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Thread: Volume problem

  1. #1 Volume problem 
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    4. "The base of a solid object is a circle with a radius of 4. Every verticle cross section is an equilateral triangle. What is the volume of the solid?"


    Ok I'm not sure about this 1, can some1 tell me if I'm doing it right

    area of triangle = (1/2) x base x height

    so I take the integral of the area of the triangle

    the formula for a circle with radius 4 is (x'2/16) + (y'2/16) = 1

    solved for y which is y = sqrt(16-x'2)= base

    To solve for the height remembered that the cross section is an equilateral triangle

    so all sides are equal to 8 since the radius is 4,

    then divided the equilateral triangle in half to make 2 right angle triangles

    so using a'2 + b'2 = c'2

    4'2 + b'2 = 8'2
    b = 4sqrt3

    side b of the triangle is the height

    so the intergal is {4 integral (2sqrt3)sqrt(16-x'2)dx} with an upper bound of 4 and a lower bound of 0

    I put a 4 infront of the integral since the intergal is only solving for the 1st quadrent and I need to take account for the other quadrents.

    Am I doing this right and can someone post the solution please. Thanks


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  3. #2  
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    Let be the equation of the circular base.

    At horizontal distance x from the origin, the length of each side of the equilateral triangle is , so the cross-sectional area is . (The area of an equilateral triangle with sides of length a is ) Hence



    which I believe should be .


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  4. #3  
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    Once again thanks you a life saver.

    Just wondering why couldn't i do the problem by taking the triangle in the 1st quad like I was attempting. I would have multiplied my answer by four to account for the other quads. I understand your answer but just am curious why what I was attempting was wrong. Thanks
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  5. #4  
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    Okay, let us do it as you suggest.

    The area of half of the equilateral-triangular cross-section at horizontal distance x from the origin is . (NB: The height of the triangle is .)

    So the volume of one quarter of the solid is

    which is . If that is easier for you, then fine. I normally prefer to do the whole lot at once.
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  6. #5  
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    Thank you, I just wanted to make sure it was possible my way as well or if it was wrong.
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    Quote Originally Posted by JaneBennet
    Okay, let’s do it as you suggest.

    The area of half of the equilateral-triangular cross-section at horizontal distance x from the origin is . (NB: The height of the triangle is .)

    So the volume of one quarter of the solid is

    which is . If that’s easier for you, then fine. I normally prefer to do the whole lot at once.
    Couldn't you just get the height of the cone? Using a side of the equilateral triangle? And half the base. One side of the triangle squared would be 64, half the base squared would be 16, so the square root of 48 would be 6.928

    Then just use the cone formula. 1/3 times pi times radius^2 times height.

    1/3 * 3.14 * 16 * 6.928 = 116.92522666

    Or a better way to remember it pi R^2 times height divided by three.

    Sincerely,


    William McCormick
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  8. #7  
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    Quote Originally Posted by William McCormick
    Couldn't you just get the height of the cone? Using a side of the equilateral triangle? And half the base. One side of the triangle squared would be 64, half the base squared would be 16, so the square root of 48 would be 6.928

    Then just use the cone formula. 1/3 times pi times radius^2 times height.

    1/3 * 3.14 * 16 * 6.928 = 116.92522666
    That formula is for a cone with a circular base. The shape here is not a cone; its one with triangular cross sections. (A cone would have a hyperbolic cross sections.)

    This shape is how shall I describe it? Something like an admirals hat perhaps. Like this (but with a much rounder base and pointier top):

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  9. #8  
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    Quote Originally Posted by JaneBennet
    Quote Originally Posted by William McCormick
    Couldn't you just get the height of the cone? Using a side of the equilateral triangle? And half the base. One side of the triangle squared would be 64, half the base squared would be 16, so the square root of 48 would be 6.928

    Then just use the cone formula. 1/3 times pi times radius^2 times height.

    1/3 * 3.14 * 16 * 6.928 = 116.92522666
    That formula is for a cone with a circular base. The shape here is not a cone; its one with triangular cross sections. (A cone would have a hyperbolic cross sections.)

    This shape is how shall I describe it? Something like an admirals hat perhaps. Like this (but with a much rounder base and pointier top):


    Maybe I am misunderstanding vertical cross section?

    A cone creates a triangular cross section.


    If the cone is standing up and you vertically dissect it.

    Sincerely,


    William McCormick
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  10. #9  
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    Quote Originally Posted by William McCormick
    A cone creates a triangular cross section,
    I should also mention how the cross sections are obtained. They are obtained by cutting the object with planes perpendicular to the base. Cutting a cone with such planes will only give a triangle if the plane is through the vertex. Otherwise (if you slice the cone on its slant side with a cut at right angles to the base) you dont get a triangle but a curve a conic section technically called a rectangular hyperbola.

    Try it and see for yourself. If you cant find any cone-shaped gadgets in your house to experiment with, you can easily make a paper cone. :wink:
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  11. #10  
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    Quote Originally Posted by William McCormick
    Maybe I am misunderstanding vertical cross section?

    A cone creates a triangular cross section.


    If the cone is standing up and you vertically dissect it.
    That's true if you take the cross-section right at the vertex. But if you are off center the cross-section makes a hyperbola.

    http://en.wikipedia.org/wiki/Conic_section
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  12. #11  
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by William McCormick
    Maybe I am misunderstanding vertical cross section?

    A cone creates a triangular cross section.


    If the cone is standing up and you vertically dissect it.
    That's true if you take the cross-section right at the vertex. But if you are off center the cross-section makes a hyperbola.

    http://en.wikipedia.org/wiki/Conic_section
    I totally agree about the off center slice being a strange shape if cut from a cone.

    Isn't a vertical cross section done from the center line of the object. The cross meaning centered.

    The hat that Jane described would not create an equilateral triangle, from every possible vertical slice, and certainly not every vertical cross section.


    I could only get a cone out of that description.

    It said every vertical cross section in his description, is an equilateral triangle. To me only a cone could fit that description.


    Sincerely,


    William McCormick
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  13. #12  
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    Quote Originally Posted by William McCormick

    I could only get a cone out of that description.
    Bill, try this in your welding shop. Get a round grill, like out of a Weber charcoal grill, For each wire on the grill, cut two steel wires of equal length, so the three wires form an equilateral triangle. Weld up all these parallel equilateral triangles, and what shape would they form? Wouldn't it be a lot like the hat shape Jane described?
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  14. #13  
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by William McCormick

    I could only get a cone out of that description.
    Bill, try this in your welding shop. Get a round grill, like out of a Weber charcoal grill, For each wire on the grill, cut two steel wires of equal length, so the three wires form an equilateral triangle. Weld up all these parallel equilateral triangles, and what shape would they form? Wouldn't it be a lot like the hat shape Jane described?
    Sure but, you just cannot get an equilateral triangle from every vertical slice whether a true cross section, or just a vertical slice.

    I still only see the cone fulfilling that description given. I do not like the question, because it turns what should be an exacting description, into a little riddle about every cross section being an equilateral triangle.

    Only a cone can look like an equilateral triangle when a true vertical cross section is performed from any angle.

    If I am missing something blatantly, and that is always a possibility I would love to know. I just don't see any other object fitting that description.

    I do see the object you are describing though Harold. I just do not see it creating an equilateral triangle.

    Not trying to be difficult. I just don't get it.


    Sincerely,


    William McCormick
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  15. #14  
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    Quote Originally Posted by William McCormick
    I do see the object you are describing though Harold. I just do not see it creating an equilateral triangle.

    Not trying to be difficult. I just don't get it.
    Well, if you constructed this shape out of equilateral triangles, it would have to have equilateral triangles in it. But I think I see the problem. It is not every vertical cross-section that is triangular, only those which are perpendicular to the ridge line. a vertical cross-section of the hat shape through the ridge line would be semicircular.
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  16. #15  
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    Yep William I was the attempting the problem the same way you were thinking it would be a cone as well. I'm also having a hard time trying to visualize the shape. Is there any way I change this into a 3D function by putting a z into the equation. I understand where I went wrong with the slices and see why this can't be a cone but it just hard for me to visualise
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  17. #16  
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    Quote Originally Posted by 11rdc11
    Yep William I was the attempting the problem the same way you were thinking it would be a cone as well. I'm also having a hard time trying to visualize the shape. Is there any way I change this into a 3D function by putting a z into the equation. I understand where I went wrong with the slices and see why this can't be a cone but it just hard for me to visualise
    See my post re the admirals hat. Thats as close as I can get to a visual description. :?

    If you still want the 3D equation of the surface, well, this is what I make of it:



    Does this help you visualize it any better?
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  18. #17  
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    thanks jane im going 2 go try and plot it on the ti 89
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  19. #18  
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    Quote Originally Posted by 11rdc11
    Yep William I was the attempting the problem the same way you were thinking it would be a cone as well. I'm also having a hard time trying to visualize the shape. Is there any way I change this into a 3D function by putting a z into the equation. I understand where I went wrong with the slices and see why this can't be a cone but it just hard for me to visualise

    Here is a look at almost the real thing. Thanks to Alibre 3D drawing program. And Camtasia studio that was able to record it.

    http://www.Rockwelder.com/Flash/conehead/Conehead.htm


    But I do not know of any shape that fits the description of the first post, and yet can be cross sectioned into equilateral triangles. Other then a cone.

    Maybe the terminology is a bit out of whack? An equilateral triangle has three equal sides all angles are 60 degrees?

    Sincerely,


    William McCormick
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  20. #19  
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    Quote Originally Posted by Harold14370
    a vertical cross-section of the hat shape through the ridge line would be semicircular
    Almost. Its actually a semi-ellipse. Putting in the formula above, you get the equation of the curve in the perpendicular cross section through the ridge as: .
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