Let $R$ be a region in the plane, and let $P$ be a point at a height $h$ above the plane. Form a cone by drawing lines from $P$ to each point on the boundary of $R$, and define a vector field by $x \Bbb i + y \Bbb j + z \Bbb k$. Denote $D$ as the region in space that is bounded above by the cone, and bounded below by $R$.
Show that the outward flux of the vector field through the boundary of $D$ is $hA$, where $A$ is the area of $R$.
In addition, use the divergence theorem to show that $D$'s volume is $\dfrac {\pi r^2} 3$.
Best Answer
By the divergence theorem, the flux equals
$$ \phi=\iiint_E \nabla \cdot \vec{F}\; dV = 3 V(E), $$
and since $E$ is a cone with basis $A$ and height $h$: $$ \phi = 3 V(E)=3 \frac{A h}{3}=Ah. $$