The following is a problem in Apostol's Calculus, Vol II, section 12.13
- Consider the line integral $$\int_C ydx+zdy+xdz=\pi a^2\sqrt{3}\tag{1}$$
where $C$ is the curve of intersection of the sphere
$$x^2+y^2+z^2=a^2\tag{2}$$
and the plane
$$x+y+z=0\tag{3}$$
Use Stokes' theorem to show that the line integral (1) has the given value and explain how to traverse $C$ to arrive at the given answer.
Solving (3) for $z$
$$z=-x-y\tag{4}$$
and subbing into (2)
$$x^2+y^2+(-x-y)^2=a^2\tag{5}$$
we obtain
$$x^2+xy+y^2=\frac{a^2}{2}\tag{6}$$
This equation represents the values of $x$ and $y$ of points $(x,y,z)$ that are on the intersection $C$. Here is a plot of these points
This ellipse is the projection of $C$ onto the xy-plane.
Because the problem statement says to use Stokes' theorem, at this point I tried to compute the surface integral
$$\iint\limits_S (\text{curl}F\cdot \hat{n})dS\tag{7}$$
$$=\iint\limits_S \text{curl}F(r(u,v))\cdot N(u,v)dudv\tag{8}$$
where
$$N(u,v)= \frac{\partial r(u,v)}{\partial u}\times \frac{\partial r(u,v)}{\partial v}\tag{9}$$
is the fundamental vector product for a parametrization $r(u,v)$ of the surface $S$ (which is defined on some region $T$ in the uv plane that is the interior of a Jordan curve, the image under $r(u,v)$ of which is $C$), and
$$F(x,y,z)=y\hat{i}+z\hat{j}+x\hat{k}\tag{10}$$
My question is about a non-optimal choice of this surface. Ideally, to facilitate calculations, we would use the plane in between $C$ as our surface. But I would like to use the portion of the sphere that has $C$ as its boundary.
The sphere can be parametrized by spherical coordinates
$$r(\phi,\theta)=a^2(\cos{\theta}\sin{\phi}\hat{i}+\sin{\theta}\sin{\phi}\hat{j}+\cos{\phi}\hat{k})\tag{11}$$
which gives us
$$N(\phi,\theta)=\frac{\partial r(\phi,\theta)}{\partial \phi}\times \frac{\partial r(\phi,\theta)}{\partial \theta}=a^2(\sin^2{\phi}\cos{\theta}\hat{i}+\sin{\theta}\sin^2{\phi}\hat{j}+\cos{\phi}\sin{\phi}\hat{k})\tag{10}$$
and since we are interested only in the values of $\theta$ and $\phi$ that correspond to values on the surface that is above the ellipse, we can sub in the parametrization in (11) into the ellipse which should give us $\phi$ as a function of $\theta$ so that we know the limits of integration in the surface integral
$$x^2+xy+y^2=\frac{a^2}{2}\tag{11}$$
$$a^2\cos^2{\theta}\sin^2{\phi}+a^2\sin{\theta}\cos{\theta}\sin^2{\phi}+a^2\sin^2{\theta}\sin^2{\phi}=\frac{a^2}{2}\tag{12}$$
$$\sin^2{\phi}=\frac{1}{2(1+\sin{\theta}\cos{\theta}})\tag{13}$$
$$\sin{\phi}=\frac{1}{\sqrt{2(1+\sin{\theta}\cos{\theta}})}\tag{14}$$
$$\phi(\theta)=\sin^{-1}{\left (\frac{1}{\sqrt{2(1+\sin{\theta}\cos{\theta}})}\right )}\tag{15}$$
Finally,
$$\text{curl}F(x,y,z)=-\hat{i}-\hat{j}-\hat{k}\tag{16}$$
and the surface integral becomes
$$\iint\limits_S (\text{curl}F\cdot \hat{n})dS\tag{17}$$
$$=\int_0^{2\pi}\int_0^{\phi(\theta)} \langle -1,-1,-1\rangle\cdot \langle a^2\sin^2{\phi}\cos{\theta},a^2\sin{\theta}\sin^2{\phi}, a^2\cos{\phi}\sin{\phi}\hat{k} \rangle d\phi d\theta\tag{18}$$
Should we expect (18) to equal $\pi a^2\sqrt{3}$?
When I try to compute this in Maple, I don't really obtain the correct result.
If we apply Stokes theorem to the half-sphere $$S=\{(x,y,z): x+y+z\geq 0, x^2+y^2+z^2= a^2\}$$ then it is better to use tilted spherical coordinates with the $z$-axis along the line $(t,t,t)$. In these coordinates $\text{curl}F$ is $-\sqrt{3}\hat{k}$, and assuming that the orientation of $C$ is clockwise, we get $$\begin{align} \iint\limits_S (\text{curl}F\cdot \hat{n})dS&= \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/2} \langle 0,0,-\sqrt{3}\rangle\cdot \langle*,*,-\cos(\phi)\rangle\, a^2\sin(\phi)\,d\phi d\theta\\ &=2\pi a^2\sqrt{3}\,\left[\frac{\sin^2(\phi)}{2}\right]_{0}^{\pi/2}=\boxed{\sqrt{3}\pi a^2}. \end{align}$$
On the other hand, if we apply Stokes theorem to the plane surface $$S=\{(x,y,z): x+y+z=0, x^2+y^2+z^2\leq a^2\}$$ then we find $$\begin{align} \iint\limits_S (\text{curl}F\cdot \hat{n})dS &=\iint_E \langle -1,-1,-1\rangle\cdot \langle -1,-1,-1 \rangle dxdy \\ &=3|E|=\boxed{\sqrt{3}\pi a^2} \end{align}$$ where
$$|E|=\pi\cdot a \cdot \frac{a}{\sqrt{3}}=\frac{\pi a^2}{\sqrt{3}}$$ is the area of the ellipse $E=\{(x,y): x^2+xy+y^2\leq \frac{a^2}{2}\}$.