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

How do i change the order of this triple integral so i can integrate it?

$$\int_{0}^9\int_{y=\sqrt{z}}^3\int_{x=0}^y z\cdot \cos(y^6)dxdydz$$

share|cite|improve this question
What are the integration limits? And please write your integral using MathJax. – Ron Gordon Apr 30 '13 at 8:36
up vote 2 down vote accepted

Your integral is

$$\int_0^9 dz \, z \: \int_{\sqrt{z}}^3 dy \, \cos{y^6} \: \int_0^y dx = \int_0^9 dz \, z \: \int_{\sqrt{z}}^3 dy \, y \, \cos{y^6}$$

Draw a picture.


You can see from that picture how to switch the order of integration and get the following

$$\int_0^9 dz \, z \: \int_{\sqrt{z}}^3 dy \, y \, \cos{y^6} = \int_0^3 dy \, y \, \cos{y^6} \: \int_0^{y^2} dz \, z$$

You will find that integral on the RHS may be done in closed form.

share|cite|improve this answer
Thankyou so much Ron. Is there a way i can reward you points? – amanda Apr 30 '13 at 9:13
You are welcome. If you can, click the up arrow as well if you want. – Ron Gordon Apr 30 '13 at 9:14

You have to change the order according to the dependencies between integration variables:

  • $x$ depends on $y$
  • $y$ depends on $z$
  • $z$ has fixed bounds

So you actually just have to first integrate with respect to $x$, then integrate what you get with respect to $y$, and finally, integrate with respect to $z$:

$$I=\displaystyle\int_{z=0}^9 z\left(\int_{y=\sqrt(z)}^3\cos(y^6)\left(\int_{x=0}^ydx\right)dy\right)dz$$

share|cite|improve this answer
Hi Dolma, this re-states what the question is asking... – amanda Apr 30 '13 at 9:05
It restates the question because that triple integral is already in the right order to integrate. – in_wolframAlpha_we_trust Apr 30 '13 at 9:12
@in_wolfram_we_trust no, it cannot simply be integrated in this form – amanda Apr 30 '13 at 9:14
Fair enough, I was a bit hasty in my first comment. There is no nice way to integrate $y.\cos(y^6)$. – in_wolframAlpha_we_trust Apr 30 '13 at 9:19
Oh ok, my bad. I thought you were asking a method to see in which order you had to integrate such multiple integrals. – Dolma Apr 30 '13 at 9:32

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.