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I am doing sample test provided by my professor. One of the question is:

Find the flux integral of $$ \overrightarrow{F}=\left(x^3, y^3, z^3\right) $$ across the surface $\Sigma:x^2+y^2+z^2=1$, and oriented by normal vectors pointing away from the origin.

I do it by this way:

Let $\overrightarrow{r}(x,y,z)=\overrightarrow{r}\big(x(\theta,\varphi),y(\theta,\varphi),z(\theta,\varphi)\big)$ \begin{cases} x=\sin\theta \cos\varphi \\ y=\sin\theta\sin\varphi \\ z=\cos\theta\\ \theta\in[0,\pi], \varphi\in[0,2\pi] \end{cases} Thus: \begin{cases} r_\theta=(\cos\theta \cos\varphi,\cos\theta\sin\varphi, -\sin\theta) \\ r_\varphi=(-\sin\theta \sin\varphi,\sin\theta\cos\varphi, 0) \\ E= \|r_\theta\|^2=1, F=r_\theta \cdot r_\varphi=0, G=\|r_\varphi\|^2=\sin^2\theta\\ \sqrt{EG-F^2}=\sin\theta \end{cases}

The unit normal vector at point $(x,y,z)$ is just $(x,y,z)$ since the radius of the surface is $1$.

Then:

\begin{align*} \int_\Sigma \overrightarrow{F}\cdot \overrightarrow{n}\ \mathrm{d}\sigma&= \int_\Sigma \left(x^4+y^4+z^4\right)\ \mathrm{d}\sigma\\ &=8\int_{ \left\{\Sigma:\ x>0,y>0,z>0 \right\} } \left[\sin^4\theta(\cos^4\varphi+\sin^4\varphi)+\cos^4\theta\right]\ \mathrm{d}\sigma\\ &=8\iint_{[0,\frac{\pi}{2}]^2}\left[\sin^4\theta(\cos^4\varphi+\sin^4\varphi)+\cos^4\theta\right]\sqrt{EG-F^2}\ \mathrm{d}\theta\mathrm{d}\varphi\\ &=8\left(\int_{0}^{\frac{\pi}{2}}\sin^5{\theta}\ \mathrm{d}\theta\int_{0}^{\frac{\pi}{2}}(\cos^4\varphi+\sin^4\varphi)\ \mathrm{d}\varphi+\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos^4\theta\ \mathrm{d}\theta \right)\\ &=8\left( \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(3\right)}{\Gamma\left(\frac{7}{2}\right)}\cdot 2\cdot\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{5}{2}\right)}{\Gamma\left(3\right)}+\frac{\pi}{2}\cdot \frac{1}{2}B\left(1,\frac{5}{2}\right) \right)\\ &=8\left( \frac{\pi}{2} \cdot \frac{1}{\frac{5}{2}}+\frac{\pi}{4}\cdot\frac{1}{\frac{5}{2} } \right)\\ &=8\left(\frac{\pi}{5} + \frac{\pi}{10} \right)\\&=\frac{12\pi}{5} \end{align*}

I spent more than 15 minutes on this problem so I'm thinking of whether there is an easier way to do this problem so that I can save more time. Is there a simpler way? Thank you for your help!

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  • $\begingroup$ Surely, $\iint(x^4+y^4+z^4)\,d\sigma=3\iint z^4\,d\sigma$? $\endgroup$ – Lord Shark the Unknown Sep 1 '17 at 18:37
  • $\begingroup$ @LordSharktheUnknown Oh, I see. Then combined with Gauss's Law it would be even simpler. Thank you for your help! $\endgroup$ – Edward Wang Sep 1 '17 at 20:12
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HINT: Divergence (Gauss') Theorem would come pretty handy and reduce the amount of calculations.


Namely using it we have:

$$\iint_\Sigma \vec{F} \cdot \vec{n} \ d\sigma = \iiint_V \nabla \cdot F \ dV = 3\iiint_V x^2 + y^2 + z^2 \ dV$$

Now using the spherical coordinates we have:

$$\iint_\Sigma \vec{F} \cdot \vec{n} \ d\sigma = 3\int_0^{2\pi}\int_0^{\pi}\int_0^{1} r^4\sin \phi \ drd\phi d\theta = \frac 35 \int_0^{2\pi}\int_0^{\pi} \sin \phi \ d\phi d\theta = \frac 65 \int_0^{2\pi} \ d\theta = \frac{12}{5}\pi$$

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