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The integral in question is $$\iiint_{y^2-4xz \ge 0} \mathrm{d}x\mathrm{d}y\mathrm{d}z$$ $$0\le x\le 1,0\le y\le 1,0\le z\le 1$$ I realize that this is just the volume of some area, however, I have been so far unable to determine the bounds of the integral. I have tried drawing a picture but I have no idea how $4xz$ looks, much less $y^2-4xz$.
I remember from my multivariable calculus course that I needed to find the maximum possible values of the variables given the constraints of the region. To find these maxima, I thought about using Lagrange Multipliers to see if I could get the bounds as functions of the variables, but the equations I get out of the method give me no useful information.
From just playing around with the inequalities, I have gotten $2\sqrt {xz} \le y \le 1$ but when trying to use this to ascertain the remaining limits I have run into a brick wall.
I am sure that once I have the limits, I can evaluate the integral. But; how do I get the limits of integration?

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You are looking to integrate over the region bounded by;

$$\{(x,y,z):y^2-4xz \ge 0~,~ 0\le x\le 1~,~0\le y\le 1~,~0\le z\le 1\}$$

Check: We are integrating an always positive function, so Tonelli's theorem says we can choose any order to integrate.   Look for the easiest.

Inspecting $y^2-4xz \ge 0$, it appears easiest to place $y$ as the outermost integral, then choose one of $x$ or $z$.   This saves messing around with square roots.

$$\{(x,y,z):0\leq y\leq 1, 0\leq x\leq 1, 0\leq z\leq \min(1,\frac {y^2}{4x})\} $$

Now $\min(1,y^2/4x) =\begin{cases} 1 &:& x\leqslant y^2/4\\ y^2/4x&:& x>y^2/4\end{cases}$

So that gives you $${\{(x,y,z):0\leq y\leq 1, 0\leq x\leq \min (1,y^2/4), 0\leq z\leq 1\}} \cup {\{(x,y,z):0\leq y\leq 1, y^2/4< x\leq 1, 0\leq z\leq y^2/4x\}}$$

And again we look at when $y^2/4\leqslant 1$ and find it is always so for $0\leq y\leq 1$

$${\{(x,y,z):0\leq y\leq 1, 0\leq x\leq y^2/4, 0\leq z\leq 1\}} \cup {\{(x,y,z):0\leq y\leq 1, y^2/4< x\leq 1, 0\leq z\leq y^2/4x\}}$$

$$I= \int_0^1\left(\int_0^{y^2/4}\int_0^{1}\operatorname d z\operatorname d x+\int_{y^2/4}^1\int_0^{y^2/4x}\operatorname d z\operatorname d x\right)\operatorname d y$$

That is all but the actual integration.

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    $\begingroup$ So we first obtain an inequality and then use the $\min$ "function" to break the region into further smaller sections? $\endgroup$ – Guacho Perez Nov 25 '16 at 3:57

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