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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.

Question:

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

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  • $\begingroup$ 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

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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!

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  • $\begingroup$ 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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