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 upwardpointing 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: $z = h(x, \, y) = \sqrt{1x^2y^2}$ $\frac{dh}{dx} = \frac{x}{\sqrt{1  x^2  y^2}}, \, \, \, \, \frac{dh}{dy} = \frac{y}{\sqrt{1  x^2  y^2}}$ $dS = \sqrt{1 + (\frac{dh}{dx})^2 + (\frac{dh}{dy})^2}\, dx \, dy$ $\vec{N} = \frac{dh}{dx} \times \frac{dh}{dy} = \frac{dh}{dx}\vec{i} \frac{dh}{dy}\vec{j} + \vec{k}$ $\vec{N} = \sqrt{1 + (\frac{dh}{dx})^2 + (\frac{dh}{dy})^2}$ $\vec{n} = \frac{\vec{N}}{\vec{N}} = \frac{1}{\vec{N}} \, (\frac{dh}{dx}\vec{i} \frac{dh}{dy}\vec{j} + \vec{k})$ 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 
Newbie Joined: Oct 2014 From: Norway Posts: 15 Thanks: 1  
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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