Using the Divergence Theorem, evaluate int int_S F.NdS, where F(x,y,z)=(z^3i-x^3j+y^3k) and S is the sphere x^2+y^2+z^2=a^2, with outward unit normal vector N.

Using the Divergence Theorem, evaluate $\int {\int }_{S}F.NdS$, where $F\left(x,y,z\right)=\left({z}^{3}i-{x}^{3}j+{y}^{3}k\right)$ and S is the sphere ${x}^{2}+{y}^{2}+{z}^{2}={a}^{2}$, with outward unit normal vector N.
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Step 1
Divergence theorem relates surface integrals and volume integrals. By using the Gauss divergence theorem we can evaluate this surface integral.
The Gauss divergence theorem formula can be stated as follows:
$\int {\int }_{S}F.NdS=\int \int {\int }_{V}÷F.dV$
$÷F=\left(\partial \frac{{z}^{3}i}{\partial x}\right)+\left(\partial \frac{-{x}^{3}j}{\partial y}\right)+\left(\partial \frac{{y}^{3}k}{\partial z}\right)$
Step 2
div F = 0, since the partial derivative of ${z}^{3}$ with respect to x is zero and partial derivative of $-{x}^{3}$ with respect to y is zero and the partial derivative of ${y}^{3}$ with respect to z is zero.
Then $\int \int {\int }_{V}0.dV=0$
Since $\int {\int }_{S}F.NdS=\int \int {\int }_{V}÷F.dV$
Then by Gauss Divergence theorem, we can say $\int {\int }_{S}F.NdS=0$.
Step 3
So the final answer is zero by the Gauss Divergence theorem.