MHB Divergence Theorem and shape of hyperboloid

Mountain1
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Hello!

I have been doing a previous exam task involving the divergence theorem, but there is a minor detail in the answer which i can't fully understand.

I have a figur given by ${x}^{2} +{y}^{2} -{z}^{2} = 1$ , $z= 0$ and $z=\sqrt{3}$

As i have understood this is a hyperboloid going from $z=0$ to $z=\sqrt{3}$ and the radius goes from $r=1$ in the bottom circle to $r=2$ in the top circle.

I also have a vectorfield $F(x,y,z) = xi + yj + zk$

The task is then to find $\int\int F*n$ on $S$ where $S$ is the curved surface of the figur.

I then used the divergence theorem. Found $\int\int\int \operatorname{div} F dV$ to be $6\sqrt{3}\pi$ and $\int\int F dS$ of the lower to circle to be 0.

Then comes the problem. The integral of the top circle which is $\int\int z dS$ where $z=\sqrt{3}$. This means that the integral is $\sqrt{3} * \text{Area of the circle}$. Which I thought was $A= 4\pi$ with $r=2$.

The answer however suggests the radius is $\sqrt{3}$ which I find a bit hard to understand. Because when I put in $z=\sqrt{3}$ into ${x}^{2}+{y}^{2} -{z}^{2} = 1$ i get $r=2$.

Can someone please help me understand ?
 
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I get the same thing.
 
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