Is the projection of my interest on the $xy$-plane a rectangle region with four vertices $(\pm \frac{\sqrt[]{2}}{2},\pm \frac{\sqrt[]{3}}{3} )$, which can be divided into four sub-region by two diagonal line $y=\pm \frac{\sqrt[]{6} }{3} x$?
Does the volume of the set $\iiint\limits_{S}^{} dV$, therefore can be compute through the four sub-region respectively, then sum up together by below? $$2\int_{x=0}^{x=\frac{\sqrt[]{2} }{2} } \int_{y=-\frac{\sqrt[]{6}x }{3} }^{y=\frac{\sqrt[]{6}x }{3}} \left [ \sqrt[]{1-2x^2}-(-\sqrt[]{1-2x^2}) \right ]dydx \\+2\int_{y=0}^{y=\frac{\sqrt[]{3} }{3} } \int_{x=-\frac{\sqrt[]{6}y }{2} }^{x=\frac{\sqrt[]{6}y }{2}} \left [ \sqrt[]{1-3y^2}-(-\sqrt[]{1-3y^2}) \right ]dxdy $$
Thanks for any help!
The answer is yes for both questions, the volume of $S$, which is the intersection of two elliptic cylinders (see Steinmetz solid), can be evaluated in that way.
However there is an easier approach. Note that $(x,y,z)\in S$ if and only if $z\in [-1,1]$ and $$(x,y)\in R_z:=\left[-\sqrt{\frac{1-z^2}{2}},\sqrt{\frac{1-z^2}{2}}\right]\times \left[-\sqrt{\frac{1-z^2}{3}},\sqrt{\frac{1-z^2}{3}}\right].$$ Therefore the volume of $S$ is $$\begin{align}V&=\int_{-1}^1\left(\iint_{R_z}dxdy\right)dz=\int_{-1}^1|R_z|dz=\int_{-1}^14\frac{1-z^2}{\sqrt{6}}dz\\ &=\frac{8}{\sqrt{6}}\left[z-\frac{z^3}{3}\right]_0^1=\frac{16}{3\sqrt{6}}=\frac{8\sqrt{6}}{9} \end{align}$$ where $|R_z|$ denotes the area of the rectangle $R_z$.