= Solution
{c}
Use <cylindrical coordinates> $(\rho,\theta,z)$. The <right circular cone> is described by $0\leq z\leq1$ and $0\leq\rho\leq a(1-z)$, and the <divergence> of $\mathbf F=(0,0,z^2)$ is $2z$. Direct <integration> gives
$$
\begin{aligned}
\iiint_R\nabla\cdot\mathbf F\,dV
&=\int_0^1\int_0^{2\pi}\int_0^{a(1-z)}2z\rho\,d\rho\,d\theta\,dz\\
&=2\pi a^2\int_0^1z(1-z)^2\,dz\\
&=2\pi a^2\left(\frac12-\frac23+\frac14\right)
=\boxed{\frac{\pi a^2}6}.
\end{aligned}
$$
To check the boundary <flux integral> independently, first consider the base $z=0$. Its outward <unit normal vector> is $-\mathbf k$, but $\mathbf F=\mathbf0$ there, so its <flux> is zero. Parametrize the lateral <surface> by
$$
\mathbf s(\theta,z)=\bigl(a(1-z)\cos\theta,a(1-z)\sin\theta,z\bigr),
\qquad0\leq\theta\leq2\pi,\quad0\leq z\leq1.
$$
The <cross product>
$$
\mathbf s_\theta\times\mathbf s_z
=\bigl(a(1-z)\cos\theta,a(1-z)\sin\theta,a^2(1-z)\bigr)
$$
points outward: its horizontal component points away from the axis of the <cone>. This gives the <vector surface element> without separately normalizing it. Consequently the lateral <flux integral> is
$$
\int_0^1\int_0^{2\pi}
\mathbf F(\mathbf s)\cdot(\mathbf s_\theta\times\mathbf s_z)\,d\theta\,dz
=2\pi a^2\int_0^1z^2(1-z)\,dz
=2\pi a^2\left(\frac13-\frac14\right)
=\boxed{\frac{\pi a^2}6}.
$$
Adding the zero base contribution agrees with the <volume integral>, verifying the <divergence theorem>. The apex has zero <surface area>; the <vector field> is smooth there, so no additional <flux> is hidden at that point.
Back to article page
