Stokes Theorem. Where is my mistake?

Question

Use Stoke's Theorem to prove that the following line integral has the indicated value. $$ \int_\mathscr{C} y \,dx +z\,dy+x\,dz = \pi a^2 \sqrt{3}$$ where $\mathscr{C}$ is the intersection curve $\mathscr{C}: \begin{cases} x^2+y^2+z^2=a^2 \\ x+y+z=0 \end{cases}$

This is what I did. $$ \mathbf{I} = \int_\mathscr{C} y \,dx +z\,dy+x\,dz = \int_\mathscr{C} \mathbf{F}(x,y,z) \cdot d\mathbf{\overrightarrow{r}} \qquad \text{where} \,\mathbf{F}(x,y,z)=(y,z,x)$$

Then, by the Stoke's theorem $$ \mathbf{I} =\iint_S(\nabla \times \mathbf{F}) \cdot\mathrm{d}\mathbf{S}= \iint_S (-1,-1,-1) \cdot \mathrm{d}\mathbf{S} $$ I choose $S$ as the plane that contains $\mathscr{C}$.

$$ \mathbb{\alpha}(x,y)=(x,y,-x-y) \qquad(x,y)\in T \\ $$ In cylindrical coordinates $$ \mathbb{\alpha}(r,\theta)=\left( rcos\theta,rsin\theta,-r(cos\theta+ sin\theta)\right) \qquad (r,\theta)\in [0,a]\times[0,2\pi]=T^* \\ N= \frac{\partial\alpha}{\partial r} \times \frac{\partial\alpha}{\partial\theta}=(r,r,r)$$

Therefore $$ \mathbf{I}= \iint_{T^*} (-1,-1,-1) \cdot (r,r,r) \mathrm{d}A = \int_0^{2\pi} \int_0^a -3r \, \mathrm{d}r\mathrm{d}\theta = -3a^2\pi$$ Unfortunately, I can't find where is my mistake. Can you help me, please?

Answer 1

Since the plane passes through the origin, the curve of intersection is a circle of radius $a$ and so we may take our surface to be a disk of radius $a$. (Note that Stokes' Theorem allows you to take $\textit {any}$ surface whose boundary is the curve over which the line integral is to be taken. We choose the disk because it is the easiest).

Now, observe that you can read off a vector normal to the disk because it is part of the plane $x+y+z=0$; namely, $(1,1,1)$ (choosing one of two possible normals).

So your integral is $$\frac{1}{\sqrt 3}\int \int _{S}(\vec \nabla \times \vec F)\cdot (1,1,1)dS=\frac{1}{\sqrt 3}\int \int _{S}(-1,-1,-1)\cdot (1,1,1)dS=-\frac{3}{\sqrt 3}\int \int _{S}dS$$ But this last integral just computes the surface area of what we have determined is a disk of radius $a$, so we get $-\frac{3\pi a^{2}}{\sqrt 3}$ and so there was no need here to worry about parameterizations! And this is probably the point of the exercise. Trying to find a suitable parameterization is a real chore.

Answer 2

One parametrization of $\mathscr{C}$ is $$ \begin{bmatrix}x\vphantom{\frac1{\sqrt2}}\\y\vphantom{\frac1{\sqrt2}}\\z\vphantom{\frac1{\sqrt2}}\end{bmatrix} =a\cos(\theta)\begin{bmatrix}\frac1{\sqrt2}\\\lower{2pt}{0}\vphantom{\frac1{\sqrt2}}\\-\frac1{\sqrt2}\end{bmatrix} +a\sin(\theta)\begin{bmatrix}\frac1{\sqrt6}\\-\frac2{\sqrt6}\\\frac1{\sqrt6}\end{bmatrix}\tag{1} $$

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However, using Stokes Theorem, we get the integral to be $$ \int_{\mathscr{C}}y\,\mathrm{d}x+z\,\mathrm{d}y+x\,\mathrm{d}z=\iint_{\mathscr{S}}\underbrace{\,\mathrm{d}y\,\mathrm{d}x+\,\mathrm{d}z\,\mathrm{d}y+\,\mathrm{d}x\,\mathrm{d}z}_{(1,1,1)\,\cdot\,n\,\mathrm{d}\sigma}\tag{2} $$ where $\mathscr{S}$ is the intersection surface $\mathscr{S}:\left\{\begin{array}{l} x^2+y^2+z^2\le a^2\\ x+y+z=0 \end{array}\right.$

Since the unit normal to $\mathscr{S}$ is $\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$, the integral on the right side of $(2)$ equals the area of $\mathscr{S}$ times $\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)\cdot(1,1,1)=\sqrt3$. That is, the integral is $$ \underbrace{\pi a^2}_{\substack{\text{area}\\\text{of $\mathscr{S}$}}}\sqrt3\tag{3} $$