$$ \int_S x\,dy\,dz+y\,dz\,dx+z\,dx\,dy, $$ where $S=\{(x,y,z)\in \mathbb{R}^{3}: x^2+y^2+z^2=1\}.$
Please, I don't know how to proceed. I will be thankful if you give me any hint, at first I thought it was an application of the Stokes' Theorem, but the differentials confuse me D: Thanks!
On
If we interpret the integral in the OP as an ordinary surface integral, then
$$\begin{align} 3 \int_S z\,dx\,dy &=3\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\sqrt{1-x^2-y^2}\,dx\,dy+3\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\left(-\sqrt{1-x^2-y^2}\right)\,dx\,dy\\\\ &=0 \end{align}$$
where we exploited the symmetry noting that $\int_S x\,dy\,dz=\int_S y\,dz\,dx=\int_S z\,dx\,dy$.
And we are done!
However, if we interpret the integral in the OP to be equivalent to the closed surface integral expressed as
$$\int_S x\,dy\wedge dz+y\,dz \wedge dx+z\,dx \wedge dy=\oint_S \vec r\cdot \hat n\,dS \tag 1$$
then we can apply the Divergence Theorem to the right-hand side of $(1)$ to reveal
$$\begin{align} \oint_S \vec r\cdot \hat n\,dS &=\int_V \nabla \cdot \vec r\,dV\\\\ &=\int_V 3\,dV\\\\ &=3\frac{4\pi}{3}\\\\ &=4\pi \end{align}$$
Alternatively, one can evaluate the surface integral directly as
$$\begin{align} 3\oint_S z\,n_z\,dS&=3\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \sqrt{1-x^2-y^2}\,dx\,dy\\\\ &+3\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\left(- \sqrt{1-x^2-y^2}\right)\,(-1)\,dx\,dy\\\\ &=6 \int_0^{2\pi}\int_0^1 \sqrt{1-\rho^2}\,d\rho\,d\phi\\\\ &=4\pi \end{align}$$
as expected!
On
$\newcommand{\Vec}[1]{\mathbf{n}}$To complement the existing good answer: In classical notation, with $dS$ denoting the area element of the unit sphere and $\Vec{n}$ the outward unit normal field, your integral represents the flux through the unit sphere of the vector field $$ F(x, y, z) = (x, y, z). $$ This can be calculated directly by noting that on the unit sphere, $$ F \cdot \Vec{n} = (x, y, z) \cdot (x, y, z) = x^{2} + y^{2} + z^{2} = 1, $$ so the flux of $F$ is $$ \iint_{S} F \cdot \Vec{n}\, dS = \iint_{S} dS = \text{area of $S$} = 4\pi. $$
The quickest way to do this is by using the Gauss-Ostrogradski theorem (which is just another name for Stokes' theorem in $\Bbb R^3$ - you have correctly guessed that this would lead to the solution): if $V$ is a 3-dimensional region, then
$$\int \limits _{\partial V} P \ \Bbb dy \Bbb dz + Q \ \Bbb dz \Bbb dx + R \ \Bbb dx \Bbb dy = \int \limits _V \left( \frac {\partial P} {\partial x} + \frac {\partial Q} {\partial y} + \frac {\partial R} {\partial z} \right) \Bbb d x \Bbb d y \Bbb d z .$$
Concretely, if $B = \{ (x,y,z) \in \Bbb R^3 \mid x^2 + y^2 + z^2 \le 1\}$, then $S = \partial B$ and
$$\int \limits _S x \ \Bbb dy \Bbb dz + y \ \Bbb dz\Bbb dx + z \ \Bbb dx \Bbb dy = \int \limits _B \left( \frac {\partial x} {\partial x} + \frac {\partial y} {\partial y} + \frac {\partial z} {\partial z} \right) \Bbb d x \Bbb d y \Bbb d z = 3 \text{vol } (B) = 4 \pi .$$
Of course, you can also do it using the definition of surface integrals of the second type, but this would require you to use a parametrization, to compute the coefficients of the metric and the formula of the outer unit normal in this parametrization and finally to perform an integral, which is way too long and tedious.