Calculating a Volume Integral with the Divergence Theorem

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Hello I have been trying to apply the Divergence Theorem to the following problem but i seem to either missinterpert or not understand the problem.

The following integral should be calculated with the Divergence Theorem and directly.

$\int_{A}xy+yz+xz$ $dA$ where $A=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2<1\}\cup\{(x,y,z)\in\mathbb{R}^3: x,y,z>0\}$

Now im aware that A is the positive octant of the ball $B_\delta(0)$ around the origin, $\delta > 0$

Calculating the integral directly i would use spherical coordinates and the fact that $x^2+y^2+z^2=r^2<1$ to determine the boundaries of the volume integral.

Now regarding the Divergence Theorem we know that $\nabla\cdot f=xy+yz+xz$. $A\subset B_\delta (0)$ is bounded and closed since $B_\delta (0)$ is bounded and closed $\forall n\in\mathbb{N}$ and by Heine-Borel compact. $\nabla\cdot f$ is smooth on $A$ $\implies$ f is smooth on $A$.

We can use the Divergence Theorem now: $\int_{\partial A}f\cdot n$ $ds$ $=$ $\int_{A}\nabla\cdot f$ $dA$ $=$ $\int_{A}xy+yz+xz$ $dA$. Where the first integral is the surface integral over the boundary of A and n the normal vector on the surface of A. This is where I began to struggle since one can't determine the vector field from $\nabla \cdot f$.

Any tips or help would be greatly apprieciated. Thank you in advance.

edit: $dA=d(x,y,z)$

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You could say $F = xyz(\mathbf i+\mathbf j+\mathbf k)$

Then $\nabla \cdot F = xy + yz + xz$

And $\iint (xyz \mathbf i+xyz\mathbf j+xyz\mathbf k)\cdot n \ dS = \iiint xy + yz + xz \ dV$

With $S$ as the surface of the unit sphere, and $V$ as the volume of the sphere.

If the region is the portion of the sphere in the first octant, you will need to also evaluate the portion of the $xy,yz,$ and $xz$ planes that bound the solid.