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.

  • $\begingroup$ Is this figure bounded below by anything? Right now, it looks like it extends forever in the negative z-direction. $\endgroup$ – colormegone Apr 29 '13 at 17:50
  • $\begingroup$ @RecklessReckoner The problem doesn't specify so I have been assuming it does not. $\endgroup$ – jrquick Apr 29 '13 at 17:53

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.

  • $\begingroup$ The point I was making in my comment above is that, in order to integrate $z = y + 1$ over the region, there is an implicit assumption that the solid is bounded below by $z = 0$. This would generally be included in the problem statement. The volume integration is really being conducted for the difference in $z$ over each infinitesimal area on the xy-plane. Were this a surface integral for $y + 1$, it would be unnecessary to say anything, but to interpret that as a volume integral, it is understood that the xy-plane is the "bottom" of the solid. $\endgroup$ – colormegone Apr 29 '13 at 18:04
  • $\begingroup$ @RecklessReckoner From the problem statement, I interpreted the phrase "above the region ..." to mean the solid is bounded below by the plane $z=0$. More precisely (maybe), the bottom is flat and ecompasses said region. It seems unambiguous to me. $\endgroup$ – David Mitra Apr 29 '13 at 18:09

Sketch the figure. If the volume is bounded from below by the plane $z=0$, the volume is

$$V = \int_0^1 dy \: \int_0^{e^y} dx \: \int_0^{y+1} dz $$


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