Four marbles of radius #3/8# inch are placed in a cylindrical container. The first three marbles fit snuggly with the fourth marble on top. The cylinder is then filled with water up to the top of the fourth marble. What is the volume of water?
Denote the radius of a marble by
Then the radius of the cylinder is:
#r + r / cos(30^@) = r + r / (sqrt(3)/2) = r + ((2r)/sqrt(3)) = r+(2sqrt(3)r)/3 = r((3+2sqrt(3))/3) = (3+2sqrt(3))/8 "inches"#
Next, the total height of the stack of marbles will be
From some slightly messy calculations, the height of a tetrahedron with edges of length
Viewed from one side, the tetrahedron looks like a triangle with base of length
Then:
#sqrt(3)r = sqrt((sqrt(3)r)^2  h^2) + sqrt((2r)^2  h^2)#
#= sqrt(3r^2h^2) + sqrt(4r^2h^2)#
Square both sides to get:
#3r^2 = 3r^2h^2+4r^2h^2 + 2sqrt((3r^2h^2)(4r^2h^2))#
Hence:
#sqrt((3r^2h^2)(4r^2h^2)) = h^22r^2#
Square both sides again to get:
#(3r^2h^2)(4r^2h^2) = (h^22r^2)^2#
That is:
#12r^47r^2h^2+h^4 = h^44h^2r^2+4r^2#
So:
#8r^4 = 3h^2r^2#
Divide both sides by
#h = sqrt(8)/sqrt(3)r = (2sqrt(6))/3r#
So the total height of the stack of marbles is:
#2r + (2sqrt(6))/3r = (6+2sqrt(6))/3r = (6+2sqrt(6))/8 "inches" = (3+sqrt(6))/4 "inches"#
The volume of the cylinder up to this height is:
#"{area of circle}" xx "{height}"#
#=pi ((3+2sqrt(3))/8)^2 xx (3+sqrt(6))/4 "inches"^3#
#=pi ((21+12sqrt(3))/64) xx (3+sqrt(6))/4 "inches"^3#
#=pi (63+36sqrt(3)+21sqrt(6)+36sqrt(2))/256 "inches"^3#
The volume of each marble is:
#4/3 pi r^3 = 4/3 pi (3/8)^3 "inches"^3#
#= pi 9/128 "inches"^3#
So the volume of four marbles is:
#= pi 9/32 "inches"^3#
So the total volume of water will be:
#pi (63+36sqrt(3)+21sqrt(6)+36sqrt(2))/256 "inches"^3  pi 9/32 "inches"^3# =
#pi (63+36sqrt(3)+21sqrt(6)+36sqrt(2))/256 "inches"^3  pi 72/256 "inches"^3# =
#pi (36sqrt(3)+21sqrt(6)+36sqrt(2)9)/256 "inches"^3#
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The volume of water in the cylindrical container can be calculated as follows:

Calculate the volume of the three marbles that are snugly fitted at the bottom of the cylinder.
Volume of one marble = (4/3) * π * (radius)^3 = (4/3) * π * (3/8 inch)^3
Total volume of three marbles = 3 * (4/3) * π * (3/8 inch)^3

Calculate the volume of the cylindrical container above the three marbles.
Height of the cylindrical container above the three marbles = height of cylinder  diameter of marble
Volume of cylindrical container above the three marbles = π * (radius)^2 * height

Calculate the volume of the fourth marble.
Volume of the fourth marble = (4/3) * π * (radius)^3 = (4/3) * π * (3/8 inch)^3

Add the volumes of the three marbles, the volume of water in the cylindrical container above them, and the volume of the fourth marble to get the total volume of water.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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