So our calc teacher is being tricky and sending us off to fend for ourselves in the world of mathematics.

Here's the question:

We need to set up an integral to find the volume of the solid formed by the area bounded by the functions y=ln(x), x=3, and the x-axis revolved around the line x=3.

The problem?

Most of us put something similar to $\pi \int_0^{\ln 3} (e^y - 3)^2\ dy$, which is believed to be correct.

He claims it can also be written as $\pi \int_0^{e^3} (3 - e^y)^2\ dy$, but hasn't given us any idea as to why.

Many thanks to anyone who can help us out.


Multiple choice options given:

  • $\pi \int_0^{e^3} (3 - e^y)^2\ dy$
  • $\pi \int_0^{e^3} (9 - e^{2y})\ dy$
  • $\pi \int_1^{3} (\ln_e(x))^2\ dy$
  • None of These <- Not this answer!
  • $\pi \int_1^{3} (9 - (\ln_e(x))^2)\ dy$
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    $\begingroup$ Either your teacher is wrong or you have copied the integrals wrongly. Certainly $\int_0^{\ln 3} (e^y-3)^2 dy \neq \int_0^{e^3} (3-e^y)^2 dy$. To see this, note that the integrands are the same, but the limits are very different. $\endgroup$ – Fredrik Meyer Feb 10 '11 at 4:41
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    $\begingroup$ Calculus teacher being too tricky. $\endgroup$ – milcak Feb 10 '11 at 4:48
  • $\begingroup$ @Fredrik Meyer: That's what's confusing us. It doesn't make sense. Is there any way e^3 could come into play? The reversal of the signs makes sense. $\endgroup$ – Justian Meyer Feb 10 '11 at 4:49
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    $\begingroup$ Dear Justian, It is not possible for the integral of a positive function (such as your function, which is positive since it is a square) to equal an integral of the same function over a larger interval. Assuming that you have stated the problem correctly, your teacher has simply made a mistake. (Something that happens to all of us at some point!) Regards, $\endgroup$ – Matt E Feb 10 '11 at 6:38
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    $\begingroup$ @Matt E: assuming the function is continuous, of course (as it is here). $\endgroup$ – Pete L. Clark Feb 10 '11 at 7:15

From my calculations, it seems that your integral is correct. The other one is not.

The volume is indeed given by $$V=\pi \int_0^{\ln 3} (e^y-3)^2 dy=5.929$$ (We go from 0 to $\ln 3$, and the radius of the circle at height $y$ is $3-e^y$)

Alternatively, integrating the $x$-coordinate we find the volume is $$V=2\pi \int_{1}^{3} (3-x)\ln x dx=5.929$$ (Using the so called "shells" method)

Hope that helps, sometimes people just make typos.

  • $\begingroup$ Eric Naslund: Thanks for the two perspectives, but we're still as confused as ever :P. Perhaps the other multiple choice options will make mistakes/possibilities more evident. $\endgroup$ – Justian Meyer Feb 10 '11 at 5:53
  • $\begingroup$ The answer is none of those choices. Sometimes people/books make mistakes. $\endgroup$ – Eric Naslund Feb 10 '11 at 17:03

Let $f$ be some function. I'll demonstrate how to calculate the volume $V$ of the solid obtained by revolving the region under the graph $f \ge 0$ on some interval $I$ around the horizontal axis.

Let $P = \{ t_0, \cdots, t_n \}$ be any partition of $I$. Let $m_i = \inf \{ f(x) : t_{i-1} \le x \le t_i \}$ and $M_i = \sup \{ f(x) : t_{i-1} \le x \le t_i \}$, be as always.

It's easy to see, that $\pi m_i^2(t_i - t_{i-1}) $ is the volume of some cylinder that lies inside $V$ on $[t_{i-1}, t_i]$. Similarly, $\pi M_i^2(t_i - t_{i-1}) $ is some cylinder that contains a part of $V$.

We know that as we sum over the partition of $I$, the lower sum will be a lower bound for the volume of $V$ and the upper sum an upper bound. So we have:

$$\sum_{i=1}^n \pi m_i^2 (t_i - t_{i-1}) \le volume(V) \le \sum_{i=1}^n \pi M_i^2 (t_i - t_{i-1})$$

These sums are the lower and upper sums for $f^2$ on $I$, so it follows that:

$$volume(V) = \pi \int_a^b f(x)^2 dx$$

Now, we need to adapt your region to this enviroment. Here we need to rotate the region about a line that is parallel to the $y$-axis. So we can shift back the region $3$ units towards the $y$-axis. Now our region is just the area under $\ln(x+3)$ on the interval $[-2,0]$.

As we are rotating about the $y$-axis, we must express the function in terms of $y$, i.e. $x=e^y - 3$, on the interval from $0$ to $\ln(0+3) = \ln 3$, so indeed, plugging this into our formula, yields:

$$volume (V) = \pi \int_0^{\ln 3} (e^y - 3)^2\ dy$$

so yes, your first integral is correct.

  • $\begingroup$ Thanks for the confirmation, but out teacher is certain that he is correct. I will post the multiple choice answers given to us. $\endgroup$ – Justian Meyer Feb 10 '11 at 5:38

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