I have a solid V enclosed by the surface (S) : $2x^2-y^2+2z^2-1=0$ and the planes $z = -1$ and $z=1$
I'm trying to calculate the area of the cross section obtained by cutting V with de $xz$ plane and the same thing but with the plane whose equation is $x-z = 0$.
I really don't know how to proceed, I tried to say that $y=0$ and rewrite (S) as $ 2x^2+2z^2-1=0$ but then I'm stuck.
For the first case, the required is shape is just circle with equation:$$ x^2 + z^2 = \cfrac{1}{2} $$ $\Rightarrow$ Radius $= \sqrt{\cfrac{1}{2}}$ $\Rightarrow$ Area $ = \cfrac{\pi}{2}$
For the second case, the area is unbounded, as the solid is not bounded sufficiently. This can be proved as follows:
Equation of solid: $ 2x^2 - y^2 + 2z^2 < 1 $ with constraint $ |z| \le 1 \Rightarrow z^2 \le 1 \Rightarrow |2x^2 - y^2| < 1$
Which would be hyperbolic sheet and the parameters of hyperbola would depend on value of $z$, on taking $z=1$, the equation of surface of intersection of the plane $z=1$ and the solid is obtained which is $y^2 - 2x^2 < 1$ which is unbounded. And as the plane $x = z$ extends infinitely in $y$ direction, the required cross sectional area is not finite.