A metal Sphere of radius Ym is melted down and remade into a cylinder of the same volume with height Ym with circular ends of radius 3m, find Y

The sphere has radius Y, so the volume of the sphere is 4/3 (pi) Y3 , The volume is the same for the cyclinder so we know the volume of the cylinder in terms of Y. Cyclinder volume = area of circle face x height.      We know the circle has radius 3 so the area of the circle face is (pi) x 32 =9 x (pi). So we can find the volume of the cylinder in terms of Y;   9 x (pi)  x Y = 9 x (pi) x Y As the volume of the sphere and cylinder are equal we can equate them (put them as equal to one another): 9 x (pi) x Y = 4/3 x (pi) x Y3    Dividing by (pi) on each side gives; 9 Y = 4/3 Y3                           Subtract 9Y from both sides: 0    = 4/3 Y3 - 9Y                    Then Factorise: 0   =  Y (4/3 Y2 -9)                    One solution to this is Y = 0, which makes sense mathematically but in the scenario given it doesnt make sense (as                                                 Y was the radius of the sphere.  The other solution is found by saying that (4/3) x Y2   - 9  = 0 (as that is the other option for the above equation to be equal to 0) So solving that we can add 9 to both sides:        4/3  x  Y2  =  9 Then multiplying by 3                                          4 Y2           = 27 Then dividing by 4 (you could have divided by 4/3 and done it in one step but you might find you need a calculator)                                                                                Y2     =   27/4        OR  6.75

So Y = 2.60 (to 3sf) or leave as the square root of 27/4)

They might ask you to put it in a decimal form or in fraction, but they are both equal to one another.

JC

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