Computing line integral using Stokes´theorem

Question

Use Stokes´ theorem to show that $$\int_C ydx+zdy+xdz=\pi a^2\sqrt{3}$$ where $C$ is the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and the plane $x+y+z=0$

My attempt:

By Stokes´ theorem I know that $$\int_S (\nabla \times F) \cdot n \ dS=\int_c F \cdot d\alpha$$ In this case the intersection curve $C$ is a circle and $S$ is "half" of the sphere

using $r(u,v)=(acos(u)sin(v), a sin(u)cos(u), acos(v))$ $0\le v \le \pi$, $-\pi/4 \le u \le 3\pi/4$ as a parametrization of the sphere and computing $\nabla \times F=(-1,-1,-1)$ (where $F=(y,z,x)$), and $${\partial r\over \partial u}\times {\partial r\over \partial v}$$ the surface integral becomes:

$$\int_{-\pi/4}^{3\pi/4}\int_{0}^{\pi}({a^2sin(v)sin(2u)\over 2}-a^2sin(v)sin^2(u)+{a^2sin(2v)\over 2})dv\ du$$, but after computing the integral I don´t get the answer, Can you please tell me where is my mistake?

Answer 1

Your parameterization doesn't look right. In particlar, note that if $x a\cos u \sin v$, $y = a\sin u \cos u$, and $z = a\cos v$, then $x^2+y^2+z^2 = a\cos^2u\sin^2v+a^2\sin^2u\cos^2u+a^2\cos^2v$, which does not simplify to $a^2$. One correct parameterization of the sphere would be $\vec{r}(u,v) = (a\cos u \sin v, a\sin u \color{red}{\sin v}, a\cos v)$, for appropriate bounds on $u$ and $v$.

Instead of letting $S$ be the half-sphere, why not let $S$ be the flat circular disk in the plane $x+y+z = 0$. Both of these surfaces have $C$ as their boundary, but one is easier to integrate over.

You know that $\nabla \times F = (-1,-1,-1)$ is constant on this disk. Also, the unit normal to the circle is the same as the unit normal to the plane $x+y+z = 0$ which is $\hat{n} = \dfrac{1}{\sqrt{3}}(1,1,1)$. Now, it is easy to compute $(\nabla \times F) \cdot \hat{n}$, which happens to be constant on this disk.

Also, for any constant $c$, we have $\displaystyle\iint\limits_S c\,dS = c \cdot \text{Area}(S)$.

Can you figure out the integral from these hints?

Also: note that if you get $-\pi^2a\sqrt{3}$ instead of $+\pi^2a\sqrt{3}$, its ok since the problem didn't specify in which direction $C$ was traversed.

Answer 2

You did not specify the orientation of the curve $C$. But, you gave the answer. So we have to choose the orientation that will give the given solution. The curve $C$ is in the plane $x+y+z = 0$. The vector $(-1,-1,-1)$ determines one orientation of this plane. Let the curve $C$ be oriented using the right-hand rule. Let $D$ be the disk in the plane $x+y+z = 0$ bounded by the circle $C$; it is the disk cetered at the origin and with radius $a$. The unit normal to $D$ is $(-1,-1,-1)/\sqrt{3}$. By Stokes' theorem \begin{align*} \int_C (y dx+z dy + x dz) & = \iint_D (-1,-1,-1)\cdot (-1,-1,-1)/\sqrt{3} \ dA \\ & = \sqrt{3} \iint_D dA \\ & = \pi a^2 \sqrt{3}, \end{align*} since $\iint_D dA$ is the area of the disk $D$.