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I am trying to verify Stoke's theorem for the surface given by $W=\{(x,y,z):x^2+y^2+z^2=1, z\geq0\}$. Its boundary is the unit circle on the $z=0$ axis. The vector field is $F=(z^2,x,y^2$)

I identified that the surface is the positive half of a sphere of radius one. I calculated the curl of my vector field, and I got $(2y,2z,1)$ and switching to spherical coordinates i get $(2\sin\theta\sin\phi,2\cos\theta,1)$.The normal vector for the half sphere is $(x,y,z)$ or $(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\theta)$ in spherical coordinates. I am having trouble computing the surface integral and the line integral in order to verify Stoke's theorem, I am also having trouble with the parameterization. Could anyone show me how to verify Stoke's theorem for this surface over this vector field ?

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    $\begingroup$ Stokes', not Stoke's. It's named for 19th century mathematician George Stokes. $\endgroup$ – anomaly Nov 26 '17 at 23:57
  • $\begingroup$ @anomaly I am sorry, for spelling it incorrectly. $\endgroup$ – Viktor Raspberry Nov 26 '17 at 23:59
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Stokes’ theorem says that for a open surface $S$ with boundary $c$ positively oriented,

$$\iint_{S} \text{curl} \vec F \cdot d\vec S=\oint_{c} F \cdot d\vec r$$

In our case the boundary of our surface is $c=\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=1, z=0 \}$.


We may parametrize our surface with $\vec r(\theta)=\langle \cos \theta, \sin \theta,0 \rangle$ and $\theta \in [0,2\pi]$ so that $d \vec r=\langle -\sin \theta, \cos \theta,0 \rangle d\theta$.

On the other hand $\vec F=\langle z^2,x,y^2 \rangle$ and on $c$ we have $\vec F=\langle 0,\cos \theta, \sin^2 \theta \rangle$ as $z=0,x=\cos \theta$, and $y=\sin \theta$.

$$\oint_{c} \vec F \cdot dr$$

$$=\int_{0}^{2\pi} \langle 0,\cos \theta, \sin^2 \theta \rangle \cdot \langle -\sin \theta, \cos \theta, 0 \rangle d\theta$$

$$=\int_{0}^{2\pi} \cos^2 \theta d\theta$$

$$=\pi$$


On the other hand we can directly compute,

$$\iint_{S} \text{curl} F \cdot n dS=\iint_{D} \text{curl} F \cdot n |\vec r_{\theta} \times \vec r_{\phi}| dA$$

Parametrize the surface $S=W$ with,

$$\vec r(\theta,\phi)=\langle \sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi \rangle$$

And $D=\{(\theta,\phi) : (\theta,\phi) \in [0,2\pi] \times [0,\frac{\pi}{2}] \}$.

On this surface $S$ we have,

$$\text{curl} \vec F=\langle 2\sin \phi \sin \theta, 2 \cos \phi,1 \rangle$$

Also, $|\vec r_{\theta} \times \vec r_{\phi}|=\sin \phi$. (It is is standard practice to show that when a sphere of radius $\rho$ is considered this quantity turns out to be $\rho^2 \sin \phi$).

A normal to a sphere centered at the origin is $\langle x,y,z \rangle$, hence the outward pointing unit normal to a sphere of radius $\rho$ is,

$$\vec n=\frac{\langle x,y,z \rangle}{\rho}$$

Since $\rho=\sqrt{x^2+y^2+z^2}$. In our case $\rho=1$, so the outward unit normal is,

$$\vec n=\langle x,y,z \rangle$$

$$=\langle \sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi \rangle$$

Putting everything together,

$$\text{curl} F \cdot \vec n |\vec r_{\theta} \times \vec r_{\phi}|=2\sin^3 \phi \sin \theta \cos \theta+2 \sin^2 \phi \cos \phi \sin \theta+\sin \phi \cos \phi$$


And the surface integral we want to compute is equal to,

$$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \left(2\sin^3 \phi \sin \theta \cos \theta+2 \sin^2 \phi \cos \phi \sin \theta+\sin \phi \cos \phi \right) d\phi d\theta$$

We can safely ignore the term with $\sin \theta$ being the only part with $\theta$ because $\int_{c}^{d} \int_{a}^{b} f(x)g(y) dx dy=\left(\int_{c}^{d} g(y) dy \right) \left( \int_{a}^{b} f(x) dx \right)$ and $\int_{0}^{2\pi} \sin \theta d\theta=0$. Similarly we can ignore the term with $\sin \theta \cos \theta$ being the part with $\theta$ for pretty much the same reason. We end with,

$$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \sin \phi \cos \phi d\phi d\theta$$

$$=2\pi \int_{0}^{\frac{\pi}{2}} \sin \phi \cos \phi d\phi$$

$$=\pi \int_{0}^{\frac{\pi}{2}} \sin (2 \phi) d\phi$$

$$=\pi$$

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  • $\begingroup$ I think you have Stokes theorem reversed ? $\endgroup$ – Viktor Raspberry Nov 27 '17 at 1:37
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    $\begingroup$ Can you please explain what you mean by that @ViktorRaspberry $\endgroup$ – Ahmed S. Attaalla Nov 27 '17 at 2:23
  • $\begingroup$ I think the curlF component of the integrand should be in the surface integral. So you basically reversed your integrands. As you can see here : ocw.mit.edu/courses/mathematics/… $\endgroup$ – Viktor Raspberry Nov 27 '17 at 2:29
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    $\begingroup$ Your absolutely right, will fix! @Viktor Raspberry $\endgroup$ – Ahmed S. Attaalla Nov 27 '17 at 2:31
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    $\begingroup$ Please see my edit @ViktorRaspberry $\endgroup$ – Ahmed S. Attaalla Nov 27 '17 at 3:35

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