Vector Calculus (5-12-2020)

Consider the plane:

$x+3y+2z=1$

This plane's intersection with the sphere is a circular closed curve. A unit vector normal to it is:

$\hat{n} = \frac{1}{\sqrt{14}}\left(\hat{i} + 3 \ \hat{j} + 2 \ \hat{k}\right)$

The line along this normal direction through the origin can be derived to be:

$x = t$ $y = 3t$ $z = 2t$

Using this parameterization, the point of intersection between the line and the plane is found to be:

$C = \left(\frac{1}{14},\frac{3}{14},\frac{2}{14}\right)$

This point $C$ is the center of the circle of the intersection of the plane and the sphere. Having found the center, consider another point on the plane. I chose:

$P = \left(1,0,0\right)$

A vector joining points $P$ and $C$ directed towards $P$ is:

$\vec{n}_1 = \vec{P}-\vec{C}$

The unit vector along this direction is:

$\hat{n}_1 = \frac{\vec{n}_1}{\lvert \vec{n}_1 \rvert}$

Now, a vector on the plane perpendicular to both $\hat{n}$ and $\hat{n}_1$ is:

$\hat{n}_2 = \hat{n} \times \hat{n}_1$

The radius of the circle of intersection can be found using Pythagoras' theorem as such:

$R = \sqrt{1 - \lvert \vec{C}\rvert^2}$

Finally, any point on the circle of intersection is:

$\vec{r} = \vec{C} + R\cos{\theta} \ \hat{n}_1 + R\sin{\theta} \ \hat{n}_2$

$d\vec{r} = \left(-R\sin{\theta} \ \hat{n}_1 + R\cos{\theta} \ \hat{n}_2\right)d\theta$

$dI = \vec{F} \cdot d\vec{r}$

This dot product evaluates to some function of $\theta$ times $d\theta$ , say $f(\theta) \ d\theta$

$\implies I = \int_{0}^{2\pi} f(\theta) \ d\theta$

This integral can also be done using's Stokes's theorem as such. Having parameterised the circle of intersection, it's elementary surface area can be parameterised as well. This gives:

$\oint_{C} \vec{F} \cdot d\vec{r} = \int_{S} \left (\nabla \times \vec{F}\right)\cdot d\vec{S}$

Where $d\vec{S} = r \ dr \ d\theta \ \hat{n}$ where $r$ varies from $0$ to $R$ . Note that:

$\nabla \cdot \left (\nabla \times \vec{F}\right)=0$