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Calculusvolume with cross sections

  1. Apr 16, 2008 #1
    I need help please on a calculus problem. I am trying to teach myself and am doing fair. I am stuck on this though.

    Question:find the volume using CROSS SECTIONS that are squares.
    base is bounded by
    y = x + 1
    y = x squared - 1
    cross sections are squares
    perpendicular to the x axis

    I know to use the area = side squared due to it being a square
    I set the 2 equations equal to find the integral goes from -1 to 2

    I took the top minus bottom to get equation of (x + 1 ) - ( x squared - 1)
    which becomes all squared due to the definition of area of a square

    so I get x^4 - x^3 -2x^2 -x^3 +x^2 +2x -2x^2 +2x +4
    then I combine like terms and integrate?
    I don't get the answer
    answer should be 8.1 it says.

    thanks in advance.
     
  2. jcsd
  3. Apr 16, 2008 #2
    If my head hasn't fallen off you should have:

    [tex][(x+1)-(x^2-1)]^2=-3x^2+4x-2x^3+4x+x^4[/itex]

    [tex]\int_{-1}^2 -3x^2+4x-2x^3+4x+x^4\;dx=?[/tex]

    What do you get for that?
     
    Last edited: Apr 16, 2008
  4. Apr 16, 2008 #3
    yes, I got that except one of the 4x I had 4 instead

    so, if that is correct....
    then I do the integral math, find that
    then I plug in 2
    then I plug in -1
    and subtract the two answers?
     
  5. Apr 16, 2008 #4

    HallsofIvy

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    Was that a typo (or has your head fallen off?)
    [tex][((x+1)- (x^2-1))^2= (-x^2+ x+ 2)^2= x^4-2x^3+ 4x+4[/tex]
    You don't need "4x" twice!

     
  6. Apr 16, 2008 #5
    Yeah I think it was apologies it should have been just +4.

    Puts head back on. :redface:

    [tex]\int_{-1}^2 -3x^2+4x-2x^3+4+x^4\;dx=?[/tex]

    Yes that's correct, sorry for the mistake. :frown:
     
    Last edited: Apr 16, 2008
  7. Apr 16, 2008 #6
    ok, thanks everyone.

    I guess I must have done something wrong but get the answer now.
    Of course I should have asked the next question first as now I am stuck on it. Can you post more than one question a day or is that considered a 'no can do' thing?
     
  8. Apr 16, 2008 #7
    Certainly, as long as you show working, or, at least indicate where you're totally flummoxed, you can post as many as you like. You might not always get a quick reply but be patient.
     
  9. Apr 16, 2008 #8
    oh wow that is great.

    Do I start a new post then or continue with this one?
    I guess you can tell it is my first day here at ye old physics forum
     
  10. Apr 16, 2008 #9
    Welcome to PF.

    And yeah just tack it on the end, is probably easier in this case, people will notice the bump generally.
     
  11. Apr 16, 2008 #10
    ok, here goes:

    Cross sections for volumes method only:
    question:
    given
    y = x cubed
    y = 0
    x = 1
    cross sections are equilateral triangles
    perpendicular to the y axis

    so first I think (watch out) that when it says perpendicular to the y axis you use y values on your integral but have to solve any equations using x = to calculate your area.

    so I solved for x = third root of y
    (the top of the graph is this equation then.)

    NOW, I get muddled big time.
    It seems I have a triangle with the side of it defined as the third root of y.
    so I found the height of an equilateral given that as my side
    I got x on the square root of 3 all divided by 2.

    I would then use this value as my 'x' in the area = 1/2 b*h
    so
    1/2 (third root of y) (third root of y)(square root of 3) = area?

    am I totally off here?
     
  12. Apr 16, 2008 #11

    Dick

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    Not totally off, but if b is a base of an equilateral triangle then the height is b*sqrt(3)/2. So (1/2)bh=?. I think you are missing a factor of 2, aren't you?
     
  13. Apr 16, 2008 #12
    ok, I see that height is b times square root of 3 all over 2
    so area is (square root of 3 )divided by four, times b squared.
    is my side of the triangle equal to that top graph which is x = cube root of y?

    do I set b = cube root of y and use those values in the area calculation?
    Obviously I have to get the equation in there.
    I have a hard time with the visual here.
     
  14. Apr 16, 2008 #13

    HallsofIvy

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    No. One side of the base figure is y= x3 or x= x1/3. The other end of that base is x= 1. The length of the base of a triangle is 1- x1/3, NOT "x1/3".

     
  15. Apr 16, 2008 #14

    Dick

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    I can't tell by reading the problem statement. All I see is that it says that the triangle is perpendicular to the y axis. It doesn't say that y^(1/3) is the side of the triangle, or the height of the triangle or whatever. Unless it says more you'll have to guess or ask for clarification.
     
  16. Apr 16, 2008 #15

    Dick

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    Uh, good point, if you read it that way. I took the x=1 as a misprint for y=1. calchelpplz, how exactly did you quote that problem?
     
  17. Apr 16, 2008 #16
    oops
    I was totally seeing it wrong.
    Is it that the base is 1 minus (y to the 1/3)? (you had x to the 1/3)
    If so I think I see it now.

    thanks!
     
  18. Apr 16, 2008 #17
    Cross sections for volumes method only:
    question:
    given
    y = x cubed
    y = 0
    x = 1
    cross sections are equilateral triangles
    perpendicular to the y axis

    no typo............
    so is the base equal to 1-(cubed root of y)
    and my height is that base (1 - cubed root of y) , times the (cubed root of 3 )over 2?
     
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