Wikipedia:Reference desk/Archives/Mathematics/2013 July 1

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July 1

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Deriving volume of a cone formula with double integration

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I am trying to derive   using the following double integral:

 

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

 

The first term easily integrates to   because I recognize it as the area of a circle times the height, while the last term similarly integrates to (after distributing the  )  . But I cannot see how the second term will integrate to the missing  . 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)[reply]

Try integration by parts. Sławomir Biały (talk) 20:12, 1 July 2013 (UTC)[reply]
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)[reply]
The log term should be the u in integration by parts, the   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)[reply]

Perhaps I did not try hard enough though, as it turns out that   and  . Oh well.--Jasper Deng (talk) 20:58, 1 July 2013 (UTC)[reply]

Still, however, I can't see how the missing   will come out of all of that.--Jasper Deng (talk) 21:04, 1 July 2013 (UTC)[reply]
Reset. Try again.  ,  , so   and  . Sławomir Biały (talk) 21:51, 1 July 2013 (UTC)[reply]
OK, so with that, we have  , which doesn't seem very encouraging, but seems solvable with a trigonometric substitution.--Jasper Deng (talk) 22:14, 1 July 2013 (UTC)[reply]
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)[reply]
I obtained the erroneous result   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  ,   - eventually the expression reduced to  ).--Jasper Deng (talk) 23:12, 1 July 2013 (UTC)[reply]
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).[reply]

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  , the second term of the integration by parts becomes  , which however still returns that erroneous result. I'm missing a key term, namely   - 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)[reply]

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)[reply]
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  ).--Jasper Deng (talk) 19:28, 3 July 2013 (UTC)[reply]

Perhaps the fact that   and   might prove itself a bit useful in simplifying some of the intermediary formulas ? — 79.113.211.75 (talk) 21:11, 3 July 2013 (UTC)[reply]

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)[reply]
I beg to differ. Ultimately, I arrive at
 
where
 
whose calculation is trivial. — 79.113.211.75 (talk) 21:59, 3 July 2013 (UTC)[reply]
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)[reply]
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   to arrive at the same result from your own previous calculations. — 79.113.211.75 (talk) 22:42, 3 July 2013 (UTC)[reply]
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)[reply]
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) = t2F(t) =   and G(x) = Ln(1 + √1 - t2), ultimately arriving at
 
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)[reply]
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)[reply]
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)[reply]