Volume of a body bounded by various functions How would you calculate the volume of a body bounded by surfaces $z=\log y, z=0, y=e, x=2, x=4$?
I'm having trouble figuring out the limits of integration.
So, $x$ and $z$ are pretty obvious, but $y$ is not.
I was thinking that $y$ needs to  go from $0$ to $e$ since the graph of the function $\log y$ gets arbitrarily close to zero.
So my integral would look like $$\int_2^4\int_0^e\int_2^{\log y}dzdydx$$
Is this correct?
 A: You saw how simple is the part for $x$: two parallel planes and all other surfaces perpendicular. But this reduces the problem to an usual one in dimension two, but in the plane $yz$. In this kind of problems the idea in to find the intersection of the several curves defining the surface.
$z=0$ and $z=\log y$, intersecting at $z=0;\;y=1$. The line $y=e$ cuts somewhere the curve $z=\log y$ ($z=1$, indeed), so, this is the upper limit for $y$. Then,
$$A=\int_2^4\int_1^e\int_0^{\log y}dzdydx$$
A: I hope the folowing sketch clarifies your doubts.

The body is limited below by $ z=0 $ and above by $ z=\log y $. On the right by the plane $ y=e $ as shown in the picture. So the limits of integration with respect to $ y $ are $ y=1 $ ( due to the condition $ 0\le z$), $ y=e $; and with respect to $ z $, $ z=0, z=\log y $. The limits with respect to $ x $ are correct. 
ADDED. The computation is easy and yields a positive value, as it should be, and not $-4e$ as per your integral (see my comment). The volume is
\begin{equation*}
V=\int_{2}^{4}\left( \int_{1}^{e}\left( \int_{0}^{\log y}dz\right) dy\right)
dx=\int_{2}^{4}\left( \int_{1}^{e}\log y\;dy\right)
dx=\int_{2}^{4}1\;dx= 2.
\end{equation*}
