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Consider the bounded surface S that is the union of $x^2 + y^2 = 4$ for $−2 \le z \le 2$ and $(4 − z)^2 = x^2 + y^2 $ for $2 \le z \le 4.$ Sketch the surface. Use suitable parametrisations for the two parts of S to verify Stokes’s Theorem for for $\mathbf{F} = (yz^2,0,0)$.

Herein, I ask only about computing line integrals instead of $ \iint_S (\nabla × F )· d\mathbf{S}$, by virtue of Stokes's Theorem. Denote the $2 \le z \le 4$ cone P, and the $-2 \le z \le 2$ cylinder C.

A boundary curve for $P$ is $x^2 + y^2 \le 4$ on the $z = -2$ plane. By virtue of the quotes on orientation here, we need the normal vector to be downward, in the direction of $(0, 0, -1).$ So the circle must be oriented clockwise when viewed from above $z = -2$ (as in my picture here). Thus, parameterise with $\mathbf{r}(t) = ( 2\sin t, 2\cos t, -2 )$ for all $0 \le t \le 2\pi$.

Then $\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int^{2\pi}_{0} (8 \cos t, 0, 0,) \cdot ( 2\cos t, ♦ ,♦ ) \, dt = 16\pi$

♦ denote objects that don't need to be computed because they're dot-producted with 0.

$1.$ My answer differs by a negative sign from that without Stokes's Theorem. What's wrong?

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    The parametrization $(2\sin t, 2\cos t, -2)$ goes clockwise, so it has the area of the circle on the right... – geodude Apr 19 '14 at 15:03
  • @geodude: This entire comment applies to a view from above the $z = -2$ plane. Yes, then the parameterisation $(2\sin t, 2\cos t, -2)$ is clockwise. Then the area of the circle would be to the right. What's wrong with this? I deduced all this from the need for a downward normal vector? –  May 04 '14 at 15:31

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There was only a small mistake, which lead to misconceptions, but aside from the mistake your theory was completely correct.

If we consider the cone $P$, the boundary curve is $x^2+y^2\le 4$

ON THE $z=2$ PLANE :)

So when you reconsider this, you are looking from "below" the $z=2$ plane, so we take:

$r(t)=(2\cos(t),2\sin(t),2)$, so $r'(t)=(-2\sin(t),2\cos(t),0)$ and

$F=(8\sin t,0,0)$.

Thus $\int_{C_1}F\cdot dr=\int_0^{2\pi}F\cdot r'(t)dt=\int_0^{2\pi}(8\sin(t),0,0)\cdot(-2\sin(t),2\cos(t),0)dt$

$=\int_0^{2\pi}-16\sin^2(t)dt=\int_0^{2\pi}-16(\frac{1}{2}-\frac{1}{2}\cos(2t))dt=\int_0^{2\pi}-8+8\cos(2t)dt$

$=-8t+4\sin(2t)|_0^{2\pi}=-16\pi$.

I hope this helps :)

edit we are considering the cone, where $2\le z\le 4$ the boundary of this is the circle lying in the $z=2$ plane, which is why we "look from below" the $z=2$ plance instead of the $z=-2$ plane.

Ellya
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  • Thank you effusively again! Would you please explain whether we need to consider the $z = -2$ boundary curve? Moreover, regarding the $z = 2$ plane, why did you consider the view from above $z = 2$? Don't we have to consider the view from below $z = 2$ ? –  May 15 '14 at 10:14
  • @LePressentiment is this sufficient? – Ellya May 20 '14 at 20:09
  • Thanks again. I'm returning to this now. I still don't apprehend why we only consider $x^2 + y^2 = 4$ while looking from below $z = 2$ ? What about $x^2 + y^2 = 4$ while looking from above $z = 2$ And what about $x^2 + y^2 = 4$ while looking from below $z = -2$ ? Aren't we computing the double integral over this combined cone and cylinder? So we need to consider the boundary curves of both? –  May 30 '14 at 14:34
  • I posted a new question about this http://math.stackexchange.com/q/814966/53259 and hope you can answer! –  May 30 '14 at 14:44
  • The boundary curve of cone $P$ is $x^2 + y^2 = 4 \wedge z = 2$. – mvw May 30 '14 at 15:37