Showing that $\oint_C(y, z, x)\cdot \mathbf{dr} = \sqrt{3}\pi a^2$ with the use of Stokes' theorem

Question

I have to show that $\oint_C(y, z, x)\cdot \mathbf{dr} = \sqrt{3}\pi a^2$ when $C$ is the curve of intersection of $x^2 + y^2 + z^2 = a^2$ and $x + y + z = 0$. I currently confused as at the end I have derived that $\oint_C(y, z, x)\cdot \mathbf{dr} = 3\pi a^2$, instead of $\sqrt{3}\pi a^2$.

I got the result by 1.) Calculating the curl $\nabla \times F = (-1, -1, -1)$ 2.) Finding the normal vector to $G(x, y, z) = x + y + z = 0$, $\nabla G(x, y, z) = (1, 1, 1) = \mathbf{n}$, so normal to the direction of $\nabla \times F$ is $-\mathbf{n}$. Hence $\int\int_C(\nabla \times F)\cdot \mathbf{n} dS = \int\int_C(\nabla \times F)\cdot \frac{\mathbf{n}}{||\mathbf{n}||} ||\mathbf{n}||dA = \int\int_C(\nabla \times F)\cdot \mathbf{n}dA$

$ \implies \int\int_C(-1, -1, -1)\cdot(-1, -1, -1)dA = 3\int\int_CdA$. The area is just the are of a circle with radius $a$, so that $3\int\int_CdA = 3\pi a^2$.

Where did I go wrong?

Answer 1

Please note that surface defined by $x^2+y^2+z^2 \leq a^2, x+y+z = 0$ is a disk of radius $a$ (it is a great circle).

The unit normal vector to the surface is,

$\hat{n} = \frac{1}{\sqrt3} (-1, -1, -1)$

So the double integral should be

$\displaystyle \iint_S (-1, -1, -1) \cdot (-\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}) \ dS = \sqrt3 \pi a^2 $,

(as the surface area of the disk is $ \ \pi a^2$).

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EDIT: though it is more tedious and unnecessary but to complete the discussion in comments, if you wanted to evaluate the integral using area of projection in $XY$ plane,

$\displaystyle I = \iint_A (-1, -1, -1) \cdot (-1, -1, -1) \ dA = 3 A$, where $A$ is the area of the projection in $XY$ plane.

The projection (in $XY$ plane) of the surface at the intersection, is $x^2+y^2 + (-x-y)^2 \leq a^2 \ $ i.e. $x^2+y^2+xy \leq \frac{a^2}{2}$. You can plot to see this is an ellipse with major axis along $y=x$ and minor axis along $y = - x$. I will leave it for you to rewrite its equation and find its area, which comes to $\displaystyle \small \frac{\pi a^2}{\sqrt3}$.

Answer 2

Notice that: $$x^2+y^2+z^2=a^2$$ is a sphere centre $(0,0,0)$ and radius $a$. And: $$x+y+z=0$$ is a flat plane passing through $(0,0,0)$ and so it bisects the sphere. By definition, this means that the points of intersection form a circle and in fact it is the great circle, so also of radius $a$.

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Now for a plane $x+y+z=0$ we know that the normal vector will be: $$n=\begin{pmatrix}1\\1\\1\end{pmatrix}$$ but we want the unit normal vector, which will just be: $$\hat{n}=\frac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}$$ Can you take it from here?