the last term similarly integrates to (after distributing the $\frac{h}{2r}$) $\frac{\pi h{r}^2}{2}$. But I cannot see how the second term will integrate to the missing $-\frac{\pi h{r}^2}{6}$. I know the volume is readily obtained using disk integration and/or polar coordinates. However, it still should be possible to obtain it using plain rectangular coordinates too.--Jasper Deng (talk) 19:34, 1 July 2013 (UTC)

Try integration by parts. Sławomir Biały (talk) 20:12, 1 July 2013 (UTC)

I did. But that natural log-based expression alone has no elementary anti-derivative I could think of. Trigonometric substitutions didn't help either.--Jasper Deng (talk) 20:41, 1 July 2013 (UTC)

The log term should be the u in integration by parts, the $x^2$ is the dv. So you shouldn't need to find an antiderivative of the log factor. (See, for instance, the ILATE rule.) Sławomir Biały (talk) 20:54, 1 July 2013 (UTC)

Perhaps I did not try hard enough though, as it turns out that $\int \ln (r+\sqrt{r^2-x^2})dx=x\ln (r-\sqrt{r^2-x^2})+r\arctan \frac{x}{\sqrt{r^2-x^2}}-x$ and $\int \ln (r-\sqrt{r^2-x^2})dx=x\ln r-\sqrt{r^2-x^2}-r\arctan \frac{x}{\sqrt{r^2-x^2}}-x$. Oh well.--Jasper Deng (talk) 20:58, 1 July 2013 (UTC)

Still, however, I can't see how the missing $-\frac{\pi h{r}^2}{6}$ will come out of all of that.--Jasper Deng (talk) 21:04, 1 July 2013 (UTC)

Reset. Try again. $u=\ln \left|\frac{r+\sqrt{r^2-x^2}}{r-\sqrt{r^2-x^2}}\right|$, $dv=x^{2}dx$, so $du=?$ and $v=?$. Sławomir Biały (talk) 21:51, 1 July 2013 (UTC)

OK, so with that, we have $\frac{1}{3}{x}^3\ln \left|\frac{r+\sqrt{r^2-x^2}}{r-\sqrt{r^2-x^2}}\right|-\int \frac{x^{3}r(r-\sqrt{r^2-x^2})dx}{\sqrt{r^2-x^2}(\sqrt{r^2-x^2}+r)(r\sqrt{r^2-x^2}+\frac{x^2}{2}-r^2)}$, which doesn't seem very encouraging, but seems solvable with a trigonometric substitution.--Jasper Deng (talk) 22:14, 1 July 2013 (UTC)

You'll probably want to apply partial fractions now. (Or, cleverly manipulate the original logarithm factor.) Sławomir Biały (talk) 22:31, 1 July 2013 (UTC)

I obtained the erroneous result $2r\sqrt{r^2-x^2}$ using trigonometric substitution. I don't really know how to decompose this expression because it contains square roots in the denominator. It seems like a trigonometric substitution would be most obvious, but it gave me that erroneous result (the substitution was $x=r\sin u$, $dx=r\cos (u)du$ - eventually the expression reduced to $\int -2r^2\sin (u)du$).--Jasper Deng (talk) 23:12, 1 July 2013 (UTC)

The special case r=1 and h=1 simplifies the expressions and is rather easily generalized afterwards. Bo Jacoby (talk) 07:04, 2 July 2013 (UTC).

I just realized that by expressing the logarithm expression as a single term, I was really making life harder for myself. By separating the natural log into $\ln (r+\sqrt{r^2-x^2})-\ln (r-\sqrt{r^2-x^2})$, the second term of the integration by parts becomes $\int \frac{-2xr}{\sqrt{r^2-x^2}}dx$, which however still returns that erroneous result. I'm missing a key term, namely $\frac{1}{3}{r}^3\arctan \frac{x}{\sqrt{r^2-x^2}}$ - and I know that's the key term that will give me the missing piece of volume.--Jasper Deng (talk) 19:12, 3 July 2013 (UTC)

The second term of which integration by parts? You shouldn't get something that looks like this, IMO, but if you did it would just integrate away. Sławomir Biały (talk) 19:23, 3 July 2013 (UTC)

With v the natural log expression, as above (in my comment "OK, so with that, we have...."), dv is much simpler if I separate the quotient inside the natural log (for clarity, I had $du=x^{2}dx$).--Jasper Deng (talk) 19:28, 3 July 2013 (UTC)

Perhaps the fact that $\underset{-\sqrt{r^2-x^2}}{\overset{\sqrt{r^2-x^2}}{\int }}=2\underset{0}{\overset{\sqrt{r^2-x^2}}{\int }}$ and $\underset{-r}{\overset{r}{\int }}=2\underset{0}{\overset{r}{\int }}$ might prove itself a bit useful in simplifying some of the intermediary formulas ? — 79.113.211.75 (talk) 21:11, 3 July 2013 (UTC)

It doesn't really help, though, because I still wind up with that nasty natural log-based expression that's been a quandry for me.--Jasper Deng (talk) 21:14, 3 July 2013 (UTC)

I beg to differ. Ultimately, I arrive at

$4\underset{0}{\overset{r}{\int }}\underset{0}{\overset{\sqrt{...}}{\int }}...dydx=\pi r^{2}h-2\frac{h}{r}I$

where

$I=\underset{0}{\overset{r}{\int }}[x^2\ln (r+\sqrt{r^2-x^2})+r\sqrt{r^2-x^2}-x^2\ln x]dx$

whose calculation is trivial. — 79.113.211.75 (talk) 21:59, 3 July 2013 (UTC)

But how did you arrive at that last integral? That's the step I'm a bit stuck on.--Jasper Deng (talk) 22:11, 3 July 2013 (UTC)

Sorry, I wrote a wrong expression for I earlier. Please check again to see the corrected version. The idea is that when the lower integration limit is 0 instead of a square root, the expression thus obtained is simpler. Or you might just use the fact that $\ln \frac{a}{b}=\ln a-\ln b$ to arrive at the same result from your own previous calculations. — 79.113.211.75 (talk) 22:42, 3 July 2013 (UTC)

To me, the first term in your integrand above isn't trivial. As you can see above, I came to the same wrong antiderivative for that term when trying to integrate it by parts.--Jasper Deng (talk) 23:27, 3 July 2013 (UTC)

Then what exactly seems to be the problem ? We have I = I₁ + I₂ - I₃. The last two are indeed trivial, and the first one is calculated through integration by parts, after making first the substitution t = rx, with F'(t) = t² ⇔ F(t) = $\frac{t^3}{3}$ and G(x) = Ln(1 + √1 - t²), ultimately arriving at

$I_1=r^2\underset{0}{\overset{1}{\int }}{t}^2[\ln r+\ln (1+\sqrt{1-t^2})]dt=r^2{[\frac{t^3}{3}(\ln r+\ln (1+\sqrt{1-t^2})-\frac{1}{3})+\frac{\arcsin t-t\sqrt{1-t^2}}{6}]}_0^1=...$

Then we add all three of them together to calculate the value of I, which we then replace into our initial formula above. — 79.113.211.75 (talk) 00:24, 4 July 2013 (UTC)

Yeah, sorry, I was having difficulties performing that integration by parts - somehow my trigonometric substitution has to be flawed for the integration by parts. The antiderivative I was shooting for is slightly different from yours, but it does the job. Still, I'd like to know what went wrong with the trigonometric substitution (which tempts me whenever I have that kind of square root expression in the integrand) that caused a whole term (the one having the inverse trigonometric function in it) to disappear.--Jasper Deng (talk) 00:52, 4 July 2013 (UTC)

My guess is that you probably made several small mistakes along the way, due to a crowded working-style. My advice would be to divide and conquer. — 79.113.211.75 (talk) 01:16, 4 July 2013 (UTC)