Parameterization of a curve equal to the intersection of a sphere with a plane

Question

In the task it is necessary to calculate the curvilinear integral. But I was faced with the fact that I can not parameterize this curve in any way. How can I parameterize a curve? $x^2 + y^2 + z^2 = a^2 , x+y+z=a$

Answer 1

Start with spherical coordinates

$$x = r\sin\theta\cos\phi$$

$$y = r\sin\theta\sin\phi$$

$$z = r\cos\theta$$

Then apply $x^2+y^2+z^2=a^2 \implies r = a$

$$x = a\sin\theta\cos\phi$$

$$y = a\sin\theta\sin\phi$$

$$z = a\cos\theta$$

Next use

$$x+y+z = a \implies \sin\theta(\cos\phi + \sin\phi)+\cos\theta = 1$$

$$ \cos\phi + \sin\phi = \frac{1-\cos\theta}{\sin\theta} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right)$$

which makes

$$\theta = 2\cot^{-1}\left(\frac{1}{\cos\phi+\sin\phi}\right)$$

The reason we found $\theta$ as a function of $\phi$ instead of the other way around is because $\phi$ will wrap around the fourth and second quadrants and there is no inverse trig function that smoothly encompasses both. On the other hand, $\theta\in[0,\pi]$ which can be covered by one inverse trig.

Using the following facts

$$\begin{cases} \sin\left(\tan^{-1}\left(\cos\phi+\sin\phi\right)\right) = \frac{\cos\phi+\sin\phi}{\sqrt{2+\sin2\phi}} \\ \cos\left(\tan^{-1}\left(\cos\phi+\sin\phi\right)\right) = \frac{1}{\sqrt{2+\sin2\phi}} \end{cases}$$

$$ \implies \begin{cases} \sin\left(2\tan^{-1}\left(\cos\phi+\sin\phi\right)\right) = \frac{\cos\phi+\sin\phi}{1+\sin\phi\cos\phi} \\ \cos\left(2\tan^{-1}\left(\cos\phi+\sin\phi\right)\right) = -\frac{\sin 2\phi}{2+\sin2\phi} \end{cases}$$

we can have our final parametrization

$$x = \frac{a\cos\phi(\cos\phi+\sin\phi)}{1+\sin\phi\cos\phi}$$

$$y = \frac{a\sin\phi(\cos\phi+\sin\phi)}{1+\sin\phi\cos\phi}$$

$$z = -\frac{a\sin\phi\cos\phi}{1+\sin\phi\cos\phi}$$

where $\phi\in[0,\pi]$ (since the parametrization is $\pi$-periodic)