Surface Integrals of Vector Fields — Question 2

Question 2 – Solution

Strategy. Use the upward graph element $\langle-g_x,-g_y,1\rangle dA$ and then exploit radial symmetry.

Step 1: Vector area element For $g(x,y)=x^2+y^2$, $$\boxed{\mathbf n\,dS=\langle-2x,-2y,1\rangle\,dA}.$$ Thus $$\mathbf F\cdot\mathbf n\,dS =(-2x^2-2y^2+1)\,dA=(1-2r^2)\,dA.$$

Step 2: Integrate $$\begin{align*} \iint_S\mathbf F\cdot\mathbf n\,dS &=\int_0^{2\pi}\int_0^1(1-2r^2)r\,dr\,d\theta\\ &=2\pi\left[\frac{r^2}{2}-\frac{r^4}{2}\right]_0^1 =\boxed{0}. \end{align*}$$

Step 3: Locate the cancellation The normal component is positive for $r<1/\sqrt 2$ and negative for $r>1/\sqrt 2$. The inner contribution is $$2\pi\left[\frac{r^2}{2}-\frac{r^4}{2}\right]_0^{1/\sqrt 2}=\frac\pi 4,$$ while the outer annulus contributes $-\pi/4$.

Verification The exact cancellation explains why zero flux does not mean the field is tangent everywhere.