MHB Divergence Theorem and shape of hyperboloid

Mountain1
Messages
1
Reaction score
0
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 ?
 
Last edited by a moderator:
Physics news on Phys.org
I get the same thing.
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
Back
Top