I haven't done a surface integral in a while so I am asking to get this checked.

enter image description here

$\mathbf{F} = \langle x, y, z\rangle$ and the surface is $z = xy + 1$ where $0\leq x,y\leq 1$.

$\hat{\mathbf{n}} = \nabla f/ \lvert\nabla f\rvert = \frac{1}{\sqrt{3}}\langle 1, 1, 1\rangle$

$dS = \frac{\lvert\nabla f\rvert dxdy}{\frac{\partial f}{\partial z}} = \sqrt{3}dxdy$

$\mathbf{F}\cdot\hat{\mathbf{n}} = \frac{1}{\sqrt{3}}(x+y+z) = \frac{1}{\sqrt{3}}(x+y+xy + 1)$

$$ \int_0^1\int_0^1(x + y + xy + 1)dxdy = \frac{9}{4} $$

However, when I used the divergence theorem, I obtained:

$$ \int_S(\mathbf{F}\cdot\hat{\mathbf{n}})dS = \int_V(\nabla\cdot\mathbf{F})dV $$ and $\nabla\cdot\mathbf{F} = 3$ so $$ \int 3dV = 3V = 3\frac{5}{4} = \frac{15}{4} $$

Which one is wrong or are both incorrect? If so, what is wrong?


I'm afraid neither is correct. You confused derivatives of $\mathbf F$ with calculating the surface normal. The Divergence Theorem does not apply, as you do not have a closed surface (so there's no volume $V$ enclosed). And to compute the surface integral you need to compute $$\int_0^1\int_0^1 \mathbf F(\mathbf g(x,y))\cdot \left(\frac{\partial\mathbf g}{\partial x}\times\frac{\partial\mathbf g}{\partial y}\right) \,dx\,dy\,,$$ where $\mathbf g(x,y) = (x,y,xy+1)$. This becomes $$\int_0^1\int_0^1 (1-xy)\,dx\,dy = \frac34\,.$$

  • $\begingroup$ Well, perhaps you didn't fully state the problem. In your terse statement, you gave us $\mathbf F$ and you said "and the surface is ..." :) $\endgroup$ – Ted Shifrin Sep 7 '13 at 16:40
  • $\begingroup$ Forget the cube. Make a closed surface by adding in three faces (bottom, right, back in your picture). The total flux outward across all four faces is $3$ times the volume enclosed. So you now need to subtract the flux across the three faces you've appended to answer your original question. $\endgroup$ – Ted Shifrin Sep 7 '13 at 17:16
  • $\begingroup$ What exactly do you not understand? $\endgroup$ – Ted Shifrin Sep 7 '13 at 17:44
  • $\begingroup$ The integral of divergence over the region, yes. But this gives the flux across all of the bounding surface. Your top surface is only part of what needs to be a closed surface. $\endgroup$ – Ted Shifrin Sep 7 '13 at 18:09

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