Express the triple integral three different ways I need to rewrite the integrals like $dx\,dy\,dz$, $dy\,dz\,dx$, and $dz\,dx\,dy$ of the solid bounded by $$x=2, y=2, z=0, x+y-2z=2$$
I do not fully understand how to rewrite the integral different ways by changing the limits. If someone could help me out, that would be superb! 
 A: Your question is fairly general.  What you should understand is that integrals like
$$
\int \int \int \cdots\cdots\cdots dx\,dy\,dz
$$
are nested, i.e. the notation above should be read as
$$
\int \left(\int \left(\int \cdots\cdots\cdots dx\right)\,dy\right)\,dz.
$$
So suppose the outer integral is $\displaystyle\int_a^b \cdots\cdots\,dz$.  Then the one immediately inside it can be
$$
\int_{a(z)}^{b(z)} \cdots\cdots\,dy
$$
i.e. the bounds of integration can depend on $z$.
In the outer integral, the variable $z$ runs from $a$ to $b$, and for any fixed value of $z$ between $a$ and $b$.  Then in the next integral inside it, for any fixed value of $z$ between $a$ and $b$, the variable $y$ runs from something that may depend on $z$ to something else that may depend on $z$.  Then for the next integral inside that, the variable $x$ runs from something that may depend on both $z$ and $y$ to something else that may depend on both $z$ and $y$.
Now look at the triangle bound by the $x$-axis, the $y$-axis, and the line $x+(y/2)=1$.  That last equation $x+y=1$, can be written as $x=1-(y/2)$ or as $y=2-2x$.  To integrate over the triangle, draw the picture and see the three vertices at $(0,0)$, $(1,0)$, and $(0,2)$.  Notice that $x$ can run from $0$ to $1$, but given some fixed value of $x$ between $0$ and $1$, the other variable $y$ can run only from $0$ to $2-2x$.  Thus we have
$$
\int_0^1 \int_0^{2-2x} \cdots\cdots dy\,dx.
$$
But we could also say $y$ runs from $0$ to $2$, and then for any fixed value of $y$ between $0$ and $2$, the other variable $x$ runs from $0$ to $1-(y/2)$.  So we have
$$
\int_0^2 \int_0^{1-(y/2)} \cdots\cdots  dx\,dy.
$$
Both integrals should evaluate to the same number.  But in some cases one integral is easy to evaluate directly and the other is not.  Thus if you have the one that's not, you transform it into the one that is.
