# Verify divergence theorem

im trying to divergence theorem by direct calculation. There is a cube with vertices $(0,0,0),(R,0,0),(0,R,0),(0,0,R),(R,R,0),(0,R,R),(R,0,R),(R,R,R)$ and we know that its volume is $R^3$ (trivial) which is one part of the divergence theorem. Now I have to calculate:

$\int\int_{S_R}v\cdot n$ $dA$ where $v(x,y,z) = (3x + z^2, 2y, R-z)$. The cube will also have unit normals, i.e $(\pm1,0,0), (0,\pm1,0) ,(0,0,\pm1)$.

Now I calculate the flux through the surface where $x$ is constant and I get:

$$\int_0^R\int_0^R 3R+z^2dydz = \frac{R^3}{2} + \frac{R^4}{3}$$ where $x = R$, and

$$\int_0^R\int_0^R -z^2dydz = -\frac{R^4}{3}$$ where $x=0$ and these two integrals sum to $\frac{R^3}{2}$.

When $y$ is constant, i.e. when $y=R$:

$$\int_0^R\int_0^R 2Rdxdz = R^3/3$$ (corresponding integral is $0$ when $y=0$). But when $z=0$:

$$\int_0^R\int_0^R -Rdxdy=-\frac{R^3}{6}$$

where I should get $\frac{R^3}{6}$ as the sum of the integrals then gives me $R^3$, but i cannot seem to get this answer.

• Your integrals are incorrect. E.g., when $y=R$ one has $\int_0^R\int_0^R 2R \ dx\>dz=2R^3$. – Christian Blatter Nov 2 '13 at 16:19

We have ${\bf v}(x,y,z):=(3x+z^2,2y, R-z)$. Let's compute the six facet-integrals:
(i) Facet $S_1:\ x=0$, ${\bf n}=(-1,0,0)$: $$\int_{S_1}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_0^R\int_0^R -z^2\ dz\>dy=-{R^4\over3}\ .$$ (ii) Facet $S_2:\ x=R$, ${\bf n}=(1,0,0)$: $$\int_{S_2}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_0^R\int_0^R (3R+z^2)\ dz\>dy=3R^3+{R^4\over3}\ .$$ (iii) Facet $S_3:\ y=0$, ${\bf n}=(0,-1,0)$: $$\int_{S_3}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_0^R\int_0^R 0\ dx\>dz=0\ .$$ (iv) Facet $S_4:\ y=R$, ${\bf n}=(0,1,0)$: $$\int_{S_4}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_0^R\int_0^R 2R\ dx\>dz=2R^3\ .$$ (v) Facet $S_5:\ z=0$, ${\bf n}=(0,0,-1)$: $$\int_{S_5}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_0^R\int_0^R -R\ dx\>dy=-R^3\ .$$ (vi) Facet $S_6:\ z=R$, ${\bf n}=(0,0,1)$: $$\int_{S_6}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_0^R\int_0^R 0\ dx\>dy=0\ .$$ Adding it all up we obtain $$\int_{\partial C}{\bf v}\cdot{\bf n}\ {\rm d}\omega=4R^3\ .$$ On the other hand $${\rm div}\>{\bf v}(x,y,z)=3+2-1=4\qquad\forall\ (x,y,z)\in{\mathbb R}^3\ .$$ It follows that $$\int_C {\rm div}\>{\bf v}(x,y,z)\ {\rm d}(x,y,z)=4{\rm vol}(C)=4R^3\ ,$$ as before.