Calculating the differential line element/parameterization I was wondering if someone could please help me with the following question, I've been stuck for a while now,
If there is a contour defined by let's say the following
-A spherical shell of radius, $A$, so $x^2+y^2+z^2=A^2$
-A tilted plane formed by rotating the x-y plane by 45 degrees about the x-axis (so y-z=0)
-A semi-circular contour, C, formed from the intersection, such that y>0
How do I parameterize this in Cartesian coordinates and find the differential line element?
I originally went about this question like this: $x=u$, $y=z$ so: $x^2+2y^2=A^2$
I'm not sure if this is 100% correct, I have been stuck on the differential line element could someone please, please help with the differential line element? How should I approach it with the parameterized parameters above (if they're correct)?
 A: As the curve is in the plane $y = z$,
For any $x \in (-A, A), ~ x^2 + y^2 + z^2 = A^2$
$ \implies x^2 + 2y^2 = A^2$
$y = \sqrt{ \frac{A^2 - x^2}{2}}$, as $y \gt 0$
That leads to parametrization of the curve as,
$ \displaystyle \left(x, \sqrt{ \frac{A^2 - x^2}{2}}, \sqrt{ \frac{A^2 - x^2}{2}} \right), - A \leq x \leq A$
Now find line element using the arc length formula,
$ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2 + \left(\frac{dz}{dx}\right)^2} ~ ~dx$
A: Hint :
It's worth to try by yourself using for example this method :
Step 1 - Make a parametric representation of a circle of radius A in the x-y plane
\begin{equation}
\mathbf r\left(\lambda\right)\boldsymbol=
\begin{bmatrix}
x\left(\lambda\right)\boldsymbol=\cdots\\
y\left(\lambda\right)\boldsymbol=\cdots\\
z\left(\lambda\right)\boldsymbol=\:\:0\:\: 
\end{bmatrix}
\tag{a}\label{a}   
\end{equation}
It would be convenient to use as parameter the angle with the $\:x\boldsymbol-$axis $\:\lambda\boldsymbol\equiv\omega\in[0,\pi]$.
Step 2 - Apply a 3D$\boldsymbol-$rotation around the $\:x\boldsymbol-$axis through an angle $\:\theta\boldsymbol=\pi/4\boldsymbol=45^\circ$
\begin{equation}
\mathbf r'\left(\omega\right)\boldsymbol=
\begin{bmatrix}
x'\left(\omega\right)\\
y'\left(\omega\right)\\
z'\left(\omega\right)
\end{bmatrix}
\boldsymbol=
\mathcal R\left(\mathbf n, \theta\right)
\begin{bmatrix}
x\left(\lambda\right)\boldsymbol=\cdots\\
y\left(\lambda\right)\boldsymbol=\cdots\\
z\left(\lambda\right)\boldsymbol=\:\:0\:\: 
\end{bmatrix}
\tag{b}\label{b}   
\end{equation}
...where $\:\mathcal R\left(\mathbf n, \theta\right)\:$ is the $\:3\times 3\:$ matrix which represents the rotation around the unit vector $\:\mathbf n\:$ through angle $\:\theta$. In our case
\begin{equation}
\mathbf n\boldsymbol=\mathbf e_{\texttt x}=\left(1,0,0\right) \quad \texttt{and} \quad \theta\boldsymbol=\pi/4\boldsymbol=45^\circ
\tag{c}\label{c}   
\end{equation}
For the $\:3\times 3\:$ matrix $\:\mathcal R\left(\mathbf n, \theta\right)\:$ see my answer here Rotation of a vector.

