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Surface area of a square and a tube

  1. Oct 22, 2012 #1
    Ok, so if I have a square that is exactly 10 inches by 10 inches, then the surface area is 100 square inches exactly.
    But if I roll up that square into a tube and calculate its surface area, it's 2∏r times the length. And since the calculation involves ∏, that means I won't get an exact answer. How is it that I get two different answers from calculating the surface area of two shapes that should have the exact same surface area?
     
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  3. Oct 22, 2012 #2

    pwsnafu

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    Did you calculate what r is?
     
  4. Oct 22, 2012 #3

    Mentallic

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    Because when you roll up the square, the circumference [itex]2\pi r[/itex] was originally 10, so we have that
    [tex]2\pi r= 10[/tex]
    [tex]r=\frac{10}{2\pi} = \frac{5}{\pi}[/tex]

    So if you ever want to get to a value N from some "crazy" number x, all you need to do is multiply by N/x.
     
  5. Oct 22, 2012 #4
    No, but would that matter? I mean, the pi is still there making the answer non-exact.
    How can 2∏r equal exactly 10 though?
     
  6. Oct 22, 2012 #5
    If you mean how can we measure it to be that, we can't. But we also can't measure the square to be exactly 10 inches.
     
  7. Oct 22, 2012 #6

    pwsnafu

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    Firstly, either this problem is physical or abstract. If it is physical we can't obtain an exact 10 in by 10 in square, so pi has nothing to do with it. If you could measure the length to (say) 100 significant figures, I can calculate pi to one million significant figures. The error of the instrument completely dwarfs any error through calculating pi.

    However, if this is an abstract problem, then pi is exact.

    He showed ##r = \frac{5}{\pi}## therefore ##2\pi r = 2 \times \pi \times \frac{5}{\pi} = 10## exactly.
     
  8. Oct 22, 2012 #7

    Integral

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    Why do you think that a factor of [itex] \pi [/itex] makes it "inexact"? We know way more digits of pi then you have the ability to measure. So as far as your ruler is concerned pi is exact.
     
  9. Oct 22, 2012 #8

    Mentallic

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    So [itex]2\pi \approx 6.28[/itex] and [itex]5/\pi \approx 1.59[/itex] and when we multiply these approximations together, we get [itex]2\pi*5/\pi \approx 6.28*1.59 \approx 9.99[/itex] which is nearly 10, and the only reason we can't get 10 exactly is because if you tried to measure the length of [itex]\pi[/itex], you can never get it exact, although you can try and get it as close to the exact answer as possible which is 3.14159...

    But this doesn't have anything to do with the fact that it's an irrational number and it has an infinite string of digits. We also can't measure 10 exactly! Instead, we would end up getting as close as possible to 10 within our error bounds, so we'd have 9.999 or 10.0001 for example.
     
  10. Oct 22, 2012 #9

    micromass

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    Obviously, the irrationality of [itex]\pi[/itex] makes it inexact (or rather, not constructive). The point is that r is also irrational, and these two cancel out.
     
  11. Oct 22, 2012 #10

    HallsofIvy

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    You are using the words "exact" and "inexact" is a very peculiar way. Do you have any reason to think that irrational numbers are "inexact"? What do you mean by that?

    (The question is about numbers, not "numerals" so the the fact that the decimal form of a number has an infinite string of digits is irrelevant.)
     
  12. Oct 22, 2012 #11

    micromass

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    Well, I should have said transcendental instead of irrational. But the idea is that numbers such as [itex]\pi[/itex] aren't constructible ( http://en.wikipedia.org/wiki/Constructible_number ). So I interpret that as being inexact. For example, given a line of length 1, I can always construct lines of length 1/3 or [itex]\sqrt{2}[/itex] (which is irrational, I know). So in this way, I consider this to be more exact than [itex]\pi[/itex]. But yes, I should have specified.

    Indeed it is.
     
  13. Oct 22, 2012 #12

    HallsofIvy

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    Ah, so that [itex]\sqrt{2}[/itex], which is also an irrational number, is "constructible". But being "transcendental" also doesn't form the "divide". Obviously, any transcendental number is not costructible but [itex]\sqrt[3]{2}[/itex] is algebraic but not constructible. In fact, the numbers that are constructible are precisely the numbers that are "algebraic of order a power of 2".
     
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