Divergence Theorem — Question 7

Question 7 – Solution

Strategy. The divergence equals $1$, so the net flux is the volume; the direct check tests the downward orientation on the paraboloid.

Step 1: Bounds and volume The surfaces meet at $r=2$, and $$0\le\theta\le 2\pi,\qquad 0\le r\le 2, \qquad r^2\le z\le 4.$$ Since $\nabla\cdot\mathbf F=1$, $$\begin{align*} \Phi_{\mathrm{net}} &=\int_0^{2\pi}\int_0^2\int_{r^2}^{4}r\,dz\,dr\,d\theta =\boxed{8\pi}. \end{align*}$$

Step 2: Top plane On $z=4$, the outward normal is $\mathbf k$, so $$\Phi_{\mathrm{top}}=\iint_{r\le 2}4\,dA=16\pi.$$

Step 3: Paraboloid The solid lies above the paraboloid, so its outward vector element is downward: $$\mathbf n\,dS=\langle 2x,2y,-1\rangle\,dA.$$ Thus $$\Phi_{\mathrm{paraboloid}}=\iint_{r\le 2}-r^2\,dA =-2\pi\int_0^2r^3\,dr=-8\pi.$$ The direct total is $16\pi-8\pi=8\pi$.

Verification The sign of the lower contribution is negative because the field points upward while the lower outward normal points downward.