Verifying Stokes's Theorem

Question

I am trying to verify Stokes's theorem if $\vec{v} = z\vec{i} + x\vec{j} + y\vec{k}$ is taken over the hemispherical surface $x^2+y^2+z^2=1, \; z>0$

I have finished the left hand size of Stokes's theorem, and the answer was $\pi$.

I am working on the RHS. I first calculate the unit normal which I go to be $x\vec{i} + y\vec{j} + z\vec{k}$.

Then I calculate the curl of $\vec{v}$ which I Got to be $\vec{i} + \vec{j} + \vec{k}$

When I multiply these together, I get $x + y + z$; however, this makes no sense as I will only have a double integral, but $3$ variables. Can someone point out where I may have gone wrong?

I think I am suppose to get $x+y+1$ and then integrate, but not sure. Any help would be greatly appreciated!

Your unit normal is not unit. You need to divide by $\sqrt{x^2+y^2+z^2}$. — Mark Fantini, Sep 08 '14 at 23:10
actually, it is normal. If you divide by that, it all reduces out — Jackson Hart, Sep 08 '14 at 23:12
If byy reducing out you mean $|\vec{n}| = 1$, then yes. It is definitely normal, but not unitary. — Mark Fantini, Sep 08 '14 at 23:15
It is actually. You end up getting (x/zi +y/zj+k)/(1/z) The magnitude reduces to 1/z since z=sqrt(1-x^2-y^2) — Jackson Hart, Sep 08 '14 at 23:16
Additioanlly I would divide by the gradient of that, not sqrt(x^2+y^2+z^2) — Jackson Hart, Sep 08 '14 at 23:18
I get what you're saying now. But if you noticed you're using $z = \sqrt{1-x^2-y^2}$ then you can notice that $x+y+z$ is actually $x+y+\sqrt{1-x^2-y^2}$. Integrate that over $x^2+y^2 \leq 1$. — Mark Fantini, Sep 08 '14 at 23:19
You're on a hemisphere, so use the equation to eliminate z in your curl : $$x^2+y^2+z^2=1$$ — AgentS, Sep 08 '14 at 23:22

Mark Fantini · Answer 1 · 2014-09-09T00:45:29.073

0

Using $z= \sqrt{1-x^2-y^2}$ then $x+y+z = x+y+\sqrt{1-x^2-y^2}$. This means you have to compute

$$\int\limits_{x^2+y^2\leq 1} x+y+\sqrt{1-x^2-y^2} \, dA.$$

You can use polar coordinates: $$x = r \cos (\theta), y = r \sin (\theta) \text{ and } dA = r \, dr \, d \theta.$$ It follows that

$$\int\limits_{x^2+y^2\leq 1} x+y+\sqrt{1-x^2-y^2} \, dA = \int_0^{2 \pi} \hspace{-5pt} \int_0^1 (r \cos (\theta) + r \sin (\theta) + \sqrt{1-r^2})r \, dr \, d \theta.$$

The integrals of $r \cos (\theta)$ and $r \sin (\theta)$ vanish, and you are left to compute

$$\int_0^{2 \pi} \hspace{-5pt} \int_0^1 r \sqrt{1-r^2} \, dr \, d \theta.$$

edited Sep 09 '14 at 00:45

answered Sep 08 '14 at 23:22

Mark Fantini

5,523

x would go from -1 to 1, what about y? – Jackson Hart Sep 08 '14 at 23:25
I am not sure how to integrate the 2nd term – Jackson Hart Sep 08 '14 at 23:26
If I use the limits of x=-1..1 and y =-((1-x^2)^(1/2)).. ((1-x^2)^(1/2)), i get the wrong answer. Any ideas? – Jackson Hart Sep 08 '14 at 23:32
I am pretty sure that sqrt (1-x^2-y^2) = 1, does anyone know why? – Jackson Hart Sep 08 '14 at 23:37
could it have something to do with the fact that int of dS = int of dxdy/n dot k? – Jackson Hart Sep 08 '14 at 23:42
I've just seen your comments now, I'll address them in the answer. – Mark Fantini Sep 09 '14 at 00:03
@JacksonHart I've added the details of the computation. I think we're all done. – Mark Fantini Sep 09 '14 at 00:45

Verifying Stokes's Theorem

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