So in this exercise I am asked to find the center of mass of a certain solid defined by 2 elipsoids that have the following parametrizations:
Elipsoid A:
$$x=\cos\alpha \sin\theta\\ y=\frac{1}{2}\sin\alpha \sin\theta\\ z=\frac{1}{3}\cos\theta$$ Ald elipsoid B
$$x=\cos\alpha \sin\theta\\ y=\frac{1}{2}\sin\alpha \sin\theta\\ z=\frac{1}{3}(\cos\theta+1)$$
for $0\leq\alpha\leq2\pi$, $0\leq\theta\leq\pi$
Now I dont know how to get a parametrization of the intersection, however, I can define two separate parametrizations that together define the intersection, changing the range values of the variables $\alpha$ and $\theta$
If I let the 1st same parametrization of A but I let
$0\leq\alpha\leq2\pi$, $0\leq\theta\leq\frac{\pi}{3}$;
and the second parametrization B with
$0\leq\alpha\leq2\pi$, $\frac{2\pi}{3}\leq\theta\leq\pi$.
Then this two parametrizations define the intersection (I have plotted it and it does give the intersection)
However, for calculating the center of mass I need the total mass, that says is proportional to the distance to the $xy$ plane, so $\rho=Kz$.
My idea to compute the total mass (correct me if I’m wrong) is to, separately in each parametrization, integrate the mass of the “shells” inside, so it would be something like this: $$\int\int\int K z dxdydx=K\int_{\theta=0}^{\frac{\pi}{3}}\int_{\alpha=0}^{2\pi}\int_{r=0}^1 r\frac{1}{3} \cos\theta \cdot \frac{r^2}{6}\sin\theta dr d\alpha d\theta$$
And the same thing with the other integral, changing the limits of $\alpha$ and the other value of $z$; where the term $\frac{r^2}{6}\sin\theta$ is the determinant of the jacobian of the change of coordinates...
Can you help me and tell me if this method is fine for calculating the total mass, or if you find any other way of doing it!, thanks in advance