How to set up the integral for finding the volume of a solid in three dimensions? While studying for my Calculus 3 exam I have gotten stuck on this particular problem, primarily in the set up. 
Find the volume of the solid under the surface z=y+1 and above the region bounded by y=ln(x), y=0, x=0, and y=1.
I know the problem requires a double integral around the given equation z=y+1 but I'm not sure of what points to use for the integrals and at which time. 
My assumption is to integrate from x=0 to x=e^y and then integrate a second time from y=0 to y=1. 
I got x=e^y from y=ln(x) => x=e^y.
 A: Your limits of integration are correct. 
To see this, sketch the given region of the $x$-$y$ plane. It is contained in the first quadrant and has a trapezoidal shape with vertices $(0,0)$, $(0,1)$, $(e,1)$, and $(1,0)$.  Note that the region is best described by horizontal slices (with vertical slices, you'd need to divide the region into two parts). Thus, integrating with respect to $x$ first is appropriate. 
So, you're thinking of the region as being a bunch of horizontal strips stacked on top of each other. In the inner integral, you integrate along a fixed strip in the $x$ direction (so the inner integral is with respect to $x$).  Then, in the outer integral, you integrate in the vertical direction from where the first strip is located to where the last one is.
The horizontal strips range from $y=0$ to $y=1$. With $y$ fixed, a horizontal strip has left edge $x=0$ and right edge $x=e^y$.
In the end, you wind up having to evaluate $\int_0^1\int_0^{e^y} y+1\,dx\,dy$. (Note you're integrating the function that gives the height to the top of the solid at the point $(x,y)$. Here, that's $z=y+1$.)
Calculating this is routine; though, an integration by parts is needed in the calculation of the outer integral.
A: Sketch the figure.  If the volume is bounded from below by the plane $z=0$, the volume is
$$V = \int_0^1 dy \: \int_0^{e^y} dx \: \int_0^{y+1} dz $$
