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Homework Help: Infinite Sum Theorem - Help Understanding

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Ok so I am going to go a bit outside of the normal procedure to post a question (again) because I don't understand the example nor the first problem fully. I will post the example first and any questions I have with what is being done will be posted as the example progresses, ok? ok

    So the example states:

    Find the volume of a pyramid of height h whose base has area B, as in Figure 6.1.7 which can be found http://images2e.snapfish.com/232323232fp537;4>nu=52::>379>256>WSNRCG=32<:7;9279347nu0mrj"

    2. Relevant equations

    Here is the entire theorem as given in the book:

    Let B(u,w) be a real function of two variables that has the addition property in the interval [a,b] i.e.,

    [tex] B(u,w) = B(u,v) + B(v,w) \;\; for \;\; u<v<w \;\; in \;\; [a,b] [/tex]

    Suppose h(x) is a real function continuous on [a,b] and for any infinitesimal subinterval:

    [tex] [x, x +\Delta x] \; of \; [a,b], [/tex]

    [tex] \Delta B = h(x) \Delta x [/tex]

    Then B(a,b) is equal to the integral:

    [tex] B(a,b) = \int^{b}_{a} h(x) dx [/tex]

    (I suppose right here I should state that I have not had any experience with a function of two variables. I have done this book chapter by chapter and this was never introduced, so I am kind of flying blind here already.)

    3. The attempt at a solution

    Ok so the example goes on to say:

    Place the pyramid on its side with the apex at x=0 and the base at x=h.

    (First problem, they seem to have switched where the apex and base are because from the picture you can clearly see that the base is set to 0)

    We use the fact that at any point x between 0 and h, the cross section has area proportional to x^2, so that:

    (Second problem, how do we know that [0,h] cross sections have area proportional to x^2? this is not mentioned in early chapters or prior to this example in any way)

    [tex] \frac {A(x)}{x^{2}} = \frac {B}{h^{2}} [/tex]

    [tex] A(x) = \frac {Bx^{2}}{h^{2}} [/tex]

    (Third problem, why did we set this up this way? Did I forget something crucial from geometry?)

    Then the volume is:

    [tex] V = \int^{h}_{0} \frac {Bx^{2}}{h^{2}} dx = \frac {1}{3} * \frac {Bh^{3}}{h^{2}} = \frac {1}{3} Bh [/tex]

    (Thankfully I understand what they did here completely, its just getting to this point...)

    _____________________________________________________________________________

    So the first problem of the chapter asks:

    The base of a solid is the triangle in the x,y plane with vertices at (0,0), (0,1), and (1,0). The cross sections perpendicular to the x-axis are squares with one side on the base. Find the volume of the solid.

    For some reason I really can't fathom how to draw this, I will post this now and attempt to draw a picture of what they are talking about. So if you reply before I post some kind of picture for the problem above just focus on my example questions.

    Thanks for any and all help and understanding.

    EDIT: http://images2e.snapfish.com/232323232fp53838>nu=52::>379>256>WSNRCG=32<:7;<;58347nu0mrj" is a picture of what I think they are saying in problem one, am I close?
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Aug 22, 2010 #2

    LCKurtz

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    Your picture isn't correct. Draw an xyz axis in the standard position with x towards you, y to the right, and z up. Plot the three points in the xy plane that you are given and the triangle they form. That triangle is the base of your solid so no part of the solid lies below the xy plane. Then draw a few of the cross sections described.
     
    Last edited by a moderator: Apr 25, 2017
  4. Aug 23, 2010 #3
    Ok LCKurtz, does http://images2e.snapfish.com/232323232fp53835>nu=52::>379>256>WSNRCG=32<::952:8347nu0mrj" look any better? If so I am still confused with all of the question I posted about the example. And also, if this is better how is it really any different that the first one I drew? Besides the actual labeling of the z-axis..

    Note that I realize that A(x) is suppose to be a square but my drawing is just a rough example of what is being explained.
     
    Last edited by a moderator: Apr 25, 2017
  5. Aug 23, 2010 #4

    LCKurtz

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    No, that isn't what it looks like even as a rough sketch. Cross sections perpendicular to the x axis are squares. It looks like this:

    crosssection.jpg
     
  6. Aug 25, 2010 #5
    Alright man but you havent address any of my questions in the example I posted, can you please help me there too? Otherwise you drawing the diagram doesn't actually help me...
     
  7. Aug 30, 2010 #6

    LCKurtz

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    Sorry for the delay in answering but I'm not at my usual location at the moment. Let x be the distance from the origin along the x axis to the representative little square at position x. Since you have the equation of the diagonal side of the base triangle, you can get the length of the side of the little square in terms of x. Then you have its area A(x). All you have to do to get the value of the required volume is integrate the cross sectional area A(x) for the values of x for which the little squares generate the volume. This same idea works in all "area by cross-section" problems.
     
  8. Sep 12, 2010 #7
    Hey, LCKurtz, I am sorry for not replying but I had a bunch of stuff to do to get ready for school. If you are still willing to answer another question then here it goes.

    How is it that the little cross section of A(x) when intergrated will equal the volume of the solid? I ask because the value of A(x) is obviously going to be constantly changing so there is going to be a large error rate unless I am missing something important.
     
  9. Sep 12, 2010 #8

    Mark44

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    The definite integral [tex]\int_a^b A(x) dx[/tex] in essence adds together all of the typical volume elements, A(x)[itex]\Delta x[/itex], as x ranges from a to b. All of these typical volume elements have this volume, which obviously depends on x. A(x) is, after all, a function whose value depends on x.

    It's the same idea as when you use a definite integral to represent the area under a curve.
     
  10. Sep 12, 2010 #9
    Haha of course it does, sorry for the misunderstanding! I got it now. Thanks for the help LCKurtz and Mark44
     
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