# Is there a way to find the bounds of integration for triple integrals without drawing pictures (analytically)?

I'm able to solve integrals and draw graphs in 2D, but I have a lot of trouble drawing 3D graphs. As such, I'm seeking a method of finding the bounds of integration for triple integrals without having to graph them. It seems reasonable to me that there is such a method, since we also need a way to solve higher dimensional integrals (4D, 5D, 6D, etc), without the ability to graph them.

I found this (relevant) question from 3 years ago, but it was left unanswered: Is there a way to find limits of integration numerically for triple integrals?

I would greatly appreciate if someone could please explain or direct me to a method for solving triple integrals (finding their bounds of integration) without having to graph them.

• What kind of information do you have about the domain of integration? For instance, in two dimensions, if you don't have the explicit bounds of integration, you might have some kind of description of the geometry of the domain instead. Alternately, you might be trying to transform from one coordinate system to another. Commented Mar 24, 2017 at 2:55
• @Kajelad For instance, for a question that I am currently working on, I am told that $z = x^2 - 1$, $z = 0$, $z - 3y = 0$, $2z + y = 0$. I am looking for a general method that I can use to find the bounds of integration for triple integral questions, so that I do not have to graph anything. Commented Mar 24, 2017 at 2:57
• @Kajelad And for the record, we haven't learned any other coordinate systems, so I'd appreciate an answer in this context. Commented Mar 24, 2017 at 3:09
• Coming up with a general way of dealing with bounding surfaces without looking at the geometry of the domain is a bit tricky. Commented Mar 24, 2017 at 3:21
• @Kajelad Surely it can't be more difficult than drawing the graph of every group of functions you encounter? And there must be such a way; otherwise, how would people solve anything more than triple integrals? It seems that drawing everything would be the most difficult way to solve these things. Surely there must be an algebraic method for finding the bounds of integration? Commented Mar 24, 2017 at 3:23

This is an answer prompted by the comments, to show how to handle the specific example of a solid bounded by the surfaces $z=x^2-1$, $z=0$, $z-3y=0$ and $2z+y=0$.
These equations don't give the points $(x,y,z)$ which belong to the solid, they only give the points on the surfaces which form the boundary of the solid. So we need to figure out how to make them into inequalities that really describe the solid.
For this, I certainly find it easier to think in terms of pictures: draw the curves $z=x^2-1$ and $z=0$ in an $(x,z)$ coordinate system, and the lines $z=3y$ and $z=-y/2$ in a $(y,z)$ coordinate system, and make a mental picture of the corresponding surfaces in $(x,y,z)$ space (three planes and one curved surface). Then it's pretty clear that the only inequalities which give a bounded solid are: $$z \ge x^2-1 ,\quad z \le 0 ,\quad z-3y \le 0 ,\quad 2z+y \le 0 .$$ But in principle it should be possible to figure it out purely algebraically. For example, it can't be $z \le x^2-1$, since then there's nothing to stop $|x|$ from getting arbitrarily large, so that would give an unbounded domain. And if it were $z \ge x^2-1$ but $z \ge 0$, then for each fixed $x$, you could make $(y,z)$ go arbitrarily far away from the origin without violating whatever choice you have made for the last two inequalities; for example, with $z-3y \ge 0$ and $2z+y \le 0$, the points $(x,y,z)=(s,-t,t/4)$ will satisfy your inequalities for and $s$ and any $t>0$, hence that would give an unbounded domain too. (Except that for the choice $z-3y \le 0$ and $2z+y \le 0$, the only point which satisfies all inequalities is $(x,y,z)=(0,0,0)$, so not much of a solid in this case.) And so on... Very tedious, but doable in principle.
So now we have the description of the solid in terms of inequalities; let's write them as $$x^2-1 \le z \le 0 ,\quad z/3 \le y \le -2z .$$
Now it seems easiest (to me) to take cross sections for fixed $x$. With $x \in [-1,1]$ fixed, the inequalities above describe a region $D_x$ in the $(y,z)$ plane, which you can handle algebraically just by inspection: $$\iint_{D_x} dy dz = \int_{z=x^2-1}^0 \left( \int_{y=z/3}^{-2z} dy \right) dz = \dots ,$$ although I find it even easier with a picture: draw the lines $z=0$, $z=x^2-1$ (below the line $z=0$ of course), $z=3y$ and $z=-y/2$, and it should be clear that the region $D_x$ is a triangle (or just a point, if $x=\pm 1$), and $$\iint_{D_x} dy dz = \operatorname{Area}(D_x) = \frac12 \cdot \mathrm{base} \cdot \mathrm{height} = \frac12 \Bigl( -2(x^2-1)-\tfrac13(x^2-1) \Bigr) \Bigl( 0 - (x^2-1) \Bigr) .$$ And finally, $$\operatorname{Volume}(D) = \iiint_D dxdydz = \int_{x=-1}^1 \left( \iint_{D_x} dy dz \right) dx = \int_{-1}^1 \frac76 (x^2-1)^2 dx = \dots$$
• @Allure: You need to look at the whole inequalities, not just the cases with equality. If $y$ is right and $z$ is up, then the inequality $x^2-1 \le z \le 0$ says that we're in a strip in the the lower half plane, and $z/3 \le y \le -2z$ says that we're to the right of the line $y=z/3$ and to the left of the line $y=-2z$. Commented Jul 17, 2018 at 10:50