Representing a region as limits of a volume integral

msslowlearner
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Homework Statement



i have the region given as being bounded by x2+y2=4 and z=0 and z=3.
this problem asks to prove gauss divergence theorem for a given vector F

Homework Equations


The Attempt at a Solution


As for the volume integral, i had no problem. But for the surface integral, how many surfaces are there actually? How are the normals represented?


I assumed it is a cylinder(Is it??).. and i got the normal vectors to the top surface and the bottom surfaces to be k and -k respectively. But these cancel out effectively. The problem here is i don't know how to represent any other surface, if any.. i really do not know to find the others. could you give me an idea on how to go about this ? i mean, finding the surfaces and limits..
 
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msslowlearner said:

Homework Statement



i have the region given as being bounded by x2+y2=4 and z=0 and z=3.
this problem asks to prove gauss divergence theorem for a given vector F

Homework Equations


The Attempt at a Solution


As for the volume integral, i had no problem. But for the surface integral, how many surfaces are there actually? How are the normals represented?


I assumed it is a cylinder(Is it??).. and i got the normal vectors to the top surface and the bottom surfaces to be k and -k respectively. But these cancel out effectively. The problem here is i don't know how to represent any other surface, if any.. i really do not know to find the others. could you give me an idea on how to go about this ? i mean, finding the surfaces and limits..

Yes, the side surface is a cylinder. You might parameterize the surface x2+y2 = 4 for z between 0 and 3 like this:

R(θ,z) = <2cos(θ), 2sin(θ),z>, 0 ≤θ≤ 2pi, 0≤z≤3
 
There are three "sides", the cylindrical x^2+ y^2= 4 and two circular ends at z= 0 and z= 3.
For those good parametric equations would be x= r cos(\theta), y= r sin(\thet), with r from 0 to 2 and \theta from 0 to 2\pi, z= 0 or 3.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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