My Math Forum Flux of a sphere

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 March 4th, 2016, 08:13 AM #1 Newbie   Joined: Oct 2014 From: Norway Posts: 15 Thanks: 1 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.
 March 4th, 2016, 10:13 AM #2 Newbie   Joined: Sep 2015 From: New york Posts: 17 Thanks: 4 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.
March 5th, 2016, 01:31 AM   #3
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Quote:
 Originally Posted by davidmoore63 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.

 March 5th, 2016, 01:48 AM #4 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 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
 March 5th, 2016, 03:04 AM #5 Newbie   Joined: Oct 2014 From: Norway Posts: 15 Thanks: 1 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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