<s>[INST] surface area of $x^2+y^2+z^2-14z=0$, where $0\leq x^2+y^2\leq3z$ What is the simplest way to compute the surface area of
$$x^2+y^2+z^2-14z=0,$$
where
$$0\leq x^2+y^2\leq3z\tag*{?}$$
I am having problems with various parameterizations.
 [/INST] Well, $z^2-14z+3z\geq x^2+y^2+z^2-14z\geq 0$ combined with $z\geq 0$ gives $z\geq 11$.
Another form of the equation is $x^2+y^2+(z-7)^2=49$, which is a sphere centered at $(0, 0, 7)$ with radius 7.
So we need to calculate a part of the area of sphere, from $z=11$ to $z=14$, or the area of sphere $x^2+y^2+z^2 = 49$ from $z=4$ to $z=7$. We even don't need to parameters in this calculation: notice that a sphere is a surface of revolution.
$S=\int_{4}^{7}2\pi\sqrt{49-z^2}\sqrt{1+(\sqrt{49-z^2})^\prime} \mathrm{d}z= 14\pi z|_4^7=42\pi$
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