 Solution

The divergence of a cross product identity and the stated static field equations giveFor an outward-oriented closed surface, the divergence theorem therefore gives the flux of the specified Poynting vector normalization:No open-surface conclusion follows from this equation without including its closing pieces.

For the particular fields, direct calculation givesThe fields also have $\nabla \times \mathbf{E}=0$, $\nabla \cdot \mathbf{B}=0$, and $\nabla \cdot \mathbf{J}=0$, consistently with the required equations. Close the outward spherical patch by the two planar half-disks on $x=0$ and $y=0$, forming the quarter unit ball $V$. Its total closed flux iswhere $r$ in this integral is cylindrical radius. The $x=0$ face has outward normal $-{\mathbf{e}}_x$ and zero flux because $P_x=0$ there. The $y=0$ face has outward normal $-{\mathbf{e}}_y$, so its flux isSubtracting these planar contributions from the closed flux gives the requested open-patch answerAs a direct check, on the unit sphere the outward normal is $(x,y,z)$ and $\mathbf{P}\cdot \mathbf{n}=xy(x^2-2z^2)$. With spherical angles $0\le \vartheta \le \pi$, $0\le \phi \le \pi /2$, its surface integral is $\frac{1}{4}{\int }_0^{\pi }{\sin }^5\vartheta d\vartheta -{\int }_0^{\pi }{\sin }^3\vartheta {\cos }^2\vartheta d\vartheta =\frac{1}{4}(16/15)-4/15=0$. The curved patch's vanishing flux is a cancellation, not an inference that its flux density vanishes everywhere.