Surface Integrals of Vector Fields — Question 5

Question 5 – Solution

Strategy. The normal pointing horizontally away from the axis also has a downward vertical component, which determines the sign of the flux.

Step 1: Oriented cross product Use $$\mathbf r(u,v)=\langle u\cos v,u\sin v,u\rangle, \quad 0\le u\le 2,\quad 0\le v\le 2\pi.$$ The ordered product $$\boxed{\mathbf r_v\times\mathbf r_u =\langle u\cos v,u\sin v,-u\rangle}$$ has a radial component pointing away from the axis.

Step 2: Integrate On the cone, $\mathbf F=\langle 0,0,u\rangle$, so $$\mathbf F\cdot(\mathbf r_v\times\mathbf r_u)=-u^2.$$ Therefore $$\begin{align*} \iint_S\mathbf F\cdot\mathbf n\,dS &=\int_0^{2\pi}\int_0^2-u^2\,du\,dv\\ &=\boxed{-\frac{16\pi}{3}}. \end{align*}$$

Step 3: Vertex The parametrization is singular at $u=0$, but that single surface point has zero area. Equivalently, integrate over $\varepsilon\le u\le 2$ and let $\varepsilon\to 0^+$.

Verification The chosen normal has negative $z$-component while the field points upward, so a negative flux is required.