Parametrizing boundary for Stokes' Theorem

Question

Evaluate the integral $\iint_S(\nabla \times \mathbf{F}) \cdot d\mathbf{S}$, where $S$ is the portion of the surface of a sphere defined by $x^2 + y^2 + z^2 = 1$ and $x + y + z \ge 1$, and where $\mathbf{F} = \mathbf{r} \times (\mathbf{i} + \mathbf{j} + \mathbf{k}), \mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$.

Is the correct approach for this problem to use Stokes' Theorem and calculate $\int_C\mathbf{F} \cdot d\mathbf{s}$, where $C$ is the boundary of $S$? I tried doing this, but I had trouble parametrizing $C$. In particular, I plugged in $z = 1 - x - y$ into $x^2 + y^2 + z^2 = 1$ and was stuck when I had to parametrize the resulting region in the xy-plane. Where did I go wrong? What would I need to do differently if I had to solve this problem going the line integral route?

Answer 1

No doubt you’re familiar with the parameterization in $\mathbb R^2$ of a circle with radius $r$ centered at $\mathbf c=(x_c,y_c)$ as $(x_c+r\cos t,y_c+r\sin t)$. We can rewrite this in vector form as $\mathbf c+(r\cos t)\mathbf e_1+(r\sin t)\mathbf e_2$. It should be fairly clear that we can use any pair of orthogonal unit vectors $\mathbf u$ and $\mathbf v$ instead of the standard basis vectors $\mathbf e_1$ and $\mathbf e_2$. This amounts to changing the starting point on the circle of the parameterization and possibly reversing its orientation. Observe now that $\mathbf c+(r\cos t)\mathbf u+(r\sin t)\mathbf v$ works as a parameterization of a circle in an ambient space of any dimension $\ge2$.

In your problem, the circle lies on the plane $x+y+z=1$ with normal $\mathbf n=\mathbf i+\mathbf j+\mathbf k$, which also happens to be the constant vector in the definition of $\mathbf F$. Its center is at $\frac13(\mathbf i+\mathbf j+\mathbf k)$ and its radius can be found via the Pythagorean theorem to be $\sqrt{2/3}$. To complete the parameterization, we just need a pair of orthogonal unit vectors that are perpendicular to $\mathbf n$. An obvious choice is $\mathbf u = \frac1{\sqrt2}(\mathbf i-\mathbf j)$ and the other is then found by computing $\mathbf n\times\mathbf u$ and normalizing. The resulting parameterization of the circle is $$\frac13\left(\sin t+\sqrt3\cos t+ 1\right)\mathbf i + \frac13\left(\sin t-\sqrt3\cos t+1\right)\mathbf j + \frac13\left(1-2\sin t\right)\mathbf k.$$

You can save yourself a lot of work in computing $\mathbf F\cdot d\mathbf s$ with the following observations:

• $\mathbf F$ is orthogonal to both $\mathbf n$ and $\mathbf r$, so along $C$ it is parallel to $\dot{\mathbf r}$, but points in the opposite direction. Thus, $\mathbf F\cdot d\mathbf s = -\lVert\mathbf F\rVert \lVert\dot{\mathbf r}\rVert dt$.
• Along $C$, $\lVert\mathbf F\rVert = r\lVert\mathbf n\rVert$.
• $\lVert\dot{\mathbf r}\rVert = {2\pi r\over 2\pi} = r$.

It sounds like you tried something like solving $x+y+z=1$ for $z$ and then substituting into the equation of the sphere, which produces $x^2+xy+y^2-x-y=0$. Using standard methods such as setting the partial derivatives to zero, you can find that the center of this conic is at $\left(\frac13,\frac13\right)$. The symmetry of the quadratic part suggests a rotation through $\pi/2$: setting $x={x'+y'\over\sqrt2}+\frac13$ and $y={y'-x'\over\sqrt2}+\frac13$ reduces the equation to $\frac32x'^2+\frac92y'^2=1$. This ellipse can be parameterized as $\left(\frac{\sqrt2}3\cos t,\sqrt{\frac23}\sin t\right)$. If you work back through the transformations, you’ll end up with the same parameterization as above.