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This comes from Apostol's Calculus, Vol. II, Section 10.9 #14:

A uniform wire has the shape of that portion of the curve of intersecion of the two surfaces $x^2+y^2=z^2$ and $y^2=x$ connecting the points $(0,0,0)$ and $(1,1,\sqrt 2)$. Find the $z$-coordinate of its centroid.

The $z$-coordinate of the centroid is defined as $\dfrac {\int_C z\, \mathrm ds}{\int_C \mathrm ds}$, where $s(t)$ is the arc length. (That is, if $\vec\alpha(t)$ is a parametrization of $C$, $s(t)=\int \lVert \vec \alpha \,' (t) \rVert \mathrm d t$.)

One valid parametrization of the curve is $\vec \alpha(t)=\left(t^2,\, t,\, t\sqrt {t^2+1}\right)\,,$ however this becomes quite difficult to work with. In fact, Mathematica can only numerically integrate $\int_C \mathrm ds= \int_0^1 \lVert \vec \alpha\, '(t) \rVert\, \mathrm dt$. I thought perhaps using a substitution like $t=\tan(\theta)$ would be helpful, but obviously if this was viable Mathematica would have performed this simple substitution in the first place.

My question is, specifically for this problem, is there a nicer parametrization that I am missing? Hints appreciated. I have tried solving $x$ and $y$ in terms of $z$ and using the resulting parametrization, but it is no better.


In case it is helpful, the book's answer is $\frac {600-36\sqrt 2 - 49 \log(9-4\sqrt 2)}{64[6\sqrt 2 + \log (3+2\sqrt 2)]}\approx0.747018$ While Mathematica returns $0.710276$ using my parametrization, so perhaps there is a mistake (either with my parametrization, the answer in the book, or the question in the book).

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    I should also remark that a preliminary check indicates that integration by parts in order to cancel the need to actually integrate $\int_C \, \mathrm ds$ doesn't look promising, but I suppose it is a possibility...2012-09-03

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In cylindrical coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$ the equation of the cone simply is $z=r$ and $y^2=x$ yields $r^2\sin^2(\theta)=r\cos(\theta)$ or $r=\cot(\theta)$. The metric in cylindricals simplifies once we substitute the curve equations: $ds^2 = dr^2+r^2d\theta^2+dz^2 = 2dr^2+\cot^2(\theta)d\theta^2=(\bigl[\frac{dr}{d\theta}\bigr]^2+\cot^2(\theta))d\theta^2=(\csc^2(\theta)+\cot^2(\theta))d\theta^2$ So, if I haven't made a silly mistake somewhere, $ s = \int_I \sqrt{\csc^2(\theta)+\cot^2(\theta)} \ d\theta $ I'm not sure if this helps or hurts.

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    a curve was specially important to a particular class of subsitutions. Surely someone will elaborate on this in response to your next question about this. Sadly I must go for now...2012-09-03