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?


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.