Results 1 to 6 of 6

Math Help - Verifying the divergence theorem

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Verifying the divergence theorem

    I must verify the divergence theorem in the following case :
    F(x,y,z)=x \vec i + y \vec j + z \vec k. And S is the surface of the sphere x^2+y^2+z^2=4.
    My attempt :
    \iint _S F \bold n dS=\iiint _S div FdS.
    I calculate the right hand side : div F=3.
    So \iiint _S div FdS = 3 times the volume of a sphere with radius 2, hence the triple integral is worth 32 \pi.
    Now comes the problems. I ain't able to calculate \iint _S F \bold n dS.
    A normal unit vector to the sphere is \bold n =\frac{2x \vec i+2y \vec j +2z \vec k}{2\sqrt 3}.
    So the double integral is worth \frac{1}{2 \sqrt 3} \iint _S (x \vec i + y \vec j + z \vec k)(2x \vec i+2y \vec j +2z \vec k) dS=\frac{1}{2\sqrt 3} \iint _S 2x^2+2y^2+2z^2 dS. And I'm stuck here because I have 3 variables ( x, y and z) but a double integral instead of a triple one.
    I know I could solve the problem faster using spherical coordinates, but for the sake of my comprehension let's do it via Cartesian coordinates please.
    Do you have any idea how could I continue, in order to reach the 32 \pi?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by arbolis View Post
    My attempt :
    \iint _S F \bold n dS=\iiint _S div FdS.
    \iint _S F \cdot \bold n \ dS=\iiint _D div F \ dV is correct, where D is the region bounded by the sphere.


    A normal unit vector to the sphere is \bold n =\frac{2x \vec i+2y \vec j +2z \vec k}{2\sqrt 3}.
    incorrect! \bold n =\frac{2x \vec i+2y \vec j +2z \vec k}{\sqrt{4x^2+4y^2+4z^2}}=\frac{x \vec i+y \vec j +z \vec k}{2}.

    So the double integral is worth \frac{1}{2 \sqrt 3} \iint _S (x \vec i + y \vec j + z \vec k)(2x \vec i+2y \vec j +2z \vec k) dS=\frac{1}{2\sqrt 3} \iint _S 2x^2+2y^2+2z^2 dS. And I'm stuck here because I have 3 variables ( x, y and z) but a double integral instead of a triple one.
    I know I could solve the problem faster using spherical coordinates, but for the sake of my comprehension let's do it via Cartesian coordinates please.
    so \iint _S F \cdot \bold n \ dS=\frac{1}{2} \iint_S (x^2+y^2+z^2) \ dS=2 \iint_S dS = 2(16 \pi) = 32 \pi.

    over the surface of the sphere we have x^2+y^2+z^2 =4. also at the end i used the fact that \iint_S dS is the surface area of the sphere, which is 16 \pi.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Oh thanks for all!
    also at the end i used the fact that is the surface area of the sphere, which is
    Actually I wasn't aware that \iint _S dS was the surface area of S.
    Is there a formula to calculate the surface area of any given closed surface S?
    I'm thinking about the formula A(S)=\frac{1}{2} \oint -ydx+xdy but I don't think it is appropriated.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,790
    Thanks
    1531
    dS is the "differential of surface area" \int\int dS= S, the area of the region integrated over! In general \int du= u- that's the "fundamental theorem of calculus".

    The formula you give is for the area of a region in the xy-plane.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by arbolis View Post
    Oh thanks for all!
    Actually I wasn't aware that \iint _S dS was the surface area of S.
    Is there a formula to calculate the surface area of any given closed surface S?
    I'm thinking about the formula A(S)=\frac{1}{2} \oint -ydx+xdy but I don't think it is appropriated.
    if z=f(x,y) is the equation of your surface, where x,y come from some region R in xy plane, then the area of of that part of your surface whose image on xy plane is R is equal to

    \iint_R \sqrt{1+f_x^2+f_y^2} \ dx dy. so in general: \iint_S g(x,y,z) \ dS= \iint_R g(x,y,f(x,y))\sqrt{1+f_x^2+f_y^2} \ dx dy. you'll certainly find this formula in your textbook!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks to both.
    Quote Originally Posted by NCA
    you'll certainly find this formula in your textbook!
    Indeed, it was the first thing written in the "Surface area" part.
    I didn't pay attention to it before.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Verifying Green theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 1st 2011, 08:14 AM
  2. Replies: 3
    Last Post: May 14th 2010, 10:04 PM
  3. Replies: 2
    Last Post: April 3rd 2010, 04:41 PM
  4. Flux integrals - verifying Stokes Theorem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 20th 2009, 08:58 AM
  5. The Divergence Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 28th 2009, 08:53 AM

Search Tags


/mathhelpforum @mathhelpforum