The base of $S$ is an elliptical region with boundary curve $9x^2+4y^2=36$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with hypotenuse in the base. The base of $S$ is an elliptical region with boundary curve $9x^2+4y^2=36$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with hypotenuse in the base.
$9x^2+4y^2=36 \implies \dfrac{x^2}{2^2}+\dfrac{y^2}{3^2}=1$. Based on the definition of the cross-section, the area of the triangular cross section ($T$) is $T=y^2=\frac{3}{2}\sqrt{4-x^2}$. By the symmetry of the object, the volume of $S$ is $3\int_{0}^{2}\sqrt{4-x^2}dx = V (\text{units}^3)$.
Is this true?
You are right that the area of each triangle is $y^2$, but that area is $y^2=\frac{9}{4}(4-x^2).$ Then $$V=\frac{9}{4}\int_{-2}^2(4-x^2)\,dx.$$ (Notice the $-2$ bound on the integral, the ellipse goes from $x=-2$ to $x=2$.)