Integral over hemiellipsoid

Question

I have to solve the the following integral $\int_{D}^{} A \cdot dS$, where $D$ is the surface of the hemiellipsoid $(\frac{x}{a})^2+ (\frac{y}{b})^2 + (\frac{z}{c})^2= 1$ a,b and c are known. It is also given from the task that $A$ can be written as $A=(B \cdot r)C$, where $B$ and $C$ are constant vectors, and $r$ is the position vector. From what I see, it seems easiest to use Gauss law for integration, where we instead of integrating over the surface of the hemiellipsoid, we integrate over the volume. However, I don't seem to know what to do with the expression $\nabla \cdot A = \nabla \cdot ((B \cdot r)C)$.

EDIT

I've tried to expand the expression above to $\nabla \cdot A = \nabla \cdot ((B \cdot r)C)= \nabla \cdot (BC \cdot rC)= (\nabla \cdot BC)\cdot rC = 0$, since $BC$ is a constant vector, which should mean that the divergence is 0. But that means that the entire integral will be $0$, which seems odd.

Answer 1

As Ted Shifrin noted, your expression for the divergence of A is incorrect. Resolving it into components, $$r=\begin{bmatrix}x\\y\\z\end{bmatrix}\\A=(B_xx+B_yy+B_zz)\begin{bmatrix}C_x\\C_y\\C_z\end{bmatrix}\\\nabla\cdot{A}=C_xB_x+C_yB_y+C_zB_z=C\cdot{B}$$ Thus, $$\int_DA\cdot{dS}=C\cdot{B}\iiint_VdV$$ Using the formula for the volume of an ellipse and dividing by two, you get: $$\int_DA\cdot{dS}=C\cdot{B}\frac4 6\pi{abc}$$

Answer 2

As an alternative to fiddling with components, use the dyadic product. $$\eqalign{ A &= (CB)\cdot r \\ dA &= (CB)\cdot dr \quad&\big({\rm differential}\big) \\ \nabla A &= CB \quad&\big({\rm gradient}\big) \\ \nabla\cdot A &= C\cdot B \quad&\big({\rm divergence}\big) \\ }$$ Then apply the divergence theorem over the hemi-ellipsoidal region $\Omega$. $$\eqalign{ \oint_{\partial\Omega}A\cdot dS &= \int_\Omega\left(\nabla\cdot A\right)\;dV \\ &= \int_\Omega(C\cdot B)\;dV \\ &= (C\cdot B)\int_\Omega dV \\ &= (C\cdot B)\left(\frac{2\pi abc}{3}\right) \\ }$$