Surface integral with unit sphere

Question

Let S be the surface defined by $$z^2+y^2+x^2=1.$$ Compute the surface integral $$\int_S -z^2-y^2-x^2dS.$$

How can I approach this? I've taken a look at multiple examples, but I don't think I've understood them properly. Thank you in advance!

Answer 1

This question actually allows a nice quick solution. Since our integrand can be written as $-(x^2+y^2+z^2)$, and this equals $-1$ on $S$, we can rewrite the integral as $-\int_{S}dS$ which is minus the surface area of the unit sphere and so our integral equals $-4\pi$.

More generally, we would look to parametrise the surface $S$ in terms of some function $\mathbf{r}(u,v)$, and then substitute $dS=|\mathbf{r}_u \wedge \mathbf{r}_v|dudv$. In this particular case, spherical polars would be a natural choice.

Answer 2

On $S$, $x^2+y^2+z^2=1$, thus, $$\int_S-(x^2+y^2+z^2)=-\int_S 1$$ Where $\int_S 1$ is simply the area of the unit sphere, $8\pi^2$. $$\int_S -z^2-x^2-y^2=-8\pi^2$$