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Another volume problem

  1. Sep 14, 2006 #1

    tony873004

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    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
    y=1/x, x=1, x=2, y=0; about the x-axis

    About halfway down, I diverge into two methods. The one on the left I use the method from FrogPad's post in my other thread (or at least my effort to understand the method he describes, btw, thank you). But it gives me 3pi/2. The one on the right gives me the same answer as the back of the book, pi/2. Maybe neither is right, and I coincidentlly got the right answer using the method on the right :bugeye:

    Also, am I using the right symbols, etc.?

    [​IMG]
     
  2. jcsd
  3. Sep 14, 2006 #2

    TD

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    Why do you start off with a summation? Also: the limits are unclear (what is the variable?).

    The volume of the solid you obtain by rotating f(x) about the x-axis between x = a and x = b is given by:

    [tex]\pi \int\limits_a^b {f\left( x \right)^2 dx} [/tex]

    In your specific case, with f(x) = 1/x and the limits given; this comes down to:

    [tex]\pi \int\limits_1^2 {\frac{1}{{x^2 }}dx} [/tex]

    Can you follow? Can you now compute this integral?
     
  4. Sep 14, 2006 #3

    J77

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    ...and

    [tex](x^{-1})^2\ne x[/tex]

    but

    [tex](x^{-1})^2= x^{-2}[/tex]

    with the limits, your left-hand column would be right (assuming the integrand is right, as above or in prev. post)
     
  5. Sep 14, 2006 #4

    HallsofIvy

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    You started off fine. I don't know why you "diverged" as you put it.
    (It would also, by the way,be better to write more than just math symbols in your work: explain what you are doing and what each formula represents.)

    Using the "disk" method you can imagine the volume divided into many thin disk with center on the x-axis, thickness [itex]\Delta x[/itex], and radius equal to the y value= [itex]\frac{1}{x}[/itex]. The area of each disk is [itex]\pi r^2= \pi \frac{1}{x^2}[/itex] and so the volume is [itex]\pi \frac{1}{x^2} \Delta x[/itex]. The approximate volume of the whole figure then is the Riemann Sum [itex]\Sigma \pi \frac{1}{x^2}\Delta x[/itex].
    Now, for some reason, you have [itex]\Sigma \pi x \Delta x[/itex].
    Surely you know that 1/x2 is not x!
    (Are you subtracting exponents- that's for multiplication, not repeated powers. xmxn= xm+n but (xn)m= xmn. (x-1)2= x-2.

    Now take the limit as the number of disks goes to infinity: the Riemann Sum [itex]\Sigma \pi \frac{1}{x^2}\Delta x[/itex] becomes the integral
    [tex]\pi \int_1^2 \frac{1}{x^2}dx= \pi \int_0^1 x^{-2} dx[/itex]
    That should be easy to integrate.
     
    Last edited: Sep 15, 2006
  6. Sep 14, 2006 #5

    VietDao29

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    Others have pointed out the errors in your OP. If you are about to integrate:
    [tex]\pi \mathop {\int} \limits_{1} ^ 2 x \ dx[/tex], the way in the left is correct.
    You should note that:
    [tex]\mathop {\int} \limits_{a} ^ b x \ dx = F(b) - F(a)[/tex], where F is the antiderivative of f.
    And in this case, you'll have: [tex]F(x) = \frac{1}{2} x ^ 2 + C[/tex].
     
    Last edited: Sep 14, 2006
  7. Sep 14, 2006 #6

    tony873004

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    The first few problems from the notes were done that way. I guess to show us why the integral is set up that way. After the first few examples, the teacher skips that step.

    Point well taken, especially when I give up for the night and try to pick it up in the morning.

    I re-did it and got the correct answer. I'll post my solution anyway, just in case I did something wrong and coincidentally got the right answer.

    Thanks for all your explanations. This is starting to become clear. :smile:
    [​IMG]
     
  8. Sep 15, 2006 #7

    J77

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    Nice LaTex style :smile:

    For the limits, you can just put the numbers.
     
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