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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:


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

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


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

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

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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I haven't checked the details of your calculations, but in the first case I have two comments: First, the region you're integrating over is the inside of the unit circle, so you should now switch to polar coords. But you've also done only the upper hemisphere, so you need to add another integral. Be careful with the way the normal is pointing!

The second integral is doable, using symmetry of trig functions.

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 $$\iint_S x^3 dA = 0\,.$$

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