Calculation of the volume of Steinmetz solid

Question

According to https://mathworld.wolfram.com/SteinmetzSolid.html, the volume of the Steinmetz solid with radii 1 referencing (11) is $V_2(1,1)=\displaystyle\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} dz dy dx = \frac{16}{3}$
My problem is I cannot get to this result. Here is my attempt:
$V_2(1,1)=\displaystyle\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2\sqrt{1-y^2} dy dx$
$Using \displaystyle\int 2\sqrt{1-t^2} dt = [arcsin(t)+t \sqrt{1-t^2}]+C$ we get
$V_2(1,1)=2\displaystyle\int_{-1}^{1} arcsin(\sqrt{1-x^2})+x \sqrt{1-x^2} dx$
The function $arcsin(\sqrt{1-x^2}$ is symmetric to y-axis and the function $x \sqrt{1-x^2}$ is symmetric to 0, hence
$V_2(1,1)=4\displaystyle\int_{0}^{1} arcsin(\sqrt{1-x^2}) dx$
Using $\displaystyle\int arcsin(\sqrt{1-t^2}) dt = [t\times arcsin(\sqrt{1-t^2})-\sqrt{1-t^2}+C]$ for $0\le t\le 1$ we get $V_2(1,1) = 4 (1\times 0-\sqrt{1-1}-(0\times arcsin(1)-\sqrt{1}) = 4$
Where is my mistake?

Answer 1

Upon integrating w.r.t. $y$, you should have ended up with

$$2 \int_{-1}^1 \left(\arcsin \sqrt{1-x^2} + \sqrt{1-x^2}\color{red}{\sqrt{x^2}}\right) \, dx$$

since $\sqrt{1-y^2} \bigg|_{y=\pm\sqrt{1-x^2}} = \sqrt{1-\left(1-x^2\right)} = \sqrt{x^2}$.

Now, $\sqrt{x^2}=|x|$ which is not odd (I assume that's what you meant by "symmetric to $0$"), so you should have arrived at

$$4 \int_0^1 \arcsin \sqrt{1-x^2} \, dx + 2 \left(\int_{-1}^0 \color{red}{-x} \sqrt{1-x^2} \, dx + \int_0^1 \color{red}{x}\sqrt{1-x^2} \, dx\right) \\ = 4 \int_0^1 \left(\arcsin \sqrt{1-x^2} + x \sqrt{1-x^2}\right) \, dx$$