Evaluate $\iint _{{S}}(y^{2}z^{2}i+z^{2}x^{2}j+x^{2}y^{2}k)\cdot \,ds$ where $S$ is the part of the sphere $x^{2}+y^{2}+z^{2}=1$\, above the xy-plane.
Answer to this question is $\pi/24$ as given in this link but when I try to evaluate it using Gauss's Divergence theorem it fails. Clearly all partial derivatives are zero. Hence I get the answer is zero. What is the reason that divergence theorem does not provide the answer or fails here?
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As pointed out above the divergence theorem requires a closed surface. We can create a closed surface by adding to $S$ a surface $S'$ defined by $x^2+y^2\leq 1$ for $z=0$ (see plot below; the arrow denotes the surface normal for $S'$).
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We can apply the divergence theorem to the closed surface $S\cup S'$ to get
$$\int_{S\,\cup \,S'} \vec{F}\cdot {\rm d}\vec{S} = \int_{S} \vec{F}\cdot {\rm d}\vec{S} + \int_{S'} \vec{F}\cdot {\rm d}\vec{S}' = \int_V \nabla\cdot \vec{F}\,{\rm d}V$$
where $V$ is the region $x^2+y^2+z^2\leq 1$ for $z\geq 0$. With $\vec{F} = (y^2z^2,\,x^2z^2,\,x^2y^2)$ we have as you have found $\nabla\cdot \vec{F} = 0$ so
$$\int_{S} \vec{F}\cdot {\rm d}\vec{S} = -\int_{S'} \vec{F}\cdot {\rm d}\vec{S}'$$
To evaluate the surface integral over $S'$ we have $\vec{d}S' = -\vec{k}\,{\rm d}x{\rm d}y$ giving us
$$\int_{S} \vec{F}\cdot {\rm d}\vec{S} = \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} x^2y^2\,{\rm d}x{\rm d}y = \frac{4}{3}\int_{0}^1x^2(1-x^2)^{3/2}\,{\rm d}x$$
which can be evaluated (using for example the $\beta$-function) to get the desired result of $\frac{\pi}{24}$.
According to Gauss's Divergence theorem, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface.
As mentioned in your question, S is the part of the sphere $$x^2+y^2+z^2=1$$ above the xy-plane. That is, S is a hemisphere with no bottom plane surface. Hence it is not a closed surface. Divergence theorem does not apply here.