# Stokes' Theorem Help

Evaluate the flux integral $$\iint_S\operatorname{curl}(\vec{F}) \cdot d\vec{S}$$ for the vector field $\vec{F}(x,y,z) = \langle(x^9 + y^7)z^5, x, y \rangle$, where $$S: \frac{x^2 + y^2}{16} + z^8 = 1, \ z\ge 0$$ and is oriented upwards.

So, we're going to use Stokes' theorem here. The first step is to parametrize the boundary curve as $\vec{r}(t) = \langle \cos(t), 4\sin(t), 0 \rangle , 0\le t \le 2\pi$.

However, after that, I have NO clue whatsoever. Can someone walk me through it?

• Looks like you have a missing parameterization. – jdods Nov 21 '17 at 0:51
• Have you tried writing down the line integral that you need to evaluate? – Zach Boyd Nov 21 '17 at 1:11

Guide:

• Compute $F(r(t))$.

• Compute $r'(t)$.

• Compute the inner product and use the following:

$$\iint_S \operatorname{curl}(F) \cdot dS = \int_0^{2\pi} F(r(t)) \cdot r'(t) \, dt$$

The boundary of your curve will be when $z=0$. Plugging in $z=0$ to the equation of your surface gives a circle of radius $4$ which can be parameterized as $$\vec r(t)=\langle 4\cos(t), 4\sin(t), 0\rangle$$ for $0\leq t\leq 2\pi$.

Thus we get $\vec F(\vec r(t))=\langle 0,4\cos(t), 4\sin(t)\rangle$. Note that the tangent vector to the curve also lies in the $xy$-plane and thus, the dot product with zero out the $z$ component totally.

On the curve we get $\vec r'(t)=\langle -4\sin(t), 4\cos(t), 0\rangle$. So $\vec F(\vec r(t)) \cdot \vec r \ '(t)=16\cos^2(t)$.

So we have with Stoke's Theorem that $\iint_S \text{curl}\ \vec F \cdot d\vec S=\int_C\vec F \cdot d\vec r$. The right side becomes

\begin{aligned} \int_0^{2\pi}16\cos^2(t)dt&=8\int_0^{2\pi}(1+\cos{2t})dt\\ &=16\pi \end{aligned}

Note that we can ignore the $\cos(2t)$ portion of the integral since it is over complete periods and everything cancels out for it.