a surface integral problem

Question

$F=(y-z,z-x,x-y)$ $S$ be the portion of surface defined by $x^2+y^2+z^2=1$ and $x+y+z\ge 1$. We want to evaluate $\int_S curl(F)\cdot dS$.

I have found $curl(F)=(-2,-2,-2)$ and the normal vector of $S$ is $(x,y,z)$. Thus the integral becomes $\int_S -2(x+y+z) dS$. However, if I use the polar coordinate of $S$, then the boundary of $\phi$ and $\theta$ is a big problem.

I still consider using Stoke's Theorem. It then should be $\int_A F\cdot dr$ where $A$ is the circle defined by the unit ball and the plane $x+y+z=1$. However I can not find a way to paramettrize this circle. That is a big problem.

Answer 1

Here is a rather direct computation. Let $(P),(S),(C)$ denote resp. the plane, the sphere and the circle. Let $R$ be the radius of $(C).$

Let $H$ denote the projection of origin point $O$ onto $(P)$. Its coordinates are $(1/3,1/3,1/3)$.

Let $A(1,0,0) \in (P)\cap(S)$. Pythagoras theorem applied to right triangle OHA gives $(3/9)+R^2=1$: hence $R=\sqrt{\dfrac23}$.

Let us now consider operation:

$$\tag{1}\underbrace{\begin{pmatrix}x\\y\\z\end{pmatrix}}_{(C)}=\underbrace{\begin{pmatrix}-a&b&c\\-a&-b&c\\2a&0&c\end{pmatrix}}_{R}\underbrace{\begin{pmatrix}R\cos{t}\\R\sin{t}\\c\end{pmatrix}}_{(\Gamma)}$$

$$\iff \ \ \begin{cases}x&=&-aR \cos t +bR \sin t+1/3\\y&=&-aR \cos t-bR \sin t + 1/3\\z&=& \ 2aR \cos t+1/3\end{cases}$$

where $a:=1/\sqrt{6}, b:=1/\sqrt{2}, c:=1/\sqrt{3}$.

Explanation: we ''prepare'' a horizontal circle $(\Gamma)$ with center in $(0,0,c)$, with the same radius $R$ as circle $(C);$ then circle $(\Gamma)$, by rotation $R$ is placed onto circle $(C)$.

I will not explain how this matrix has been obtained: without being difficult, it would take a too long time; I advise instead to do a quick verification that it is the rotation matrix that ''does the job it is intended for'', in particular that the $z$ axis comes onto the axis defined by $OH$.

Now, the final computation is, with expressions given in (1):

$$\int_0^{2 \pi}((y(t)-z(t))x'(t)+(z(t)-x(t))y'(t)+(x(t)-y(t))z'(t))dt=-4 \pi/\sqrt{3}.$$

as desired.

Answer 2

It is not too hard to change coordinates for this problem. Effectively we can rotate the vector field and plane. That sounds complicated, but in this case it is simple.

First we make a sketch. We know the plane intersects the sphere at a circle, which we will call $C$. From the equation for the plane, we locate the points (1,0,0), (0,1,0) and (0,0,1) on our sketch. Those points are on the plane and on the sphere, so the circle, $C$, goes through those points. The center of the circle must be at (1/3,1/3,1/3), because the center is on the plane and the center is equidistant to the 3 points. We can calculate the radius of circle as $a=\sqrt{2/3}$. And the distance from the origin to the center of the circle is $b=\sqrt{3}/3$, but we will not even need that.

We want to integrate $\vec{g}=\vec\nabla\times\vec{F}=(-2,-2,-2)$ over part of the surface of the sphere. We note that $\vec{g}$ is perpendicular to the plane and has magnitude $g=2\sqrt{3}$.

Let's set this up in our spherical $(r,\theta,\phi)$ coordinate system, with $\theta$ the azimuthal (0-2$\pi$) angle and $\phi$ the co-latitude (0-$\pi$). We choose this system so that the plane is at $z$=constant. In this system $\vec{g}$ is still perpendicular to the plane, but now $\vec{g}=-2\sqrt{3}\hat{k}$. The surface normal is $\hat{r}$, so the dot product is $\vec{g}\cdot\hat{r}= -2\sqrt{3}\cos\phi$. The surface area element in spherical coordinates is $r^2 \sin\phi \, d\theta \, d\phi$, and we have $r=1$ on the surface. So, the integral becomes, in our spherical coordinate system, $$I=-2\sqrt{3} \int \int \cos\phi\sin\phi \,d \phi \,d \theta \;.$$

The limits of the $\theta$ integration are zero to $2\pi$. The lower limit of the $\phi$ integration is $\phi_1=0$ and the upper limit, $\phi_2$, is where the radius of the circle comes in. Since we have a unit sphere, the radius $a$ of circle, C, is $a=\sin\phi_{2}$, as can be seen from sketch in, say, the $xz$-plane.

The integral is actually quite easy, when we use $\sin\phi_2=a=\sqrt{2/3}$.
$$I=-4\pi \sqrt{3} \int \cos\phi\sin\phi \,d \phi = -4\pi\sqrt{3} \big[ \frac{1}{2}\sin^2\phi \big]_{\phi_1}^{\phi_2} = -2\pi\sqrt{3}\sin^2\phi_2 = -\frac{4\pi}{\sqrt{3}}\;.$$

Answer 3

You can also use the divergence theorem here, it will save you the heavy computations you are not comfortable with:

If $S_2$ denotes the part of the plane $x+y+z=1$ inside the sphere $x^2+y^2+z^2=1$, and $E$ the 3D-region bounded by $S$ and $S_2$, then $$ \iint_S \nabla \times F\cdot dS+ \iint_{S_2} \nabla \times F\cdot dS= \iiint_E \nabla\cdot \nabla \times F\; dV = 0 $$ Since the plane $x+y+z=1$ has normal unitary vector $n=-\frac{\pmatrix{1\\1\\1} }{\sqrt{3}}$, it follows that $$ \iint_S \nabla \times F\cdot dS= - \iint_{S_2} \nabla \times F\cdot dS = - \iint_{S_2}\pmatrix{-2\\-2\\-2}\cdot\pmatrix{-1\\-1\\-1}\frac{dS}{\sqrt{3}}=\frac{-6}{\sqrt{3}}\;A(S_2)=\frac{-6}{\sqrt{3}}\; \frac{2\pi}{3}=\frac{-4 \pi}{\sqrt{3}} $$

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Note: Perhaps the most difficult part is finding the area of $S_2$ ($A(S_2)$). The projection of $S_2$ in the $xy$ plane is the ellipse $x^2+y^2+(1-x-y)^2=1$, which has area $A'=\frac{2\pi}{3\sqrt{3}}$. It is not difficult to see that $A(S_2)=\sqrt{3}A'$. Alternatively, you can use the arguments described here to show that the disc $S_2$ has radius $\sqrt{\frac{2}{3}}$.