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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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 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. Tags flux, sphere Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Kinroh Physics 4 April 26th, 2015 09:51 AM kgottsch Applied Math 0 March 14th, 2013 03:20 PM cummings123 Calculus 3 October 17th, 2012 03:41 PM 1thoughtMaze1 Calculus 14 December 27th, 2011 06:29 AM maximade007 Calculus 0 November 28th, 2011 09:41 AM

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