# Surface integral of the second type [closed]

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 $$x^2+y^2+z^2=3$$.

• sure. Have you done a surface integral before? if you have, what part of the process worked there that doesn't work here? Commented Jun 13 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 Commented Jun 13 at 13:17
• Have you tried to use the spherical coordinates to calculate the integral? Commented Jun 13 at 15:28
• if so, then I have to use the divergence theorem but I think it would be more complicated :( Commented Jun 13 at 15:35
• @ghostio Divergence theorem is probably the way to go. Commented Jun 13 at 16:16

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}
\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 $$r\int_0^{2\pi}\int_0^{\pi}\sin\theta\,d\theta\,d\varphi=r4\pi\,.$$
• But sir why that is the cross product while $F.dS$ is the scalar product Commented Jun 14 at 16:51
• @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. Commented Jun 14 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. Commented Jun 14 at 18:45