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i'm having trouble with this integral the integral is $\int_0^9\int_{\sqrt z}^3\int_0^y z\cos(y^6)\,dx\,dy\,dz$.

We aren't given any more information and i'm a bit stuck as to where to start. I don't know whether or not to change variable to cylindrical or spherical or anything. The y integration part doesn't seem to work any way i go about it. Any help would be much appreciated :)

Thanks heaps

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There is no obvious symmetry in the region, so I do not believe cylindrical or polar coordinates is the way to go. One thing to notice is that this reduces almost immediately to a double integral, since the integrand is independent of $x$ so integrating $x$ is easy. Once you're at the double integral stage, you should be able to draw the domain (now 2-dimensional) you need to integrate over, and find the right way to change the order of the iterated integral. –  kigen May 3 '13 at 6:54
    
I would guess (but haven't verified) that exchanging the order of the $y$ and $z$ integrals will be useful (carefully handling the region in doing so, of course). If you're really lucky a factor $y^5$ will magically appear so as to make the $y$ integration simpler. –  Harald Hanche-Olsen May 3 '13 at 6:56

3 Answers 3

Firstly,we have $0<y<\sqrt{z}<3$,and the integral is "indepent" form $x$ indeed, then we can write $\int^9_0\int^3_{\sqrt{z}}\int^y_0 z\cos(y^6)dxdydz=9\int_0^9dz\int^{\sqrt{z}}_0z\cos(y^6)dy$,but i don't know this indefinite integral:$\int \cos(y^6)dy$.That is my advice.

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@proximal's hint is correct. The outermost integral gives you something that is just a function of $y$ and $z$, let's call it $f\left(y,z\right)$. Then you have

$$ \int_0^9 \int_{\sqrt{z}}^3 f\left(y,z\right)dzdy. $$

In this case, $f$ is nice enough so that you can switch the order of integration (draw this on paper!) to get

$$ \int_0^3 \int_{0}^{y^2} f\left(y,z\right)dydz. $$

If you want to check your answer at the end, it is

$$ \frac{\sin\left(3^6\right)}{12} $$

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Change the order of integration in $y$ and $z$. First, write as a double integral by integrating over $x$:

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

Now, draw a picture. The integration region is really $z \in [0,y^2]$ for $y \in [0,3]$. The integral is then

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

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