Flux integral through tricky surface.

calculusgrad-curl-divintegrationVector Fields

Let $T$ be the area lying in the first octant where $x\geq0,y\geq0,z\geq0$ limited by the surfacs $z=a^2-x^2$ and $y=a^2-x^2$.

Calculate $\iint_S \vec{F}\cdot\hat{N}dS$ where $\vec{F}=(x,y,z)$ for $(x,y,z)\in\mathbb{R^3}$, $S$ is the part of the boundary $\partial T$ to $T$ that lies on the surface $z=a^2-x^2$ and $\hat{N}$ is the unit normal vector pointing out from $T$.

I have discussed the problem with a couple of friends, and we are all stuck. We where thinking about calculating $Div\vec{F}$ but then again we struggle with the limits of that triple integral. Any help would be really great! Thanks in advance

Best Answer

To make good use of the divergence theorem:

  • verify that the surface integrals through the parts of the coordinate planes that, together with the two given surfaces, bound the region in the first octant, don't contribute to the flux (their share is $0$, which is easy to compute or even simply reason);
  • note that due to symmetry in the two surfaces ($y \leftrightarrow z$) that make up $\partial T$ and with the given vector field, you expect the same flux through these two parts of $\partial T$.

By the divergence theorem you find a total flux of $\tfrac{8}{5}a^5$ so the flux through the part $S$ is half of this: $\tfrac{4}{5}a^5$.

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You can verify this by actually computing the surface integral. To do this, we parametrize $S$ (the paraboloid $z=a^2-x^2$ cut off by the paraboloid $y=a^2-x^2$) as follows: $$\vec r(u,v)=\left(u,v\left(a^2-u^2\right),a^2-u^2\right) \;,\; 0 \le u \le a\;,\; 0 \le v \le 1$$ Then: $$\vec N=\frac{\partial \vec r}{\partial u}\times\frac{\partial \vec r}{\partial v}=\left(2a^2u-2u^3,0,4-u^2\right)$$ and: $$\vec F \cdot \vec N=a^4-u^4$$ so: $$\iint_S \vec F \cdot \vec N\,\mbox{d}S = \int_0^1 \int_0^a \left(a^4-u^4\right)\,\mbox{d}u \,\mbox{d}v=\frac{4}{5}a^5$$

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