I found such an exercise among my set of exercises preparing for exams and I have no idea how to solve that.
Every point of ellipse $\{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 = 1, x + z = 1 \}$ is connected to the point $(0,0,0)$ with line segments. Set $M$ is created by all such line segments excluding the points at their ends. Prove that $M$ is a manifold and calculate the integral:$$\int_M \sqrt{x + 3z}\ d \lambda_2$$
I made such a picture. Set $M$ is denoted by green colour.
Manifold proof
In the next paragraph, I calculated the determinant of jacobian matrix and it is never equal to 0. Therefore we know that it is indeed a manifold.
Integral
On the eclipse, we can use cylindrical parametrization:
- $x = r \cos \alpha$
- $y = r \sin \alpha$
Form the first equation: $x^2 + y^2 =1 \implies r^2 \cos^2 \alpha + r^2 \sin^2 \alpha = 1 \implies r^2 = 1 \implies r = 1$.
From that and second equation we have: $x + z = 1 \implies z = 1 - \cos \alpha$.
Therefore, when we are on the elipse, we have:
- $x = \cos \alpha$
- $y = \sin \alpha$
- $z = 1 - \cos \alpha$
To facilitate movement on line segments from $(0,0,0)$ to the eclipse we put additional parameter t that belongs to $(0, 1)$. Therefore, we can parametrize set M by:
- $x = t \cos \alpha$
- $y = t \sin \alpha$
- $z = t - t \cos \alpha$
Now, it's necessary to calculate the determinant of Jacobian matrix for such parametrization:
\begin{pmatrix}cos \alpha & -t sin \alpha \\sin \alpha & t cos \alpha \\1 - cos \alpha & t sin \alpha\end{pmatrix}
It's not square, so we need to multiply by a transposition. We get:
$$\begin{pmatrix} cos \alpha & -t sin \alpha \\ sin \alpha & t cos \alpha \\ 1 - cos \alpha & t sin \alpha \end{pmatrix}^T \cdot \begin{pmatrix} cos \alpha & -t sin \alpha \\ sin \alpha & t cos \alpha \\ 1 - cos \alpha & t sin \alpha \end{pmatrix} = $$
$$\begin{pmatrix} cos \alpha & sin \alpha & 1 - cos \alpha \\ -t sin \alpha & t cos \alpha & t sin \alpha \end{pmatrix} \cdot \begin{pmatrix} cos \alpha & -t sin \alpha \\ sin \alpha & t cos \alpha \\ 1 - cos \alpha & t sin \alpha \end{pmatrix} = $$
$$\begin{pmatrix} cos^2 \alpha + sin^2 \alpha + (1 - cos \alpha)^2 & cos \alpha (-t sin \alpha) + sin \alpha (t cos \alpha) + (1 - cos \alpha)(t sin \alpha) \\ (-t sin \alpha) cos \alpha + (t cos \alpha) sin \alpha + (t sin \alpha)(1 - cos \alpha) & (-t sin \alpha)^2 + (t cos \alpha)^2 + (t sin \alpha)^2 \end{pmatrix}$$
$$ = \begin{pmatrix} 2 - 2 cos \alpha & t sin \alpha (1 - cos \alpha) \\ t sin \alpha (1 - cos \alpha) & t^2 \end{pmatrix}$$
Therefore, $\sqrt{Det(J^TJ)} = t$
We can calculate the integral:
$$\int_M \sqrt{x + 3z}\ d \lambda_2 = \int_0^{2 \pi} \int_0^1 \sqrt{t cos \alpha + 3(t - tcos \alpha)} \cdot t \ d t \ d \alpha = $$$$= \int_0^{2 \pi} \int_0^1 \sqrt{3 - 2cos \alpha} \cdot t^{\frac{3}{2}} \ d t \ d \alpha = $$$$= \frac{2}{5} \int_0^{2 \pi} \sqrt{3 - 2cos \alpha} \ d \alpha $$
And I'm stuck here. What do I do next?