Results 1 to 4 of 4

Math Help - Difficult Surface Integral...

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    13

    Difficult Surface Integral...

    In order to figure this out, I think I have to parametrize the integral correct? If I do, would this parametrization be valid?

    x=x
    y=x^2

    BTW I think the answer will be zero since the flux going in is equal to the flux coming out. This of course is assuming this flux is the same as electrical flux.
    Attached Thumbnails Attached Thumbnails Difficult Surface Integral...-screen-shot-2009-12-05-9.51.19-pm.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
     \int\int_S ( x^2 e^{-z} + y^2 e^{-z} )dS


     = \int\int_S (x^2+y^2)e^{-z}~dS


     x^2 + y^2 = 9 = 3^2 , z \in [0,10]


    if we transform in cylindrinal coordinates ,

     x^2 + y^2 becomes  r^2

     dS becomes  rd\theta dz


     \int\int_S (r^2)e^{-z}(rd\theta dz)

     = \int\int_S r^3 e^{-z} d\theta dz

    but  r = 3 for every points on the surface

    Surface integral  = (3)^3 \int_0^{2\pi} \int_0^{10} e^{-z} dz d\theta

     = 27(2\pi) (1- e^{-10} )


    do we need to calcuate the integral for the base ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    7
    Thanks for the help, one question though... Why did you switch the <br /> <br />
d\theta<br />
for dz?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by simplependulum View Post
     \int\int_S ( x^2 e^{-z} + y^2 e^{-z} )dS


     = \int\int_S (x^2+y^2)e^{-z}~dS


     x^2 + y^2 = 9 = 3^2 , z \in [0,10]


    if we transform in cylindrinal coordinates ,

     x^2 + y^2 becomes  r^2

     dS becomes  rd\theta dz


     \int\int_S (r^2)e^{-z}(rd\theta dz)

     = \int\int_S r^3 e^{-z} d\theta dz

    but  r = 3 for every points on the surface

    Surface integral  = (3)^3 \int_0^{2\pi} \int_0^{10} e^{-z} dz d\theta

     = 27(2\pi) (1- e^{-10} )


    do we need to calcuate the integral for the base ?
    first of all dS is not necessarily dxdy and so you need to be careful. second the cylinder is closed and so you also need to consider the top and the bottom of the cylinder!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. difficult integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 26th 2010, 12:42 PM
  2. Difficult integral!?!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 20th 2010, 10:33 AM
  3. Difficult Integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 16th 2010, 02:20 PM
  4. Difficult integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 22nd 2009, 07:20 AM
  5. Difficult Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 10:08 AM

Search Tags


/mathhelpforum @mathhelpforum