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Thread: Find diameter of hole with quarter volume of whole shape.

  1. #1

    Find diameter of hole with quarter volume of whole shape.

    The region bound by
    Code:
    y=(1/8)x^2
    and the y axis is rotated about the line
    Code:
    y=8
    A hole is drilled that takes up 1/4th the volume of the original volume, find the diameter to 3 decimal places.

    2pi Int_0^8(8x - (x^3)/8)dx
    ... =256pi

    Dividing by 4 yields 64pi

    Code:
    64pi = 2pi Int_0^z(8x - (x^3)/8)dx
    32 = Int_0^z(8x - (x^3)/8)dx
    32 = 4b^2 - (b^4)/32
    
    -(b^4)/32 + 4b^2 - 32 = 0
    
    By quadratic formula:
    b^2 = (256 + 16sqrt(12)) and 256 - 16sqrt(12)
    
    Take three square root and multiply by 2, yielding
    
    Diameter = 35.295 or 28.325

  2. #2
    Elite Member stapel's Avatar
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    Quote Originally Posted by Seed5813 View Post
    The region bound by
    Code:
    y=(1/8)x^2
    and the y axis is rotated about the line
    Code:
    y=8
    A hole is drilled that takes up 1/4th the volume of the original volume, find the diameter to 3 decimal places.
    I'm not understanding the "region" here. The first curve is a parabola whose axis in the y-axis. So the y-axis is "inside" the curve. How is the y-axis then a bound on the region? Also, the parabola opens upward, including through y = 8. The figure would be infinite!

    Is there maybe a picture that goes with this? Or other information? Thank you!

  3. #3
    Quote Originally Posted by stapel View Post
    I'm not understanding the "region" here. The first curve is a parabola whose axis in the y-axis. So the y-axis is "inside" the curve. How is the y-axis then a bound on the region? Also, the parabola opens upward, including through y = 8. The figure would be infinite!

    Is there maybe a picture that goes with this? Or other information? Thank you!
    Its a bullet shape. The Y axis is the flat side of the bullet and you stop the parabola at y=8 and rotate around that.

    Sent from my LGLS755 using Tapatalk

  4. #4
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Seed5813 View Post
    Its a bullet shape. The Y axis is the flat side of the bullet and you stop the parabola at y=8 and rotate around that.
    "The flat side of the bullet" is the base. The parabola and the line y = 8 could approximate this, but then the y-axis would be irrelevant. Do you perhaps mean that the "bullet" shape is cut in half, from its tip to its base? So the bounds are as follows...?

    . . . . .[tex]\mbox{a. }\, y\, =\, \frac{1}{8}x^2,\, x\, \geq\, 0[/tex]

    . . . . .[tex]\mbox{b. }\, x\, =\, 0[/tex]

    . . . . .[tex]\mbox{c. }\, y\, =\, 8[/tex]

    And the resulting region is rotated about the horizontal line y = 8...?

    Thank you!

  5. #5
    Quote Originally Posted by stapel View Post
    "The flat side of the bullet" is the base. The parabola and the line y = 8 could approximate this, but then the y-axis would be irrelevant. Do you perhaps mean that the "bullet" shape is cut in half, from its tip to its base? So the bounds are as follows...?

    . . . . .[tex]\mbox{a. }\, y\, =\, \frac{1}{8}x^2,\, x\, \geq\, 0[/tex]

    . . . . .[tex]\mbox{b. }\, x\, =\, 0[/tex]

    . . . . .[tex]\mbox{c. }\, y\, =\, 8[/tex]

    And the resulting region is rotated about the horizontal line y = 8...?

    Thank you!
    Yes. Its cut in half but after being rotated it makes a bullet shape

    Sent from my LGLS755 using Tapatalk

  6. #6
    Elite Member stapel's Avatar
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    Quote Originally Posted by Seed5813 View Post
    Yes. Its cut in half but after being rotated it makes a bullet shape
    How? You have half of a parabola:

    Code:
    shape:
    
         ^y
         | 
    ---------------- y=8   
         |####'
         |###'
         |###'
         |###'
         |##'
         |#'
         |'
    -----+------------>x
         |
    You're twirling it up, over, and around the given horizontal line. How is that process recreating the left-hand half of the bullet, while not creating anything above the given axis?

    Code:
    flipped:
    
         ^y
         | 
         |
         |'
         |#'
         |##'
         |###'
         |###'
         |###'
         |####'
    ---------------- y=8   
         |####'
         |###'
         |###'
         |###'
         |##'
         |#'
         |'
    -----+------------>x
         |
    Please show what you're doing. Thank you!

  7. #7
    Quote Originally Posted by stapel View Post
    How? You have half of a parabola:

    Code:
    shape:
    
         ^y
         | 
    ---------------- y=8   
         |####'
         |###'
         |###'
         |###'
         |##'
         |#'
         |'
    -----+------------>x
         |
    You're twirling it up, over, and around the given horizontal line. How is that process recreating the left-hand half of the bullet, while not creating anything above the given axis?

    Code:
    flipped:
    
         ^y
         | 
         |
         |'
         |#'
         |##'
         |###'
         |###'
         |###'
         |####'
    ---------------- y=8   
         |####'
         |###'
         |###'
         |###'
         |##'
         |#'
         |'
    -----+------------>x
         |
    Please show what you're doing. Thank you!
    Its a parabola bounded by x=0 and y=8.

    Sent from my LGLS755 using Tapatalk

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