Surface Integrals of Vector Fields — Question 10

Question 10 – Solution

Strategy. Decompose the closed surface into three oriented pieces and track the outward normal on each one.

Step 1: Curved side Parametrize by $$\mathbf r(\theta,z)=\langle\cos\theta,\sin\theta,z\rangle.$$ The outward vector element is $\langle\cos\theta,\sin\theta,0\rangle\,d\theta\,dz$. Its dot product with $\mathbf F$ is $1$, so $$\Phi_{\mathrm{side}}=\int_0^{2\pi}\int_0^2 1\,dz\,d\theta=4\pi.$$

Step 2: Caps On the top disk $z=2$, the outward normal is $\mathbf k$, and $$\mathbf F\cdot\mathbf k=-2z=-4.$$ Thus $\Phi_{\mathrm{top}}=-4\pi$. On the bottom disk $z=0$, the outward normal is $-\mathbf k$, but the vertical field component is zero, so $\Phi_{\mathrm{bottom}}=0$.

Step 3: Net flux $$\boxed{\Phi_{\mathrm{net}}=4\pi-4\pi+0=0}.$$

Verification The side carries $4\pi$ units outward while the top carries the same amount inward; the signed balance is zero even though neither contribution vanishes.