3

Let $$ F(x,y,z)=\frac{x{\bf i}+y{\bf j}+z{\bf k}}{(x^2+y^2+z^2)^{3/2}}. $$ How can I calculate $$ \iint_SF\cdot d{\bf S} $$ where $S$ a the "upper semi-unit-sphere" and the boundary of $S$ given by $$ \begin{cases} x+y+z=3\\ (x-1)^2+(y-1)^2+(z-1)^2=1 \end{cases}? $$

  • If I change the coordinate to make the equations of the boundary of the semisphere simpler: $$ z'=0,\quad x'^2+y'^2+z'^2=1, $$ then I messed up with $F(x',y',z')$. But if I don't change the coordinates, I messed up with the parameterization of the surface. Any idea?
  • Does the integral $$\iint_S\frac{\bf x}{|{\bf x}|^3}\cdot d{\bf S}$$ have some meaning in physics?

[Added] I didn't expect that my description of the surface in the integral is so difficult to be understood. Suppose we have a unit sphere centered at $(1,1,1)$ $$ \Omega=\{(x,y,z)\in{\Bbb R}^3:(x-1)^2+(y-1)^2+(z-1)^2=1\} $$ and the plane $$ P=\{(x,y,z)\in{\Bbb R}^3:x+y+z=3\}. $$ Geometrically, the plane $P$ would cut the sphere $\Omega$ into two pieces and $S$ is one of them while I don't specify which one so that the result would be up to the choice of these two pieces.

The confusion might due to my notation: $$ \begin{cases} x+y+z=3\\ (x-1)^2+(y-1)^2+(z-1)^2=1 \end{cases}. $$ which is equivalent to $$ \{(x,y,z)\in{\Bbb R}^3:x+y+z=3\ {\bf and}\ (x-1)^2+(y-1)^2+(z-1)^2=1\} $$ which is the boundary of $S$.

  • Did you try the divergence theorem? – Gil Bor Jan 07 '14 at 17:52
  • $\displaystyle{\large{\vec{r} \over r^{3}} = -\nabla\left(1 \over r\right)}$. – Felix Marin Jan 07 '14 at 17:58
  • 1
    $\displaystyle{{\vec{r}\cdot{\rm d}\vec{S} \over r^{3}} = {\rm d}\Omega_{\vec{r}}\ \mbox{where}\ \Omega_{\vec{r}}}$ is the 'solid angle'. – Felix Marin Jan 07 '14 at 18:01
  • To evaluate, use the divergence theorem. For physical meaning, look up Gauss law in electrostatics. Your $F$ is the electric field due to a pt charge at the origin. The divergence theorem (=Gauss law) says the surface integral of the electric field (so called the total flux) gives the total amount of electric charge inside the surface of integration. (Same idea for gravitation). – Gil Bor Jan 07 '14 at 18:04
  • Can one call the surface $S$ a "half-ball" -- a hemisphere with a circular bottom? And it's the "upper half-ball" -- meaning that you see the circular bottom if you look out from the origin? – John Jan 07 '14 at 18:35

2 Answers2

2

There are a few ways to solve this problem.

  1. You are asked to evaluate $$ \int \mathbf{F} \cdot \mathbf{dS} $$ The normal to the plane is $(1,1,1)3^{-1/2}$ , and hence the integral reduces to $$ \int_{\{(x-1)^2+(y-1)^2+(z-1)^2 \leq 1 \cap x+y+z=3\}} \frac{\sqrt{3}}{(x^2 + y^2 + z^2)^{3/2}}\,dS $$ Now the integrand is rotationally symmetric, and hence one can rotate the region and integrate over a suitable disc with center on the $z-$ axis at $(0,0,\sqrt3).$

    Specifically the integral is (in polar coordinates) $$ \int_0^1 \int_0^{2\pi} \frac{\sqrt{3}r\,dr\,d\theta}{(r^2 + 3)^{3/2}} $$

    Note this tells you in particular that the integral is not 0, since the integrand is positive.

  2. A second method is as follows. You need to evaluate the following surface integral (stokestheorem) over an appropriate portion of a sphere of radius 2, centered at the origin. I'll leave the details to you, one can evaluate this integral by switching to spherical coordinates.

  3. A third (essentially same as the second ) approach would be to look at the integral over unit disc on the plane $x+y+z=3$ centered $(1,1,1)$ as equal to (again by stoke's theorem) the integral over a cone below the said disc, above the origin, with a ball of radius $\epsilon$ around the origin punctured. The integral on the surface of the cone vanishes (why?), and you're left to evaluate the integral in (2) except on a sphere centered origin radius $\epsilon.$ The answer will turn out to be indepenedent of $\epsilon.$

achille hui
  • 122,701
Raghav
  • 1,348
  • +1 but you should mention since $\vec{\nabla}\cdot \vec{F} = 0$ when $\vec{x} \ne \vec{0}$, you can use Gauss's theorem to deform the surface to other surface with same boundary without changing the integral (assuming one never crosses the origin during the deformation). – achille hui Jan 07 '14 at 18:09
  • @Raghav, I don't understand the first method you wrote. How does the integral reduce to the one you put in the fourth line? –  Jan 07 '14 at 18:13
  • @Jack, the integral over the boundary of the half ball $(x-1)^2 + (y-1)^2 + (z-1)^2 \le 1 \cap x + y + z \ge 3$ is zero by divergence theorem. So the integral you want is equal to the integral over the disk in method $1$. – achille hui Jan 07 '14 at 18:18
  • @achillehui, ah, I see, thanks. I still don't understand the calculation part of $\int \sqrt{3}/(x^2+y^2+z^2)^{3/2}$ –  Jan 07 '14 at 18:30
  • 1
    @Jack, on the plane $x + y + z = 3$, the normal vector $\hat{n}$ is $\frac{1}{\sqrt{3}}(1,1,1)$, so $\vec{x}\cdot\hat{n} = \frac{x+y+z}{\sqrt{3}} = \sqrt{3}$ – achille hui Jan 07 '14 at 18:43
  • @achillehui, I didn't put it right: I see from your previous comment how he got $\int \sqrt{3}/(x^2+y^2+z^2)^{3/2}$. But I don't know how the polar coordinate is used here to calculate this reduced integral. –  Jan 07 '14 at 19:52
  • Use $x = r\cos \theta, y = r \sin \theta.$ Then $r$ varies from $0$ to $1,$ and $\theta$ varies on $[0,2\pi].$ Jacobian of this transformation is $r.$ On this plane, $z = \sqrt3,$ so that $z^2 = 3.$ – Raghav Jan 07 '14 at 20:04
  • What I don't get is the "rotationally symmetric" part. –  Jan 07 '14 at 21:16
  • 1
    The integrand depends only on the radial component and not on $\theta, \phi$ (say when written in spherical coordinates), (i.e. latitude and co-latitude). – Raghav Jan 07 '14 at 21:28
0

The integral is zero.

The field is a radial field, centered at the origin. (It points directly away from the origin, and depends only on the distance from the origin.)

The field arises from a source at the origin. Your surface does not enclose the origin. Hence there are no sources within your surface, and the integral vanishes. (Check out Gauss's Law.)

John
  • 26,319
  • The integral is not zero. $S$ is not a closed surface and $\vec{F}\cdot d\vec{S}$ is having same sign over the whole $S$. – achille hui Jan 07 '14 at 17:57
  • It's a semi-sphere, centered at $(1,1,1)$ with its great circle centered at $(1,1,1)$ and normal to the line $x=y=z.$ It certainly is a closed surface. – John Jan 07 '14 at 18:03
  • a semi-sphere is half of a sphere and a sphere is a surface. The surface you mention is the boundary of a half-ball. – achille hui Jan 07 '14 at 18:14
  • OK, it looks like the words say one thing and the equations say another. I asked the OP which he's after. – John Jan 07 '14 at 18:39