Find the $y_G$ coordinate of the center of mass $G$ of the solid $A=\{(x,y,z) \in \mathbb{R}^3 \ | \ 3x^2 \leq y^2+z^2 \leq 3-x^2, yz \leq 0, z \geq |y|\}$ with constant density $\rho(x,y,z)=1$.
Can someone help me with the integration intervals for $y_G$? I'm not sure they are correct.
My try is the following: since $z \geq |y|$, it is $z \geq 0$ and so $yz \leq 0 \iff y \leq 0$, from this it follows that $|y|=-y$; so the set $A$ is equivalent to $$A=\{(x,y,z) \in \mathbb{R}^3 \ | \ 3x^2 \leq y^2+z^2 \leq 3-x^2, y \leq 0, z \geq -y\}$$
Using cylindrical coordinates $x=x$, $y=r \cos \theta$ and $z=r \sin \theta$ the set $A$ becomes
$$B=\{(x,r,\theta) \in \mathbb{R}^3 \ | \ 3x^2 \leq r^2 \leq 3-x^2, \cos \theta \leq 0, \sin \theta \geq -\cos \theta\}$$
It is $\cos \theta \leq 0 \iff \frac{\pi}{2}+2k\pi \leq \theta \leq \frac{3\pi}{2}+2k\pi$, intersected with $0 \leq \theta < 2\pi$ it gives $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$.
It is $$\sin \theta \geq - \cos \theta \iff \cos \left(\theta-\frac{\pi}{4}\right) \geq0 \iff \frac{\pi}{2}+2k\pi \leq \theta-\frac{\pi}{4} \leq \frac{3\pi}{2}+2k\pi$$ $$\iff \frac{3\pi}{4}+2k\pi \leq \theta \leq \frac{7\pi}{4}+2k\pi$$ Intersected with $0 \leq \theta < 2\pi$ it is $\frac{3\pi}{4} \leq \theta \leq \frac{7\pi}{4}$.
Intersecting the two conditions on $\theta$, it is $\frac{3\pi}{4} \leq \theta \leq \frac{3\pi}{2}$.
Finally, it is $3x^2 \leq r^2 \leq 3-x^2 \iff \sqrt{3}|x| \leq r \leq \sqrt{3-x^2}$; moreover, this implies that $\sqrt{3}|x| \leq \sqrt{3-x^2} \iff 3x^2 \leq 3-x^2 \iff -\sqrt{3}/2 \leq x \leq \sqrt{3}/2$ and so it is $$B=\left\{(x,r,\theta) \in \mathbb{R}^3 \ | \ -\frac{\sqrt{3}}{2} \leq x \leq \frac{\sqrt{3}}{2}, \frac{3\pi}{4} \leq \theta \leq \frac{3\pi}{2}, \sqrt{3}|x| \leq r \leq \sqrt{3-x^2} \right\}$$ So it is $$\int_A y dxdydz=\int_B r^2 \cos \theta dxdrd\theta = \int_{-\sqrt{3}/2}^{\sqrt{3}/2}dx\int_{3\pi/4}^{3\pi/2}d\theta \int_{\sqrt{3}|x|}^{\sqrt{3-x^2}}r^2 \cos \theta dx$$ Is this correct?
Edit: Found my mistake, I've solved incorrectly $\cos \left(\theta-\frac{\pi}{4}\right) \geq 0$; solving it correctly, it leads to $-\frac{\pi}{2} \leq \theta -\frac{\pi}{4} \leq \frac{\pi}{2}$ and so I agree with the answer of @Math Lover. However, my questions remain:
Questions:
i) I was thinking about this: since $z \geq 0$ it is $z=|z|$, hence $A$ seems even in $z$ too and so I would say that $z_G=0$: however I'm suspicious about this reasoning, and I think it doesn't work because for the symmetry it is required that $(x,y,-z) \in A$ and if I specify that $z \geq 0$ and so $z=|z|$, while it's true that $|-z|=|z|$ but from $-z \geq 0$ follows $z \leq 0$ and so the set isn't even in $z$ because the conditions $z \geq 0$ and $z \leq 0$ are different. Hence $(x,y,-z) \notin A$ if I write $A$ equivalently with $z \geq 0$ and $z=|z|$; is this the reason why it is wrong to say that $z_G=0$ with a symmetry argument?
ii) From the fact that $\sqrt{3}|x| \leq r \leq \sqrt{3-x^2}$ I have deduced that $\sqrt{3}|x| \leq \sqrt{3-x^2}$ and found a stricter bound on $x$ (because if I only considered the condition on $\sqrt{3-x^2}$ to be real it would've been $-\sqrt{3} \leq x \leq \sqrt{3}$); when I have these chained inequalities of the form "$E_1 \leq E_2 \leq E_3$", I must always consider the condition $E_1 \leq E_3$ too because, if I don't, there could be values of $E_3$ "that go below $E_1$" and so I'm integrating on a bigger set, overestimating the value of the integral? Thank you.
We are talking about solid bound outside cone $3x^2 = y^2 + z^2$ and inside sphere $x^2+y^2+z^2 = 3$ along with constraints, $y z \leq 0$ and $z \geq |y|$.
First of all there is symmetry about $x = 0$, so $\overline x = 0$.
We are using $y = r \cos\theta, z = r \sin\theta$ and as we have $y z \leq 0$ and $z \geq |y|$,
we get $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{4}$
As the mass density is $1$, the integral to find mass should be,
$\displaystyle m = \int_{-\sqrt3/2}^{\sqrt3/2} \int_{\pi/2}^{3\pi/4} \int_{\sqrt3|x|}^{\sqrt{3-x^2}} r \ dr \ d\theta \ dx$
Now $ \displaystyle \overline y = \frac{1}{m} \iiint y \ dV, \ \overline z = \frac{1}{m} \iiint z \ dV \ $ (as $dm = dV$)
Please note that due to symmetry about $x = 0$, we could have as well considered the solid only for $x \geq 0$. We would get same values of $\overline y$ and $\overline z$.
Last but not least, this can be much simpler in spherical coordinates,
$y = \rho \cos\theta \sin\phi, z = \rho \sin\theta \sin\phi, x = \rho \cos\phi$
$0 \leq \rho \leq \sqrt3, \frac{\pi}{3} \leq \phi \leq \frac{2 \pi}{3}, \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{4}$