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March 4th, 2016, 08:13 AM   #1
paw
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Flux of a sphere

Evaluate the surface integral $\iint_S \vec{F} \bullet \vec{n} \, dS$, where $\vec{n}$ is the upward-pointing unit normal vector to the given surface S.

$\vec{F} = x\vec{i} + y\vec{j} + z\vec{k}$; S is the spherical surface with equation $x^2 + y^2 + z^2 = 1$.

Here's how I've tried to solve it:

$z = h(x, \, y) = \sqrt{1-x^2-y^2}$

$\frac{dh}{dx} = \frac{-x}{\sqrt{1 - x^2 - y^2}}, \, \, \, \, \frac{dh}{dy} = \frac{-y}{\sqrt{1 - x^2 - y^2}}$

$dS = \sqrt{1 + (\frac{dh}{dx})^2 + (\frac{dh}{dy})^2}\, dx \, dy$

$\vec{N} = \frac{dh}{dx} \times \frac{dh}{dy} = -\frac{dh}{dx}\vec{i} -\frac{dh}{dy}\vec{j} + \vec{k}$

$|\vec{N}| = \sqrt{1 + (\frac{dh}{dx})^2 + (\frac{dh}{dy})^2}$

$\vec{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{1}{|\vec{N}|} \, (-\frac{dh}{dx}\vec{i} -\frac{dh}{dy}\vec{j} + \vec{k})$

Derivation:

$\iint_S \vec{F} \bullet \vec{n} \, dS = \int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} (x\vec{i} + y\vec{j} + z\vec{k}) \bullet (-\frac{dh}{dx}\vec{i} -\frac{dh}{dy}\vec{j} + \vec{k}) \frac{1}{|\vec{N}|} \, |\vec{N}| \, dx \, dy = \int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} \, (-\frac{dh}{dx} \, x -\frac{dh}{dy} \, y + \sqrt{1 - x^2 - y^2}) \, dx \, dy = ... = 2 \pi $

I know that the answer should be $4 \pi$, and I suspect that what I've done wrong is that I've only evaluated the top hemisphere of the sphere. If my assumption is correct, how do I include the bottom hemisphere in the integral? Please go slow on me as this is new to me.
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March 4th, 2016, 10:13 AM   #2
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That is a very laborious method. I can think of two easier ways (excuse lack of proper scripting):

1) direct evaluation as a surface integral:

Int (r.n) dS = r Int dS = r. 4 pi r^2 = 4pi, since r.n is just r which is a constant on the surface.

2) using the divergence theorem

Int (r.n) dS = Int div(r) dV = 3 Int dV = 3. 4/3 pi r^3 = 4 Pi r^3 = 4 Pi, since the divergence of the vector function r is 3.
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March 5th, 2016, 01:31 AM   #3
paw
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Quote:
Originally Posted by davidmoore63 View Post
Int (r.n) dS = r Int dS = r. 4 pi r^2 = 4pi, since r.n is just r which is a constant on the surface.
You start with Int (r.n) dS, but where did the F go? Is r the radius? If so, why can you substitute F with r? Sorry, this is new to me.
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March 5th, 2016, 01:48 AM   #4
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The square root function only yields the positive root, ie the upper half of the x,y, z continuum.

That is why you need to double it to get the lower half as well
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March 5th, 2016, 03:04 AM   #5
paw
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So the solution says that the derivation should go like this:

$\iint_S \vec{F} \bullet \vec{n} \, dS = \iint_S \big \langle x, \, y, \, z \big \rangle \bullet \big \langle x, \, y, \, z \big \rangle \, dS$

to which I agree.

But how should I set the limits? The ones I've used above, $\int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} \, dx \, dy$, give $2 \pi$, as I showed in the first post.

Last edited by paw; March 5th, 2016 at 03:08 AM.
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