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.

  • $\begingroup$ Your integrals are incorrect. E.g., when $y=R$ one has $\int_0^R\int_0^R 2R \ dx\>dz=2R^3$. $\endgroup$ – 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.

  • $\begingroup$ I see where i went wrong.. i kept treating R as the variable. silly mistake... $\endgroup$ – user65972 Nov 2 '13 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.