Finding the limits of integration for the volume of a region inside a cube Premise
We are restricted to the region $x, y, z \in [f, 1]$ where $0 \leq f < 1$. The surface $y^2=4xz$ divides this cube into two regions. We are interested in finding the volume of the region where $y^2 - 4xz \geq 0$.
My Attempt
Let $t=\frac{y^2}{4}$ for brevity.
When $f=0$, it is relatively easy to figure out the limits. We can separate the region into two where $x \leq t$ and $x > t$. In the first case, we have $y^2 - 4xz \geq 0$ for all $z \in [0, 1]$ because $xz \leq x\cdot 1 \leq t$. Similarly, in the second case, we have $y^2 - 4xz \geq 0$ for all $z \in [0, \frac{t}{x}]$ because $xz \leq x\cdot \frac{t}{x} = t$. Applying these limits, we have
$$
\begin{align}
V &= \int_{0}^{1}{dy\int_{0}^{t}{dx\int_{0}^{1}{dz}}} + \int_{0}^{1}{dy\int_{0}^{t}{dx \int_{0}^{\frac{t}{x}}{dz}}}\\
  &= \int_{0}^{1}{dy\int_{0}^{t}{dx\cdot (1)}} + \int_{0}^{1}{dy\int_{0}^{t}{dx\cdot \frac{t}{x}}}\\
  &= \frac{1}{12} + \frac{1 + 3\log{2}}{18} \\
  &\approx 0.254
\end{align}
$$
However, when $f > 0$, I cannot seem to understand the proper limits for $z$. Is it $f$ to $\frac{t}{x}$ or is the lower limit $f^2$? What happens when $f$ is less than or equal to $\frac{y^2}{4}$?
I tried a few times, but the results ended up being negative, which is clearly wrong as volumes are nonnegative.
Specific Questions
Primary objective
I would like help in figuring out the limits in the case where $f>0$.
Secondary objectives

*

*Is this approach sensible? Perhaps there is a better way to solve this problem?

*Are there any glaring oversights?

*What books/courses/reference should I follow to be able to solve these types of problems?

 A: We need to find the volume of the cube formed by $x, y, z \in [f, 1], 0 \leq f \lt 1$, with the condition $y^2 \geq 4xz$ (outside the cone).
For the cube, the upper limit for $x, y, z$ is $1$.
For the cone, please note when $x = z = 1, y = 2$ (I am only considering first octant given our cube is in the first octant) but as we are restricted by $y = 1$ for the cube, we have $4xz \leq 1 \implies x, z \leq  \min (\frac{1}{4f}, 1)$ for $f \ne 0$.
At the same time for the points outside the cone, $y^2 \geq 4f^2 \implies y \geq 2f$ for $x = z = f$ and $y \geq 2 \sqrt f \ \gt f $ as $f \lt 1$ for $x = 1, z = f$ or $z = 1, x = f$.
These are important observations to find the right limits for our volume integral. What it indicates is that if $f \leq 0.25, $ we have following $3$ vertices of the cube outside the cone and rest $5$ are inside -
$(f,1,f),(f,1,1), (1,1,f)$
In fact at $f = 0.25$, $(f,1,1)$ and $(1,1,f)$ are on the cone and only $(f,1,f)$ is outside.
Also note that at $f = 0.5$, $(f,1,f)$ is on the cone and rest are inside and so for $f \geq 0.5$, we have not volume bound (as all points of the cube are inside the cone, $y^2 \leq 4xz$).
So here is the integral to find the desired volume
i) $0 \leq f \leq 0.25$ (first integral is zero at $f = 0.25$)
$\displaystyle  \int_{f}^{0.25} \int_{f}^{1} \int_{2\sqrt{xz}}^{1} dy \ dx \ dz + 
\int_{0.25}^{1} \int_{f}^{1/(4z)} \int_{2\sqrt{xz}}^{1} dy \ dx \ dz$
And your answer is indeed correct for $f = 0$.
ii) $0.25 \leq f \leq 0.5$ (zero for $f = 0.5$)
$\displaystyle \int_{f}^{1/(4f)} \int_{f}^{1/(4z)} \int_{2\sqrt{xz}}^{1} \ dy \ dx \ dz$
To your other specific questions, no there does not seem anything wrong with your approach. But your approach did not extend to other cases, probably because you did not visualize the surface well enough. A $3D$ sketch using an online tool helps. I literally had to keep a cube in front of me while answering the question.
