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Can anyone help me please :(

Calculate the surface integral of the $2$nd type: $$ \iint_S \dfrac{x^2\,dy\,dz + y^2\,dz\,dx + z^2\,dx\,dy}{x^3+y^3+z^3} $$ where $S$ is the outer side of a sphere $x^2+y^2+z^2=3$.

Felix Marin
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ghostio
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  • sure. Have you done a surface integral before? if you have, what part of the process worked there that doesn't work here? – Jackozee Hakkiuz Jun 13 '24 at 13:07
  • I have done some exercises but I have never seen such a difficult one like this. I've tried using the polar coordinate but the integrand is so complicated – ghostio Jun 13 '24 at 13:17
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    Have you tried to use the spherical coordinates to calculate the integral? – xpaul Jun 13 '24 at 15:28
  • if so, then I have to use the divergence theorem but I think it would be more complicated :( – ghostio Jun 13 '24 at 15:35
  • @ghostio Divergence theorem is probably the way to go. – Abezhiko Jun 13 '24 at 16:16
  • @Abezhiko does divergence theorem apply? there is a singularity at the origin, so $d\omega$ isn't defined there. – Jackozee Hakkiuz Jun 13 '24 at 23:22
  • The integrand reduces nicely (albeit with a lot of trigonometric manipulation) if you parameterize $S$ with the right set of spherical coordinates. – user170231 Jun 13 '24 at 23:23
  • @user170231 the usual spherical change of coordinates doesn't work as nicely? To be clear, I haven't done the work. – Jackozee Hakkiuz Jun 13 '24 at 23:27
  • @JackozeeHakkiuz The singularity will be "compensated" by the Jacobian of spherical coordinates. Indeed, the integrand behaves like $\frac{1}{r}$, while the Jacobian is proportional to $r^2$. – Abezhiko Jun 15 '24 at 17:52

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The vector field $$ \boldsymbol{F}=\frac1{x^3+y^3+z^3}\pmatrix{x^2\\y^2\\z^2} $$ is parallel to the unit outward normal at the sphere $$ \boldsymbol{n}=\frac1{\sqrt{x^2+y^2+z^2}}\pmatrix{x\\y\\z} $$ and $$ \boldsymbol{F}\cdot\boldsymbol{n}=\frac1{\sqrt{x^2+y^2+z^2}}=\frac1r\,. $$ Therefore, \begin{align} &\int_S\frac{x^2\,dy\wedge dz+y^2\,dz\wedge dx+z^2\,dx\wedge dy}{x^3+y^3+z^3} &=\int_S\boldsymbol{F}\cdot\boldsymbol{n}\,dS=\frac1r (4\pi r^2)=4\pi r\,. \end{align}

Alternatively, you can go the hard way and take spherical polar coordinates. This leads to

\begin{align*} &\int_S\frac{x^2}{x^3+y^3+z^3}\,dy\wedge dz=r\int_0^{2\pi}\int_0^{\pi}\frac{\cos^3\varphi\sin^4\theta}{\cos^3\varphi\sin^3\theta+\sin^3\varphi\sin^3\theta+\cos^3\theta}\,d\theta\,d\varphi\,,\\[2mm] &\int_S\frac{y^2}{x^3+y^3+z^3}\,dz\wedge dx=r\int_0^{2\pi}\int_0^{\pi}\frac{\sin^3\varphi\sin^4\theta}{\cos^3\varphi\sin^3\theta+\sin^3\varphi\sin^3\theta+\cos^3\theta}\,d\theta\,d\varphi\,,\\[2mm] &\int_S\frac{z^2}{x^3+y^3+z^3}\,dx\wedge dy=r\int_0^{2\pi}\int_0^{\pi}\frac{\cos^3\theta\sin\theta}{\cos^3\varphi\sin^3\theta+\sin^3\varphi\sin^3\theta+\cos^3\theta}\,d\theta\,d\varphi\,. \end{align*} The sum of these integrals is $$ r\int_0^{2\pi}\int_0^{\pi}\sin\theta\,d\theta\,d\varphi=r4\pi\,. $$

Kurt G.
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  • But sir why that is the cross product while $F.dS$ is the scalar product – ghostio Jun 14 '24 at 16:51
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    That is because Elie Cartan got the very good idea to invent differential forms when he was drinking a lot of absinthe one day. – Kurt G. Jun 14 '24 at 17:45
  • @ghostio Joking aside. The relationship between the differential form integral and the $dS$-integral is something fundamental that you should learn and remember. This is not the place to repeat it. Studying the literature and practicing examples is the way to go. – Kurt G. Jun 14 '24 at 17:54
  • The wedge product of differential forms is not cross product of vectors, although in $\Bbb R^3$ they are related. Your $dx,dy$ notation for a surface integral omits the wedge, although, as @Kurt wrote, it should be there for rigor. If you want to see a more detailed, careful treatment of differential forms and surface integrals, you might look at my YouTube lectures, linked in my profile. – Ted Shifrin Jun 14 '24 at 18:45