Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having some difficulty identifying the bounds of a three-dimensional region.

I am asked to evaluate $\iiint_R (xz+3z)dV$, where $R$ is the region bounded by the cylinder $x^2 + z^2 = 9$ and the planes $x+y=3$, $z=0$, and $y=0$, above the $xy$-plane.

Plotted by Grapher on MacOSX

A sketch of the region R

So now that I've a sketch of the region $R$, I am trying to find the bounds of $x$, $y$, and $z$ but I'm always confused when it comes to identifying the correct bounds.

For example, I don't know which of the following for $x$ are correct:

$-3 \le x \le 3$

$-3 \le x \le 3-y$

(are both of them wrong?)

Could someone please give me some tips on how I should go about constructing the inequalities for this 3D region?

share|cite|improve this question

First you need to decide what order you will integrate in. As the integrand has no $y$ in it, that integral is easy to do and I would do it first. In that case, you can take $x$ and $z$ to be fixed (they are supplied by the outer integrals) and you need to find the range in $y$. $y$ can't be less than $0$ as that plane is one of you boundaries and can't be greater than $3-x$. So the inner integral is $\int_0^{3-x}dy$. Then if we do $x$ next, we have that $x$ ranges from $-\sqrt{9-z^2}$ to $\sqrt{9-z^2}$ and finally $z$ ranges from $0$ to $3$ because of the $xy$ plane restriction. So our final integral becomes $$\int_{0}^3\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}}\int_0^{3-x} (xz+3z)\;dy \;dx \; dz$$

share|cite|improve this answer
I think $z$ need only range from $0$ to $3$ since the region is bounded below by the $xy$ plane. – icurays1 Dec 1 '12 at 16:45
Hmm - I really can't see how $x$ is as you have defined. From what I can see, shouldn't it be $-3 \le x \le 3$? The thing is, I'm not sure when to use which equation. – JTJM Dec 1 '12 at 16:55
@NathanWilson: Say you are at $z=1$. In that case, $x$ ranges from $-\sqrt 8$ to $+\sqrt 8$. As we are inside the $z$ integral, we choose $z$ first, then figure out the range of $x$ – Ross Millikan Dec 1 '12 at 17:36
@icurays1: correct. Fixed – Ross Millikan Dec 1 '12 at 17:38
Right. So if, assuming it's easier to start integrating with $z$ or $x$ first, will the boundaries still be the same? (I don't think so). – JTJM Dec 2 '12 at 13:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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