0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Looks like you have a missing parameterization. $\endgroup$ – jdods Nov 21 '17 at 0:51
  • $\begingroup$ Have you tried writing down the line integral that you need to evaluate? $\endgroup$ – Zach Boyd Nov 21 '17 at 1:11
2
$\begingroup$

Guide:

  • Compute $F(r(t))$.

  • Compute $r'(t)$.

  • Compute the inner product and use the following:

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.