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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!
  • $\pi \int_1^{3} (9 - (\ln_e(x))^2)\ dy$
  • 10
    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

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