Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

Cylindrical hole with radius r is drilled symmetrically through the center of a sphere with radius R.?

Cylindrical hole with radius r is drilled symmetrically through the center of a sphere with radius R. Where r ≤ R. What is the volume of the remaining material?

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  • 8 years ago
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    Radus of the Sphere = R

    Volume of the Sphere = V = (4/3) π R³

    Radius of the cylinder = r

    Height of the cylinder = h

    Draw the figure. From the figure it is observed = R² - r² = (h/2)²

    => h² = 4(R² - r²)

    => h = 2√(R² - r²)

    Volume of the cylindrical hole = v = π r² { 2√(R² - r²) }

    Volume of the remaning material = V - v = π{ (4/3) R³ - 2 r²√(R² - r²)}

    = 2π { (2/3)R³ - r²√(R² - r²) }

  • Anonymous
    8 years ago

    You can use the 'Washer Method' to find the volume.

    http://mathworld.wolfram.com/images/eps-gif/Spheri...

    (Just assume the hole goes all the way through)

    (Also suppose the axes are x and y, and z is out of the screen)

    It's a bit difficult to explain, but you can rotate the 'area' outside the cylinder and inside the sphere, about the y axis to get the answer.

    Finding this area in terms of y:

    It's the shape of a washer, outer radius (let it be r') varying with y (height), and inner radius constant as r.

    The outer radius varies with y as r' = √(R² - y²)

    So, the area of the 'washer' would be π ( r'² - r² ) = π (R² - r² - y²)

    Integrating this with respect to y:

    y varies from -√(R² - r²) to +√(R² - r²)

    of:

    ∫ π (R² - r² - y²) dy

    Integrating and substituting limits, we get

    Volume = (4/3) π (R² - r²) (√(R² - r²) )

    Source(s): Google images/Wolfram for the image.
  • 8 years ago

    you can easily derive the formula the formula of the remaining material.

    V = volume of the spherical segment - volume of the cylinder

    the formula is,

    V = 8pi/6[(R^2 - r^2)^(3/2)]

    note:

    x = multiplication sign

    sqrt = square root

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