# calculating a surface integral $\iint\vec{F}\cdot\vec{n}dA$

I am given that the vector field is $\vec{F}=[x^2, z, -y]^T$, the surface is the unit sphere, i.e $S = \{(x,y,z)\in\mathbb{R}^3|x^2 + y^2 + z^2 = 1\}$, and I need to calculate the following integral:

$$\iint_S\vec{F}\cdot\vec{n}dA$$

What I tried: I parametrized the surface $S$ in two ways, neither of which led me to a meaningful solution. Let $G(u,v)=[u,v,\sqrt{1-u^2-v^2}]^T$ be the natural parametrization of $S$, then we have that

$$\frac{\partial G}{\partial u}\times\frac{\partial G}{\partial v} = [\frac{u}{\sqrt{1-u^2-v^2}},\frac{v}{\sqrt{1-u^2-v^2}},1]^T$$

I'm not even sure what the bounds for $u$ and $v$ are: I assumed they would be $-1\leq u \leq 1$ and $-1\leq v\leq 1$, but I'm pretty sure this is wrong. So then with this parametrization we get that

$$\vec{F} = [u^2,\sqrt{1-u^2-v^2},-v]^T$$

so

$$\iint_S\vec{F}\cdot\vec{n}dA = \iint_W \vec{F}(G(u,v))\cdot\left(\frac{\partial G}{\partial u}\times\frac{\partial G}{\partial v}\right)dudv$$

$$= \iint_W\frac{u^3}{\sqrt{1-u^2-v^2}}dudv$$

where the region $W$ is in the $uv$-plane. I have no idea on how to calculate this. I tried using polar but I got no where. The other parametrization I tried was the spherical coordinates, so

$$G(\psi, \theta)=[\sin\psi \cos\theta, \sin\psi \sin\theta, \cos\psi]^T$$

since the radius of the sphere is 1. Doing the same process again gives

$$\iint_W \sin^4\psi \cos^3\theta - \cos\psi\sin\theta\sin^2\psi - \sin^2\psi\sin\theta\cos\psi d\psi d\theta$$

and I don't think I can integrate this. Anything I'm doing wrong? This isn't homework, I'm studying for a final.

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But the best solution is to exploit symmetry as much as possible. Note that $$F\cdot n = x^3 + zy -yz = x^3$$ and you should be able to convince yourself easily that $$\iint_S x^3 dA = 0\,.$$