Finding surface of the set in $\mathbb{R}^3$

Question

Given set $$M \equiv x^2 + y^2 \leq 2z,~z \in [0, 1]$$ find it's surface.

Using cylindrical coordinates I'm finding that $$r^2 = 2z \implies z = \frac{r^2}{2}.$$

Now my transformation is of the form $x = r\cos\phi$, $y = r\sin\phi$ and $z = r^2/2$. Taking my transformation as a vector $\Psi := (x, y, z) = (r\cos\phi, r\sin\phi, r^2/2)$ norm of the normal vector is then $||\partial_r \Psi \times \partial_\phi \Psi|| = ||\mathbf{n}|| = r\sqrt{1 + r^2}$. Hence the surface of the set is $$S = \int_M 1~dS = \int_{0}^{2\pi}d\phi\int_{0}^{1}dr~r\sqrt{1 + r^2} = \frac{2\pi}{3}\left(2\sqrt{2} - 1\right).$$

Is my approach correct or did I made any mistake?

Answer 1

I know this doesn't answer your question, but ...

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          ^{Mathematica: ImplicitRegion[x^2 + y^2 <= 2 z && 0 <= z && z <= 1, {x, y, z}]}

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Answer 2

Your norm of normal vector approach is correct.

However to check by the usual axial cut shells for a surface of revolution along arc I took

$$ dA= 2 \pi r ds = \frac{2 \pi r dr }{\cos \varphi} $$

when by differentiation

$$ \tan \varphi= 1/r $$

$$ A = 2 \pi \int \sqrt{1+r^2} dr $$

upper limit should be $r=\sqrt2$