- #1
kostoglotov
- 234
- 6
Homework Statement
It's a long winded problem, I'll post a picture and an imgur link
Imgur link: http://i.imgur.com/5wvbqO2.jpg
Homework Equations
Divergence Theorem
[tex]\iint\limits_S \vec{F}\cdot d\vec{S} = \iiint\limits_E \nabla \cdot \vec{F} \ dV[/tex]
The Attempt at a Solution
I'll follow my own solution up to the point that it disagrees with the solution given in the solutions manual, then I'll explain why I cannot understand/accept what the solution manual is telling me.
First off, the area of S(a) is
[tex]a^2\int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} \sin{\phi} \ d\phi \ d\theta[/tex]
Thus
[tex]| \Omega (S) | = \int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} \sin{\phi} \ d\phi \ d\theta[/tex]
I can see that for
[tex]\vec{F} = \frac{\vec{r}}{|\vec{r}|^3}[/tex]
[tex]\nabla \cdot \vec{F} = 0[/tex]
Thus
[tex]\iiint\limits_E \nabla \cdot \vec{F} \ dV = 0[/tex]
I can see that for [itex]\partial S[/itex] there are 6 surfaces, but that the lateral polar rectangular surfaces up/around the sides have normal vectors that are orthogonal to the specified vector field at all points on said surfaces, and that therefore the flux across those surfaces will be zero.
So
[tex]\iiint\limits_E \nabla \cdot \vec{F} \ dV = 0 = \iint\limits_{S} \vec{F}\cdot d\vec{S} + \iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)[/tex]
Thus
[tex]\iint\limits_{S} \vec{F}\cdot d\vec{S} = -\iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)[/tex]
Now, our S(a) is a portion of the surface of the sphere subtended by the surface S.
The negative sign of the RHS indicates that we used a normal vector in the opposite direction to the normal vector used for the flux across S on the LHS, since we are interested in a quantity without direction here, let's change the direction of the normal vector, to give us
[tex]\left|\iint\limits_{S} \vec{F}\cdot d\vec{S}\right| = \left|-\iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)\right| = \iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a)[/tex]
So if we show that
[tex]\iint\limits_{S(a)} \vec{F}\cdot d\vec{S}(a) = |\Omega (S)| = \int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} \sin{\phi} \ d\phi \ d\theta[/tex]
Then
[tex]\iint\limits_{S} \vec{F}\cdot d\vec{S} = |\Omega (S)| [/tex]
where [itex]\vec{F} = \frac{\vec{r}}{|\vec{r}|^3}[/itex], [itex]\vec{r}[/itex] being a vector from P to any point on S, which is what we set out to prove in the first place.
I will omit some of my working now. I set out to compute the surface integral of S(a). I get to here.
As S(a) is a portion of the surface of a sphere
[tex]\vec{n} = \frac{1}{a}\langle x,y,z \rangle[/tex]
Then
[tex]\iint\limits_{S(a)} \vec{F} \cdot d\vec{S}(a) = a \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \frac{1}{|\vec{r}|} \sin{\phi} \ d\phi \ d\theta[/tex]
The solutions manual takes a slightly different, but equivalent route, but is telling me, at roughly this point, that [itex]|\vec{r}| = a[/itex].
I don't understand this. I can't accept it. [itex]\vec{r}[/itex] is the position vector from P to the surface S, not the position vector connecting P to S(a)...how can [itex]|\vec{r}| = a[/itex]?
edit: wait
in order to get to here
[tex]\iint\limits_{S(a)} \vec{F} \cdot d\vec{S}(a) = a \int_{\theta_1}^{\theta_2} \int_{\phi_1}^{\phi_2} \frac{1}{|\vec{r}|} \sin{\phi} \ d\phi \ d\theta[/tex]
I need to assume that the x,y,z that I use in my normal vector, are the same x,y,z that I use in [itex]\vec{r}[/itex] anyway...so that would mean that [itex]|\vec{r}| = a[/itex], but how can we justify this?
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