Divergence Theorem — Question 4

Question 4 – Solution

Strategy. The origin is excluded from the shell, so the field is smooth there and has zero divergence; the inner normal points toward the origin.

Step 1: Apply the theorem For $r>0$, $$\nabla\cdot\left(\frac{\langle x,y,z\rangle}{r^3}\right)=0.$$ Therefore $$\boxed{\iint_{\partial E}\mathbf F\cdot\mathbf n\,dS =\iiint_E0\,dV=0}.$$

Step 2: Outer sphere At $r=b$, $\mathbf F$ points in the outward-normal direction with normal component $1/b^2$. Hence $$\Phi_{r=b}=\frac 1{b^2}(4\pi b^2)=4\pi.$$

Step 3: Inner sphere The outward normal for the shell at $r=a$ points into the cavity, namely $-\mathbf e_r$. Thus $$\Phi_{r=a}=-\frac 1{a^2}(4\pi a^2)=-4\pi.$$ The two boundary components sum to zero.

Verification The cancellation depends on reversing the normal at the inner boundary; using the radial normal on both spheres would not describe the shell’s outward orientation.