We have that
\begin{align*}
F(r(u,v)) &= F(u \cos v, u \sin v, v) \
&= u \sin v i - u \cos v j + v^2 k.
\end{align*}Thus,
\begin{align*}
F \cdot r_u &= (u \sin v) (\cos v) + (u \cos v) (-\sin v) + (v^2)(0) \
&= 0, \
F \cdot r_v &= (u \sin v) (-\sin v) + (u \cos v) (\cos v) + (v^2)(1) \
&= v^2.
\end{align*}Therefore,
[\mathbf{F} \cdot \mathbf{n} = F \cdot (r_u \times r_v) = F \cdot \mathbf{k} = v^2.]Thus,
\begin{align*}
\iint_S \mathbf{F} \cdot d\mathbf{S} &= \iint_S \mathbf{F} \cdot \mathbf{n} , dA \
&= \iint_S v^2 , dA \
&= \int_0^{4 \pi} \int_0^2 v^2 , du , dv \
&= \int_0^{4 \pi} \left. \frac{1}{3} u^3 \right|_0^2 , dv \
&= \int_0^{4 \pi} \frac{8}{3} , dv \
&= \frac{8}{3} \cdot 4 \pi \
&= \boxed{16 \pi}.
\end{align*}
Evaluate the surface integral
S
F · dS
for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x, y, z) = y i − x j + z2 k
S is the helicoid (with upward orientation) with vector equation
r(u, v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 4π
1 answer