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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.

EDIT:

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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    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.2011-02-10
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    Calculus teacher being *too* tricky.2011-02-10
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    @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.2011-02-10
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    For starters, is our first integral even correct?2011-02-10
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    @Jay Emm: The reversal of the signs makes sense, but doesn't actually matter. Note that $x^2=(-x)^2$. And yes, your integral would be the correct answer, not your teacher's. Provided you have copied everything correctly, as Fredrik mentioned.2011-02-10
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    @Brandon Carter: Some of us are online right now and can't seem to find any errors with the post. No matter how we rotate or split the graph, there's no way we can achieve equality. It just doesn't seem possible to have one integral equal another of a greater range of numbers.2011-02-10
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    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,2011-02-10
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    @Matt E: assuming the function is continuous, of course (as it is here).2011-02-10

2 Answers 2

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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.

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    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.2011-02-10
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    The answer is none of those choices. Sometimes people/books make mistakes.2011-02-10
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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.

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    Thanks for the confirmation, but out teacher is certain that he is correct. I will post the multiple choice answers given to us.2011-02-10