I would like to find the coordinates of mass centre of the the following solid $$D = \{(x, y, z) \in \mathbb{R}^3: (x^2 + y^2 + z^2)^2 \le z \}.$$
The density of this solid is constant and equal to $a$.
I know that I am to compute some triple integrals: $$M = \int \int \int _D a \, d \Omega, \\ C_{Mx} = \frac{1}{M}\int\int\int_D a x \, d \Omega,$$ similarly for $y$ and $z$.
How to deal with these integrals?
On
First, let's reduce the problem by substituting:
$$x=r \cos \phi \\ y=r \sin \phi$$
The condition becomes:
$$(r^2+z^2)^2 \leq z \\ 0 \leq r^2 \leq \sqrt{z}-z^2$$
To satisfy this condition we should also have:
$$0 \leq z \leq 1$$
The first integral is:
$$M=a \int_0^{2 \pi} d \phi \int_0^1 dz \int_0^{\sqrt{\sqrt{z}-z^2}} r dr= \pi a \int_0^1 (\sqrt{z}-z^2) dz= \frac{\pi a}{3}$$
The $x$ and $y$ integrals are:
$$C_x=\frac{a}{M} \int_0^{2 \pi} \cos \phi d \phi \int_0^1 dz \int_0^{\sqrt{\sqrt{z}-z^2}} r^2 dr=0$$
$$C_y=\frac{a}{M} \int_0^{2 \pi} \sin \phi d \phi \int_0^1 dz \int_0^{\sqrt{\sqrt{z}-z^2}} r^2 dr=0$$
They are zero because the solid defining condition is even, so the center line passes through $x=y=0$. In these variables, we are integrating $\cos \phi, \sin \phi$ over the whole period which gives us $0$ again.
Finally, the last integral:
$$C_z=\frac{2 \pi a}{M} \int_0^1 z dz \int_0^{\sqrt{\sqrt{z}-z^2}} r dr=\frac{ \pi a}{M} \int_0^1 z (\sqrt{z}-z^2) dz = \frac{3\pi a}{20 M}=\frac{9}{20}$$
The mass center of $D$ is $\left(\bar x,\bar y,\bar z\right)$, where$$\bar x=\iiint_D x\,\mathrm dx\,\mathrm dy\,\mathrm dz,\ \bar y=\iiint_D y\,\mathrm dx\,\mathrm dy\,\mathrm dz\text{, and }\bar z=\iiint_D z\,\mathrm dx\,\mathrm dy\,\mathrm dz.$$By symmetry, $\bar x=\bar y=0$. In order to compute $\bar z$, I will use spherical coordinates. So, we have $x=\rho\cos(\theta)\sin(\varphi)$, $y=\rho\sin(\theta)\sin(\varphi)$, and $z=\rho\cos(\varphi)$. In these coordinates, the condition $(x^2+y^2+z^2)^2\leqslant z$ becomes $\rho^4\leqslant\rho\cos(\varphi)$. So, $0\leqslant\varphi\leqslant\frac\pi2$, and the computation of $\bar z$ becomes\begin{multline}\int_0^{2\pi}\int_0^{\pi/2}\int_0^{\sqrt[3]{\cos(\varphi)}}\rho\cos(\varphi)\rho^2\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta=\\=2\pi\int_0^{\pi/2}\int_0^{\sqrt[3]{\cos(\varphi)}}\rho^3\cos(\varphi)\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi.\end{multline}It turns out to be equal to $\frac{3\pi}{20}$. So, the mass center of $D$ is $\left(0,0,\frac{3\pi}{20}\right)$.