Problem Statement:

The function $( f(x) = 2.5 - x $) and the function $ g(x) = \frac{1}{x} $ enclose a region in the first quadrant. Find the volume of the rotational body that is made by rotating this region with respect to the $x$-axis.

Graph of the functions

Method One:

Volume generated by $ f(x) $:

$ V_f = \pi \int_{0.5}^{2} (2.5 - x)^2 \, dx $

Volume generated by $g(x) $:

$ V_g = \pi \int_{0.5}^{2} \left(\frac{1}{x}\right)^2 \, dx $

Volume of the wanted region:

$ V = V_f - V_g = \pi \left( \int_{0.5}^{2} (2.5 - x)^2 \, dx - \int_{0.5}^{2} \left(\frac{1}{x}\right)^2 \, dx \right) = 1.125 \pi $

This method works and gives the right solution according to the book.

Method Two:

This method, which I initially thought could work, does not give the correct result. Here's the thought process:

$( f(x) - g(x) $) gives us the difference between the two functions.

If we treat $( f(x) - g(x) $) as the radius for each individual disk, then the area of one general disk should be $\pi (f(x) - g(x))^2 $.

Taking the integral from $0.5$ to $2$ of $ \pi (f(x) - g(x))^2 $ should give us the volume:

$[ \pi \int_{0.5}^{2} \left((2.5 - x) - \frac{1}{x}\right)^2 \, dx $]

Geometrically, this idea seems to make sense, but the algebra and resulting volume do not match the first method's result.


Why doesn't this method work, even though it seems to make sense geometrically? Could someone explain the discrepancy?

  • $\begingroup$ Let's simplify the idea: Take the lines $y = 1$ and $y=3$ and the interval $[0, 1].$ Rotate this region around the x-axis. Use your method, then compare with the geometry. What goes wrong? This gives you an idea. $\endgroup$ Commented Jun 17 at 13:08
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    $\begingroup$ SMK, what you are integrating (from $0.5$ to $2$) is the area of a disk with radius $f(x)-g(x)$, whereas what you should is the area of an annulus: a disk with radius $f(x)$ minus a smaller disk inside with radius $g(x)$. $\endgroup$ Commented Jun 17 at 13:50
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    $\begingroup$ "Intuitively it works, it is just finding the difference between the two functions and then integrating from a to b ... ." What intuition did you use here? If it is just integrating the difference between two functions, why is there a square power in the integral? $\endgroup$
    – David K
    Commented Jun 17 at 13:57
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    $\begingroup$ SMK, I suggest you rather elaborate more on your wrong method (directly in your post, better than in comment). And/or "see" the slices of this solid, perpendicular to the x-axis. They are annuli, not disks. $\endgroup$ Commented Jun 17 at 14:09
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    $\begingroup$ Because of the hole in the middle, no slice is a disk, and the slice at coordinate $x$ has radius $f(x)$, not $f(x)-g(x)$. So in fact, geometrically it makes no sense at all to interpret $f(x) - g(x)$ as if it were the radius of a disk. $\endgroup$
    – David K
    Commented Jun 17 at 17:24

1 Answer 1


You seem to mixing something up, and your comments haven't made it clear. Here's how to view it.

Suppose $f(x) \geq g(x)$ on some interval $[a, b].$ We want to find the volume of the solid formed when this region is rotated around the $x$-axis. As $g$ dominates $f$ on this interval, its corresponding "radius" is larger, and the volume formed by rotating the area under $g$ around the $x$-axis is $$V_f = \pi \int_a ^b f(x)^2 \ dx.$$ Likewise, the volume formed for $f$ is $$V_g = \pi \int_a ^b g(x)^2 \ dx.$$ But we want the volume formed by the region between! So we need to subtract, and the desired volume is $$V_f - V_g = \pi \int_a ^b f(x)^2 - g(x)^2 \ dx.$$ To answer your main question: the radius of the disk is NOT $f(x)-g(x)$ in this case, because that would be the radius if we rotated around $g(x)$! Keep this in mind when your axis of rotation is different. Note in the figure below, although $g(x)$ is a constant function, this may be different in other cases! Here's how I tell my students: "outside squared minus inside squared," where we measure the distance from the function to the axis of rotation. Watch this video here that I created for my university's math department while a graduate student for many more examples - and note my explanations and sketches!

But the main question is why? We need to view these as annunlar regions. Take two concentric circles of radius $r$ and $R$ where $r < R.$ What's the area of the "donut" region? Intuitively, because the smaller circle contains area in the big circle, it gets double counted, so we subtract, giving the difference $\pi R^2 - \pi r^2 = \pi(R^2 - r^2).$ We see this in action in the following figures from OpenStax Calculus, Volume 2.

Description of the "washer" method for volumes.

But I think there's also something bigger at play here: you also seem to be confusing $(x-y)^2$ and $x^2 - y^2$. We note:

$$\color{red}{\boxed{{(x-y)^2 \neq x^2 - y^2}}}$$

which is a huge misconception!

  • $\begingroup$ Hi Sean. I wanted to thank you for taking the time to answer my question. However, maybe my first explanation was not good enough to express what is on my mind. I will edit my first question with better explanations, so that hopefully everybody could understand little better about the problem I am stating. $\endgroup$
    – SMK
    Commented Jun 17 at 14:51
  • $\begingroup$ It also seems that you have subtracted the smaller valued function with the bigger function. Maybe just a typo? $\endgroup$
    – SMK
    Commented Jun 17 at 15:24
  • $\begingroup$ See the edits. The punchline is: "Be careful where you measure!" $\endgroup$ Commented Jun 17 at 15:32
  • $\begingroup$ Hi Sean. I am really sorry that I cannot really understand, but what do you mean by that the answer would have been right if I rotated with respect to g(x). Am I not rotating with respect to g(x), and how do you know that it would be right if I were to be rotating with respect to g(x)? I mean, the difference is in between f(x) and g(x)... $\endgroup$
    – SMK
    Commented Jun 17 at 15:39
  • $\begingroup$ The instructions in the question state "...rotating this region with respect to the x-axis." You're going around the line $x = 0$, so all distances MUST be measured from there. $\endgroup$ Commented Jun 17 at 15:40

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