Let S denote the solid enclosed by x^2+y^2+z^2=2z and

Let S denote the solid enclosed by ${x}^{2}+{y}^{2}+{z}^{2}=2z$ and ${z}^{2}={x}^{2}+{y}^{2}$. What is the length of of the curve determined by (x,y,z): ${x}^{2}+{y}^{2}+{z}^{2}=2z$ and ${z}^{2}={x}^{2}+{y}^{2}$? What is the surface area of S?
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primenamaqm
We have two surfaces:
${x}^{2}+{y}^{2}+{z}^{2}=2z$ and ${z}^{2}={x}^{2}+{y}^{2}$
The loci of the intersection of the surfaces is thus that of the simultaneous solution, thus:
$\left({z}^{2}\right)+{z}^{2}=2z⇒{z}^{2}-z=0⇒z=0,1$

If we examine the surfaces, then the first solution is that of a single (tangency) point, and thus the second solution is the sought solution, as such the length of the curve (using ${P}_{\text{circle}}=2\pi r$) is:
$L=2\pi$