91Ó°ÊÓ

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Plane } \mathbf{F}=2 x y \mathbf{i}+2 y z \mathbf{j}+2 x z \mathbf{k} \text { upward across the portion of }} \\ {\text { the plane } x+y+z=2 a \text { that lies above the square } 0 \leq x \leq a,} \\\ {0 \leq y \leq a, \text { in the } x y-\text { plane }}\end{array}\)

a. A torus of revolution (doughnut) is obtained by rotating a circle \(C\) in the \(x z\) -plane about the \(z\) -axis in space. (See the accompanying figure.) If \(C\) has radius \(r>0\) and center \((R, 0,0),\) show that a parametrization of the torus is $$\begin{aligned} \mathbf{r}(u, v)=&((R+r \cos u) \cos v) \mathbf{i} \\\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned}$$ where \(0 \leq u \leq 2 \pi\) and \(0 \leq v \leq 2 \pi\) are the angles in the figure. b. Show that the surface area of the torus is \(A=4 \pi^{2} R r\)