Question: Can you help me with this problem?

> 
> Evaluate 
>  $${\int \int \int\_{\Omega } xy^{2}z^{2}dxdydz}$$
>  where $\Omega = \left \{ z=xy,x=1, z=0 \right \}$
> 
> 
> 

**Edit:**
where $\Omega$ is bounded by the surfaces $z=xy$, $y= x$ , $x=1$, $z=0$.
Sorry !!!

Answer: 
Hint. Note that
$$\iiint\_{\Omega } xy^{2}z^{2}dxdydz=\int\_{x=0}^1\left(\int\_{y=0}^x\left(\int\_{z=0}^{xy} xy^{2}z^{2}dz\right)dy\right)dx$$
Can you take it from here?
Robert Z is correct, the setup is 
$$
\iiint\_\Omega xy^2z^2\;dV=\int\_0^1 \int\_0^x \int\_0^{xy} xy^2z^2 \;dz\;dy\;dx
$$
which can be easily computed from there to obtain $\dfrac{1}{198}$, you can even [confirm at this link](http://www.wolframalpha.com/input/?i=Integrate%5Bxy%5E2z%5E2,%7Bx,0,1%7D,%7By,0,x%7D,%7Bz,0,xy%7D%5D). The 'trick' for triple integrals is to understand the bounds for one variable because after that you can project down to a plane and then it is setting up the remaining double integral which is usually much easier (as it only involves plotting a picture in a plane - which we are used to). 

Notice they region has bounds $z=xy$ and $z=0$. So know we know that $0\leq z \leq xy$ and that's a bound! So we can integrate with respect to $z$ first, giving the inner integral above. Now we project the space down to the $xy$-plane and see what we have. Down in the plane we know that $y=x$ is a bound and clearly $x=1$ is a bound. Sketch this in the plane. Since $z=0$ is a bound (same thing as $xy=0$), we know at least one of $y=0$ or $x=0$ is a bound. From the picture you drew, clearly it is $y=0$. But then this gives the rest of the double integral set up. Sometimes trying to sketch the region in 3D can help if you really can't understand any of the variables to get a handle on one of them to reduce the triple integral into a double one. Your region is plotted below.

$\hskip1.5in$ [![enter image description here](https://i.stack.imgur.com/4i9HW.png)](https://i.stack.imgur.com/4i9HW.png)