# Homework Help: Limits of Integration

1. Jan 22, 2014

### vanceEE

1. The problem statement, all variables and given/known data

Let f be the function defined by $$f(x) = - ln(x) for 0 < x ≤ 1.$$ R is the region between the graph of f and the x-axis.

http://learn.flvs.net/webdav/educator_apcalcbc_v10/module08/imgmod08/08_10_01.gif

b. Determine whether the solid generated by revolving region R about the y-axis has finite volume. If so, find the volume. If not, explain why.

2. Relevant equations

$$y = -ln(x)$$
$$x = e^{-y}$$

3. The attempt at a solution

$$V = \pi \int_{x=0^+}^{x=1} [e^{-y}]^2 dy$$
$$V = \pi\int_{∞}^{0} [e^{-2y}] dy$$
$$\uparrow$$ This is my mistake.
$$V = -\frac{\pi}{2}$$

The actual solution is $$V = \pi \int_{0}^{∞}[e^{-2y}] dy = \frac{\pi}{2}$$
But why are the limits of integration flipped? For part a, (Determine whether region R has a finite area. If so, find the area. If not, explain why.) my limits of integration were [x=0,x=1] $$\int_{0^+}^{1} -ln(x) dx = 1$$, so wouldn't I just set $$e^{-y}$$ equal to 0 and 1 for part b? If not, please explain analytically why I need to flip my limits of integration, I can see from the graph that when x → 0, y → ∞ so please explain the problem analytically. The rotations about the y axes are very tricky for me and ANY advice would help :-) This is a very simple, but confusing concept.

Last edited: Jan 22, 2014
2. Jan 22, 2014

### gopher_p

Essentially, since $y=-\ln x$ is decreasing on $(0,1]$, $dy\sim-dx$. I would advise either setting up the integral entirely in $x$ and then making an obvious (?) substitution, or just set it up entirely in $y$ from the get-go. when you mix stuff up like you did (integrand in one variable, limits in another) in ends badly more often than not in my experience.

3. Jan 22, 2014

### vanceEE

I just talked to my teacher and she told me that writing the coordinates are helpful when dealing with these type of problems. In this case they will be M(0,∞) and N(1,0) therefore, since we integrate with respect to x from left to right, my limits with respect to x will be $$\int_{0}^{1} f(x)dx$$ and since we integrate with respect to y from bottom to top, my limits with respect to y will be $$\int_{0}^{∞} g(y)dy$$

4. Jan 22, 2014

### gopher_p

Yes. This is what I mean when I say to set things up entirely in one variable or the other. It's just much easier that way.

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