Results 1 to 4 of 4

Math Help - Divergence Theorem

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    7

    Divergence Theorem

    Let S the surface x^2+y^2-(z-6)^2=0 for 3\le z\le 6. Graph S and graphically indicate an orientation for S and compute the whole flux through S of the field \bold F(x,y,z)=(x(3-z),y(3-z),(3-z)^2).

    I can apply the divergence theorem but I'm struggled with the bounds, I don't know the triple integral should look like.

    How can be done by using a line integral? I'd have to parametrize the surface, but I'd like to know how to set up the integral!

    Any help will be appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Divergence Theorem

    Quote Originally Posted by Markov View Post
    Let S the surface x^2+y^2-(z-6)^2=0 for 3\le z\le 6. Graph S and graphically indicate an orientation for S and compute the whole flux through S of the field \bold F(x,y,z)=(x(3-z),y(3-z),(3-z)^2).

    I can apply the divergence theorem but I'm struggled with the bounds, I don't know the triple integral should look like.

    How can be done by using a line integral? I'd have to parametrize the surface, but I'd like to know how to set up the integral!

    Any help will be appreciated!
    First be careful with the divergence theorem as the cone is not closed and so the divergence theorem does not apply. Note however that the value of the vector field on the flat surface is zero so it wont contribute to the flux and you can close the surface and then apply the divergence theorem.

    My guess is that you having problems parameterizing this becuase it is not centered at the origin. We can fix that by shifting everything down 6 units

    z \mapsto z+6 This gives the equivilent problem

    S x^2+y^2-z^2=0 for -3 \le z \le 0

    and the vectorfield is

    \mathbf{F}(x,y,z)=<x(3-(z+6)),y(3-(z+6)),(3-(z+6))^2> \newline =<-x(z+3),-y(z+3),(z+3)^2>

    Now the parametric form of the cone can be found using Cylindrical coordinates or spherical coordinates.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    7

    Re: Divergence Theorem

    Yes, I was thinking the same thing. So I don't have a good parametrization, what could be yours? After that what's the set up for the integral?

    By applying the divergence theorem I'm struggling a bit with the bounds, can you give me a hand?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Divergence Theorem

    Quote Originally Posted by Markov View Post
    Yes, I was thinking the same thing. So I don't have a good parametrization, what could be yours? After that what's the set up for the integral?

    By applying the divergence theorem I'm struggling a bit with the bounds, can you give me a hand?
    The simplist way is to use cylindrial coordinates and you have the equation

    z^2=x^2+y^2 Just plug the definition of cylindrical coordinates into this equation and it should all work itself out nicely. Give this a shot and see what happens. Warning be very careful with the plus or minus.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 14th 2010, 10:04 PM
  2. Replies: 2
    Last Post: April 3rd 2010, 04:41 PM
  3. Divergence Theorem Help!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 4th 2009, 09:29 AM
  4. divergence theorem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 28th 2009, 08:03 PM
  5. Divergence Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 26th 2009, 09:33 PM

Search Tags


/mathhelpforum @mathhelpforum