I'm having serious problems calculating a volume of a cylindroid, defined this way: z-parallel cylinder heads, and included between $z = 0$ and the function $f(x,y)= y^2-x^4$. The vertical projection is a quadrilateral with this vertices: $(0,0) , (0,3), (3,0), (1,1)$.
Trying to understand how I can do this using triple integrals (for wires), I splitted the problems in two: when $z>0$, and $y>x^2$ , and when $y<x^2$ ($z<0$). So, I got one integral for the first case: $$\int^1_0\int^{3-2x}_{x^2}\int^{y^2-x^4}_{0}1dzdydx$$, and two (to represent the section on which to integrate) integrals foor the second part:$$\int^1_0\int^{x^2}_{0}\int_{y^2-x^4}^{0}1dzdydx$$ $$\int^3_1\int^{3-2x}_{1}\int_{y^2-x^4}^{0}1dzdydx$$
I sum all those contributes, but the result is wrong (hand checked and Wolfram-checked). Where is my solve wrong?
Your integral for the positive part is correct. For the negative part, the region is split into a curved triangle below $y=x^2$ for $x\in [0,1]$; and a triangle below the line through $(1,1)$ and $(3,0)$ for $x\in [1,3]$. You had the right formula for the line, but forgot that it should be in terms of $x$, not $y$.
$$ x = 3-2y \ \Longrightarrow \ y = \frac{3-x}{2} $$
Thus, the last integral is $$ \int_1^3\int_0^{(3-x)/2}\int_{y^2-x^4}^0 \ dz \ dy\ dx $$
Alternatively, the negative part can be combined into one integral by switching $x$ and $y$
$$ \int_0^1\int_{\sqrt{y}}^{3-2y}\int_{y^2-x^4}^0 \ dz \ dx \ dy$$