Parametric curve of intersection - line integral with respect to arc length

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... – process91 Sep 3 '12 at 4:16

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.
I believe there are a couple mistakes, but I want to check to see that I am not following it incorrectly: in the first display equation you have $2dr^2$ but you drop the 2 when you go to the second line. Also, in second display equation there is a mistake: $\frac {dr} {d\theta} = -\csc^2(\theta)$ so we get $[\frac {dr} {d\theta}]^2=\csc^4(\theta)$. Correcting these, I think we arrive at $$s=\int_I \sqrt{\cot^2 (\theta)+2\csc^4(\theta)}\, \mathrm d \theta$$. All that being said, however, when solving for $r$ I don't see how you arrived at $r=\cot(\theta)$, I get $r=\cot(\theta)\csc(\theta)$. – process91 Sep 3 '12 at 12:24
When I go with $r=\cot(\theta)\csc(\theta)$ I get the same result as before (I have to integrate over $[\pi/2,3\pi/4]$), which makes sense since this essentially is the substitution $t=\cot(\theta)$ from my original parametrization. I have a strong suspicion that, given some parametrization, if $\int\lVert \vec \alpha \,'(t) \rVert \mathrm d t$ cannot be expressed in terms of elementary functions, then no other parametrization will yield an integral in terms of elementary functions (I think this can be proved by contradiction). Does this make sense? – process91 Sep 3 '12 at 12:32