# Thread: Verifying the divergence theorem

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 $32 \pi$?

2. Originally Posted by arbolis
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 $16 \pi.$

3. 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.

4. 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.

5. Originally Posted by arbolis
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!

6. Thanks to both.
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.