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Thread: Verifying the divergence theorem

  1. #1
    MHF Contributor arbolis's Avatar
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    Verifying the divergence theorem

    I must verify the divergence theorem in the following case :
    $\displaystyle F(x,y,z)=x \vec i + y \vec j + z \vec k$. And $\displaystyle S$ is the surface of the sphere $\displaystyle x^2+y^2+z^2=4$.
    My attempt :
    $\displaystyle \iint _S F \bold n dS=\iiint _S div FdS$.
    I calculate the right hand side : $\displaystyle div F=3$.
    So $\displaystyle \iiint _S div FdS =$ 3 times the volume of a sphere with radius $\displaystyle 2$, hence the triple integral is worth $\displaystyle 32 \pi$.
    Now comes the problems. I ain't able to calculate $\displaystyle \iint _S F \bold n dS$.
    A normal unit vector to the sphere is $\displaystyle \bold n =\frac{2x \vec i+2y \vec j +2z \vec k}{2\sqrt 3}$.
    So the double integral is worth $\displaystyle \frac{1}{2 \sqrt 3} \iint _S (x \vec i + y \vec j + z \vec k)(2x \vec i+2y \vec j +2z \vec k) dS=\frac{1}{2\sqrt 3} \iint _S 2x^2+2y^2+2z^2 dS$. And I'm stuck here because I have 3 variables ($\displaystyle x$, $\displaystyle y$ and $\displaystyle z$) 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 $\displaystyle 32 \pi$?
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  2. #2
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    Quote Originally Posted by arbolis View Post
    My attempt :
    $\displaystyle \iint _S F \bold n dS=\iiint _S div FdS$.
    $\displaystyle \iint _S F \cdot \bold n \ dS=\iiint _D div F \ dV$ is correct, where $\displaystyle D$ is the region bounded by the sphere.


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

    So the double integral is worth $\displaystyle \frac{1}{2 \sqrt 3} \iint _S (x \vec i + y \vec j + z \vec k)(2x \vec i+2y \vec j +2z \vec k) dS=\frac{1}{2\sqrt 3} \iint _S 2x^2+2y^2+2z^2 dS$. And I'm stuck here because I have 3 variables ($\displaystyle x$, $\displaystyle y$ and $\displaystyle z$) 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 $\displaystyle \iint _S F \cdot \bold n \ dS=\frac{1}{2} \iint_S (x^2+y^2+z^2) \ dS=2 \iint_S dS = 2(16 \pi) = 32 \pi.$

    over the surface of the sphere we have $\displaystyle x^2+y^2+z^2 =4.$ also at the end i used the fact that $\displaystyle \iint_S dS$ is the surface area of the sphere, which is $\displaystyle 16 \pi.$
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    MHF Contributor arbolis's Avatar
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    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 $\displaystyle \iint _S dS$ was the surface area of $\displaystyle S$.
    Is there a formula to calculate the surface area of any given closed surface $\displaystyle S$?
    I'm thinking about the formula $\displaystyle A(S)=\frac{1}{2} \oint -ydx+xdy$ but I don't think it is appropriated.
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    dS is the "differential of surface area" $\displaystyle \int\int dS= S$, the area of the region integrated over! In general $\displaystyle \int du= u$- that's the "fundamental theorem of calculus".

    The formula you give is for the area of a region in the xy-plane.
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    Quote Originally Posted by arbolis View Post
    Oh thanks for all!
    Actually I wasn't aware that $\displaystyle \iint _S dS$ was the surface area of $\displaystyle S$.
    Is there a formula to calculate the surface area of any given closed surface $\displaystyle S$?
    I'm thinking about the formula $\displaystyle A(S)=\frac{1}{2} \oint -ydx+xdy$ but I don't think it is appropriated.
    if $\displaystyle z=f(x,y)$ is the equation of your surface, where $\displaystyle x,y$ 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

    $\displaystyle \iint_R \sqrt{1+f_x^2+f_y^2} \ dx dy.$ so in general: $\displaystyle \iint_S g(x,y,z) \ dS= \iint_R g(x,y,f(x,y))\sqrt{1+f_x^2+f_y^2} \ dx dy.$ you'll certainly find this formula in your textbook!
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    MHF Contributor arbolis's Avatar
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    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.
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