# Evaluating the Surface Integral $\iint_{x^3+y^3+z^3=a^3} \frac{\bf{x}}{||\bf{x}||} \cdot d\bf{S}$

Compute the surface integral: $$\int_S({x\over \sqrt{x^2+y^2+z^2}}, {y\over \sqrt{ x^2+y^2+z^2}}, {z\over \sqrt{x^2+y^2+z^2}}) \cdot \vec n \ dS$$

where $S: x^3+y^3+z^3=a^3$

The first parametrization that came to my mind was: $r(x,y)=(x,y,(a^3-x^3-y^3)^{1/3})$ but the integral becomes very hard to compute; I also gave

$$r(u,v)=(a(\cos(u)\sin(v))^{2/3},a(\sin(u)\sin(v))^{2/3},a(\cos(v))^{2/3})$$

(I was thinking about some type of spherical transformation) but then again the integral becomes vey hard to compute. Can you please help me with this problem? I would really appreciate it :)

• Oh good god. I can't help but this seems really cruel to me. I would think spherical coordinates would be best, but then your surface would be really strangely parameterized. This is one of the things I absolutely abhor about multivariable calculus. Professors think it's cute to give these really obnoxious problems that don't test understanding but rather they test your patience. – Cameron Williams May 21 '15 at 4:05
• I guess you can use divergent theorem..... (Not sure if it will be easier then) – user99914 May 21 '15 at 4:12
• I don´t know, I´m almost sure that there is a parametrization that simplifies everything (I think I´m being optimistic) – user128422 May 21 '15 at 4:17
• @copper.hat the equation has cubes, intentionally or not... – user147263 May 21 '15 at 4:27
• This surface is known as the Fermat Cubic: en.wikipedia.org/wiki/Fermat_cubic. Wikipedia gives a parameterization, but I'm not sure if it makes things easier. – JimmyK4542 May 21 '15 at 4:31

This is not a complete answer. I did some progress and convert the surface integral into an improper double integral. So anyone willing to find a solution may use the final result in this post.

Your surface for $$a=1$$ and ratio $$1:1:1$$ between the $$x$$, $$y$$, and $$z$$ axis looks like this

$$\qquad \qquad \qquad \,\,$$

so it is not a closed surface and we cannot use the divergence theorem as it was mentioned in the comments too. Next, note that

$$\begin{array}{} g(x,y,z)=x^3+y^3+z^3-a^3 \\ {\bf{F}} = \frac{\bf{x}}{\left\| {\bf{x}} \right\|}= {x\over \sqrt{x^2+y^2+z^2}}{\bf{i}} + {y\over \sqrt{ x^2+y^2+z^2}} {\bf{j}} + {z\over \sqrt{x^2+y^2+z^2}} {\bf{k}} \\ {d\bf{S}}= {1 \over {\partial{g} \over \partial z}} \nabla g \, dx dy \\ {1 \over {\partial{g} \over \partial z}}\nabla g=\frac{1}{z^2}(x^2 {\bf{i}} + y^2 {\bf{j}} + z^2 {\bf{k})} \\ {\bf{F}} \cdot {d\bf{S}} = \frac{x^3+y^3+z^3}{z^2 \sqrt{x^2+y^2+z^2}} dx dy = \frac{a^3}{z^2 \sqrt{x^2+y^2+z^2}} dx dy\\ \end{array}$$

and finally the surface integral becomes

$$\iint_{S} {\bf{F}} \cdot {d\bf{S}} = \int_{x=-\infty}^{+\infty} \int_{y=-\infty}^{+\infty} \frac{a^3}{z^2 \sqrt{x^2+y^2+z^2}}dydx \tag{*}$$

and note that $$z$$ is a function of $$x$$ and $$y$$ by the relation

$$z=(a^3-x^3-y^3)^{\frac{1}{3}}$$

Now, one may work on $$(*)$$ to obtain some result.