The question goes like this:
Evaluate $\displaystyle \iint_S (yz\,\hat{\imath}+zx\,\hat{\jmath}+xy\,\hat{k})\,dS$ where $S$ is the surface of the sphere $x^2+y^2+z^2=a^2$ in the first octant.
I saw this and my first thought was Stokes' theorem, but I considered the possibility of working through this "the hardest way possible".
The surface can be parametrized as follows: $S: f(x,y) = \begin{bmatrix} x \\ y \\ \sqrt{a^2-(x^2+y^2)} \end{bmatrix} \begin{cases}0\le x \le \sqrt{a^2-y^2} \\ 0 \le y \le a\end{cases}$
The bound for $x$ is necessary as it is to ensure the bound is precisely defined.
While working through this I found it helpful to remember that $\sqrt{a^2-(x^2+y^2)}=z$. It's part of how I made this easier than surface integrals normally are.
\begin{align} \dfrac{\partial f}{\partial x} &= \begin{bmatrix} 1 \\ 0 \\ \dfrac{-x}{\sqrt{a^2-(x^2+y^2)}} \end{bmatrix} \\ \dfrac{\partial f}{\partial y} &= \begin{bmatrix} 0 \\ 1 \\ \dfrac{-y}{\sqrt{a^2-(x^2+y^2)}} \end{bmatrix} \\ \\ \hat{n}\,d\Sigma &= \dfrac{\partial f}{\partial x} \times \dfrac{\partial f}{\partial y} \\ &= \det\begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 0 & \dfrac{-x}{\sqrt{a^2-(x^2+y^2)}} \\ 0 & 1 & \dfrac{-y}{\sqrt{a^2-(x^2+y^2)}} \end{pmatrix} \\ &= \begin{bmatrix} \dfrac{x}{\sqrt{a^2-(x^2+y^2)}} \\ \dfrac{y}{\sqrt{a^2-(x^2+y^2)}} \\ 1 \end{bmatrix} = \begin{bmatrix} \dfrac{x}{z} \\ \dfrac{y}{z} \\ 1 \end{bmatrix} \\ \\ \vec{F} \cdot \hat{n}\,d\Sigma &= \begin{bmatrix} yz \\ zx \\ xy \end{bmatrix} \cdot \begin{bmatrix} \dfrac{x}{z} \\ \dfrac{y}{z} \\ 1 \end{bmatrix} \\ &= xy + xy + xy \\ &= 3xy \end{align}
So the whole integral simplifies to $\displaystyle \int_0^a \int_0^{\sqrt{a^2-y^2}} 3xy \,dx\,dy$.
Is this correct so far?