I'm trying to calculate the mass of a bullet-shaped object; a cylinder and a curved top, where only the density and equation for the round top is known.
bullet-shaped object is formed from a cylinder with a curved top. A cylinder that has radius a is situated with its base on the (x, y)-plane with the origin located at the centre of the base. The curved top of the bullet is
The cylinder is situated with its base on the $(x, y)$-plane with the origin located at the centre of the base.
The radius of the cylinder is equal to $a$.
The curved top is given by the formula: $$z = 4a-\frac{x^2}{a}+\frac{y^2}{a}$$
Density is given by $$D \cdot \left(\frac{\rho^2}{a^2}+1 \right)$$ where $a$ and $D$ are constants.
Attempt:
I'm sure it's a matter of triple integration:
$$ \text {Mass } = \int \delta V = \int f(\rho,\phi,z) ~\rho ~\delta z ~\delta \phi ~\delta \rho$$ $$= \int D \cdot \left(\frac {\rho^3}{a^2} + \rho \right) ~\delta z ~\delta \rho $$
I think the scalar field $f$ is independent of the azimuthal angle $\phi$ so we instead have:
$$\int\delta V = 2\pi \int_0^h D \cdot \left(\frac{\rho^3}{a^2}+\rho \right) ~\delta z ~ \delta\rho$$
Now, I have the density, and I have the equation for the curved top. How do I go about calculating the density of both the cylinder and the curved top?
I'd appreciate if, someone does take a shot at shedding some light on the problem, if they used similar notation.
Notations are not very clear. If $~\rho~$ is the radial distance from z-axis to a point, parallel to xy-plane, then $~\rho = \sqrt{x^2 + y^2}, x = \rho \cos\phi, y = \rho \sin\phi$
The curved top is $~ \displaystyle z = 4a-\frac{x^2}{a} + \frac{y^2}{a} = 4a - \frac{\rho^2 \cos 2\phi}{a}~$. As it sits on top of the cylinder of radius $a$ with base of the cylinder in xy-plane i.e. $z = 0$, the triple integral for the mass of the object can be written as,
$$\displaystyle \int_0^{2\pi} \int_0^a \int_0^{(4a^2 - \rho^2 \cos 2\phi) /a} D \cdot \left(\rho + \frac{\rho^3}{a^2}\right) ~ dz ~ d\rho ~ d\phi$$