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It is given that

$M_g:=(x,y,z) in mathbb R^3:z in ]1,infty[, x^2+y^2=g(z)^2$

Let $alpha > 0$ and $g(z):=z^-alpha$

I have been able to determine that $lambda_M_g(M_g)=2piint_]1,infty[g(z)sqrt1+(g^'(z))^2dz$

Question: Dependending on $alpha$, determine wheter $lambda_M_g(M_g)$ is finite or indeed infinite.

My starting point:
$2piint_]1,infty[g(z)sqrt1+(g^'(z))^2dz=2piint_]1,infty[z^-alphasqrt1+alpha^2z^-2(alpha+1)dz$

I have attempted substitution to no avail, and I cannot find a clear trend in terms of the $alpha$ select, for example if $alpha=1$, we get:

$2piint_]1,infty[frac1zsqrt1+frac1z^4dzgeq 2piint_]1,infty[frac1zdz =infty$

First of all, notice that $lambda_M_g (M_g) ge 0$ because the integrand is positive.

From now on, I shall drop the factor $2 pi$ for convenience, since it doesn't change anything.

If $alpha > 1$ then, assuming that your formula for $lambda_M_g (M_g)$ is correct (I haven't checked it), you have that $z > 1$ implies $z^-2(alpha+1) < 1$ (because $alpha> -1$), therefore $sqrt1 + z^-2(alpha+1) < sqrt 2$, so it follows that

$$lambda_M_g (M_g) le sqrt 2 int _1 ^infty z^-alpha mathrm d z = sqrt 2 frac z^-alpha + 1 -alpha + 1 Bigg|_1 ^infty = frac sqrt 2 alpha - 1$$

which is finite, therefore in this case the integral is between $0$ and $frac sqrt 2 alpha - 1$, therefore finite.

If $alpha = 1$ then

$$lambda_M_g (M_g) = int _1 ^infty z^-1sqrt1+z^-4 mathrm d z ge int _1 ^infty z^-1sqrt1+0 mathrm d z = log z big| _1 ^infty = infty ,$$

therefore the integral is infinite in this case.

Finally, if $alpha in (0,1)$, then

$$lambda_M_g (M_g) = int _1 ^infty z^-alpha sqrt1 + z^-2(alpha + 1) mathrm d z ge int _1 ^infty z^-alpha sqrt1 + 0 mathrm d z = frac z^-alpha + 1 -alpha + 1 Bigg| _1 ^infty = infty ,$$

therefore the integral is infinite in this case, too.

To conclude, the integral is finite for $alpha > 1$ and infinite for $alpha in (0,1]$.

$$forall z >1, f(z) = z^-alphasqrt1+z^-2(alpha +1) > 0$$

Let $alpha > beta > 1$:
$$z^beta -alphasqrt1+z^-2(alpha +1) rightarrow_z rightarrow infty 0 $$

So $f(z) = mathscro_z rightarrow infty(z^-beta)$ but the integral between 1 and $infty$ of $z mapsto z^-beta$ converges, thus the integral converges if $alpha > 0 $.

Similarly, $ f(z) sim_z rightarrow infty z^-1$ so the integral does not converge.

As for when $alpha <1$, you can choose $1 > beta > alpha$ and you can show that $z^-beta = mathscro_z rightarrow infty(f(z))$, thus the integral does not converge.

If $alpha >0, zge 1,$ then $z^-2(alpha+1)le 1.$ Thus

$$ 1<sqrt1+alpha^2z^-2(alpha+1)le sqrt 1 + alpha^2.$$

Thus the integral in question is finite iff $int_1^infty z^-alpha,dz <infty.$ That, as you know, holds iff $alpha >1.$