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I'm trying to solve this exercise about Stokes Theorem: prove the validity of Stokes Theorem when $F(x,y,z)=(y,2x,2z)$ and $S$ is the part of the plane $z=y$ that lies in the cylinder $x^2+y^2=4y$ and $C$ is the area where the cylinder and the plane meet.

Solution:using stokes theorem $$\oint F.dR = \int \int_{S}(\operatorname {curl} F).n ds$$ From here I did some straight forward calculations. $$\operatorname {curl}F=(0,0,1)$$ $$n= \frac {(x,y,-2)}{\sqrt{x^2+y^2+4}}$$ $$ds= \frac{1}{2}\sqrt{x^2+y^2+4}$$

And finally we have $$\oint F.dR= -\int \int 1 dxdy$$ Here's my question: first why am I getting a negative integration? And second the last double integral means the area of what?? Or how do I determine the bounds for the integrals? Sorry if this is dumb? I'm just so confused!

Aaron Hendrickson
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Hoda Bibo
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2 Answers2

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Short answer: your normal vector is pointing in the wrong direction.

Parameterize $S$ by

$$\vec s(u,v)=2u\cos v\,\vec\imath+2(u\sin v+1)\,\vec\jmath+2(u\sin v+1)\,\vec k$$

with $0\le u\le1$ and $0\le v\le2\pi$. (If you're unsure where this is coming from, complete the square in the equation of the cylinder.) The boundary $\partial S$ can be parameterized by

$$\vec r(t)=2\cos t\,\vec\imath+2(\sin t+1)\,\vec\jmath+2(\sin t+1)\,\vec k$$

with $0\le t\le2\pi$.

By Stokes' theorem,

$$\oint_{\partial S}\vec F(x,y,z)\cdot\mathrm d\vec r = \iint_S\operatorname{curl}\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\operatorname{curl}\vec F(x,y,z)\cdot\vec n\,\mathrm dS$$

In order to preserve the positive orientation of $\partial S$ (counter-clockwise when viewed from above), you need the normal vector $\vec n$ to point upward, and this is obtained with

$$\vec n = \frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}=-4u\,\vec\jmath+4u\,\vec k$$

Then the integral of the curl is

$$\iint_S\operatorname{curl}\vec F(x,y,z)\cdot\mathrm d\vec S = \int_0^{2\pi}\int_0^14u\,\mathrm du\,\mathrm dv=4\pi$$

The line integral has the same value:

$$\oint_{\partial S}\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}(8\cos t(1+\cos t)-4\sin t(1+\sin t)+4\sin(2t))\,\mathrm dt=4\pi$$

user170231
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Stokes theorem: $\iint \nabla\times F\ dS = \oint F\cdot dr$

$S$ is the piece of the plane $z = y$ bound by the cylinder.

$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1) = (0,-1,1)$

If we stay in Cartesian, the bounds would be $-\sqrt {4y-y^2}\le x\le \sqrt {4y-y^2}, 0\le y\le 2$

In polar the bounds would be $r \le 4\sin\theta, 0\le \theta\le \pi$

However, we can short-cut around any calculations based on these bounds.

$\iint \nabla \times F\ dS = \iint dA = A$ were $A$ is the area of the cross-section of the cylinder. i.e. $4\pi$

A parameterization of the contour might be:

$x = 2\cos t, y = 2\sin t+2, z = 2\sin t+2$

Verifing Stokes theorem:

$4\pi = \int_0^{2\pi} (2\sin t+2, 4\cos t, 4\sin t + 2)\cdot (-2\sin t, 2\cos t, 2\cos t)\ dt$

When we multiply this all out, we will only need to care about $\sin^2 t$ terms and $\cos^2 t$ terms as $\int_0^{2\pi} \sin t \ dt = \int_0^{2\pi} \cos t\ dt = \int_0^{2\pi} \sin t\cos t\ dt = 0$

$\int_0^{2\pi} -4\sin^2 t + 8\cos^2 t\ dt = 4\pi$

Doug M
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