Stokes' Theorem

**PedroMinsk** · June 28th 2011, 05:40 PM

$\int\limits_C \vec{F} . dr$

C: $x^2 + y^2 + z^2 = a^2$ intersection $y + z = a$

$\vec{F} = (3y +z, x + 4y, 2x + y)$

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$\nabla_X\vec {F} = (1, -1, -2)$

$\vec{r(u,v)} = (u, v, a-v)$

$\vec{n} = (0, 1, 1)$

I made the intersection between the sphere and the plane and got:

$\frac{x^2}{a^2} + \frac{(y-a)^2}{\frac{a^2}{2}} = 1$

I know the area of this ellipse, but what is the area of this ellipse projected on the xy plane?

If I knew this could substitute an got the value:

$\iint\limits_D -3 dudv = -3A(D)$

**FernandoRevilla** · June 29th 2011, 01:29 AM

The solution is $-3A$ being $A$ the area enclosed by the circle $x^2+y^2+z^2=a^2,\;y+z=a$ .

**PedroMinsk** · June 29th 2011, 03:18 AM

I think you meant 'sphere'.

But the problem is that the intersection between the sphere and the plane give us the ellipse that I wrote in the first post.

The area of this ellipse multiply by -3 is = $\frac{\pi a^2}{\sqrt{2}}.(-3)$ = $\frac{-3\sqrt{2} \pi a^2}{2}$ .

But the answer is $\frac{-3\sqrt{2} \pi a^2}{4}$ .

Shouldn't I project this area on the xy plane?

**FernandoRevilla** · June 29th 2011, 08:53 AM

Originally Posted by PedroMinsk

I think you meant 'sphere'.

No, I exactly meant by $A$ the area enclosed by the circle $x^2+y^2+z^2=a^2,\;y+z=a$ . Then, $\iint_{S}\textrm{rot}\vec{F}\cdot \vec{n}\;dS=(-3/\sqrt {2})A$ ( notice that we forgot to write $\vec{n}=(1/\sqrt{2})(0,1,1)$ ). The area enclosed by the circle is $A=\pi a^2/2$ so, the solution certainly is $\dfrac{-3\sqrt{2}\pi a^2}{4}$ .

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