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The given is $${(x,y,z)}\mid 0\leq z\leq 4-2\sqrt{x^2+y^2}$$ The lowest $x$ and $y$ can be is zero, resulting in the inequality be $$z\leq 4$$ The height and the radius is the largest $z$ can be, which is $4$. So the radius and height are both $4$.

Am I correct?

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Let consider a section in the $x-z$ plane that is for $y=0$ to obtain

$$0\leq z\leq 4-2\sqrt{x^2} \implies 0\leq z\leq 4-2|x|$$

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then the radius at the base is equal to $2$.

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