TLDR: read the final paragraph. For my course I have to evaluate $$\int_{S^2}d\vec R\ R_iR_j\frac{\partial^2G(\vec R)}{\partial R_i\partial R_j}$$ Here the Einstein summation convention is used, $G$ is a still unknown function, $\vec R$ is the position vector in $\mathbb R^3$ and we integrate over the unit sphere so $\vec R^2=x^2+y^2+z^2=1$ and $d\vec R$ is just the surface element on the sphere.
To evaluate this the teaching assistent gave us the hint that we can assume $G$ is spherically symmetric. This means that any term with $i\neq j$ will vanish because you then integrate an odd function $R_iR_j$ over a symmetric domain. This means the integrand becomes $$R_i^2\frac{\partial^2G(\vec R)}{\partial R_i^2}=x^2\frac{\partial^2 G}{\partial x^2}+y^2\frac{\partial^2 G}{\partial y^2}+z^2\frac{\partial^2 G}{\partial z^2}$$ The final step is that because of spherical symmetry we can replace $\frac{\partial^2 G}{\partial x^2}=\frac{\partial^2 G}{\partial y^2}=\frac{\partial^2 G}{\partial z^2}=\frac 1 3\nabla^2G$ which will then greatly simplify the integral. I don't understand this last step so that's what my question is about. I tried to make sense of it but couldn't see how to do it.
$$\frac{\partial^2 G}{\partial x\partial y} = \frac{\partial^2 G}{\partial y\partial x}$$
$$\eqalign{\frac{\partial^2 G}{\partial x\partial y}&=\frac{\partial (G(\vec R+\Delta y\mathbf{\widehat j})-G(\vec R))}{\partial x}\\&=\frac{ G(\vec R + \Delta x\mathbf{\widehat i}+ \Delta y\mathbf{\widehat j})-G(\vec R + \Delta y\mathbf{\widehat j})- G(\vec R+\Delta x\mathbf{\widehat i})+G(\vec R)}{\Delta x}\\ &=\frac{G(\vec R + 2\Delta x\mathbf{\widehat i})-2G(\vec R + \Delta x\mathbf{\widehat j})+G(\vec R)}{\Delta x}=\frac{\partial^2G}{\partial^2 x} \;\; as \;\; \Delta x,\Delta y \rightarrow 0}$$
That's because of spherical symmetry that for example $G(\vec R+\Delta x\mathbf{\widehat i})=G(\vec R + \Delta y\mathbf{\widehat j})$
The same will go for $y \; \text{and} \; z$