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

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

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

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What are the integration limits? And please write your integral using MathJax. – Ron Gordon Apr 30 '13 at 8:36

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

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.

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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 $z$:

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

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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