I have a density function $\rho(x,y,z)$ and I am trying to find the mass of the shape parametrised by
\begin{equation*}\textbf{r}(u,v) = (\cos(u)\cos(v), \cos(u)\sin(v), u) , \hspace{2mm} u \in (-\frac{\pi}{2}, \frac{\pi}{2}), v \in (0, 2\pi)\end{equation*}
which is the boundary of a solid of revolution of $y = \cos(z)$ when rotated around the $z$ axis.
I think a logical way to set up this integral is \begin{equation} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{-\cos(z)}^{\cos(z)} \int_{-\cos(z)}^{\cos(z)}\rho(x,y,z)dxdydz \end{equation}
By the observation that both $-\cos(z) \leq y \leq \cos(z)$ $-\cos(z) \leq x \leq \cos(z)$ by looking at a plot (I used this site here)
Is this correct? I also considered using a surface integral but I couldn't get a good normal vector since the parametrisation is not regular or if I should have introduced a third variable $t$ to parametrise the solid shape and not the boundary.A spherical coordinate transform was almost perfect but the radius was bounded by a trigonometric function which was extremely messy, and I don't think using the divergence theorem and a vector field works nice either.
I may end up using numerical integration anyway, so tricky integrals are acceptable, I just don't want any transcendental functions or ludicrous expressions like spherical coordinates gives me.
Your approach is correct, except that your assumption that $−cos(z) \le y \le cos(z) ,−cos(z) \le x \le cos(z)$ isn't correct. that assumption would imply that (x,y) is bounded by a square with sides length $(2cos(z))$ . What you want however is that (x,y) is bounded by a circle with radius $cos(z)$.
This can be tricky if you are using Cartesian coordinates for (x,y) however integration over a circular area with radius $R$ can also be done using polar coordinates. $$\int_0^R{\int_0^{2\pi}{r\cdot f(rcos(\alpha),rsin(\alpha))d\alpha}dr}$$
As such your integral would become
$$\int_{-\frac{\pi}2}^{\frac{\pi}2}{\int_0^{cos(z)}{\int_0^{2\pi}{r \cdot \rho(rcos(\alpha),rsin(\alpha),z)d\alpha}dr}dz}$$