Divergence Theorem — Question 1

Question 1 – Solution

Strategy. Replace the closed-surface flux by a volume integral over the radius-$2$ ball.

Step 1: Divergence $$\boxed{\nabla\cdot\mathbf F=1+1+1=3}.$$ If $E$ is the ball bounded by $S$, then $$\operatorname{vol}(E)=\frac 43\pi(2^3)=\frac{32\pi}{3}.$$

Step 2: Apply the theorem $$\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =\iiint_E3\,dV=3\left(\frac{32\pi}{3}\right)=32\pi}.$$

Step 3: Direct check On the sphere, the outward unit normal is $\mathbf n=\langle x,y,z\rangle/2$. Therefore $$\mathbf F\cdot\mathbf n=\frac{x^2+y^2+z^2}{2}=2.$$ Multiplying by the surface area $4\pi(2^2)=16\pi$ gives $2(16\pi)=32\pi$.

Verification The radial field points outward everywhere, so the positive sign is required.