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Question

A cylindrical can has a base radius of 3.4 cm.
It contains a certain amount of water such that when
a sphere is placed inside the can, the water just
covers the sphere. If the sphere fits exactly inside
the can, find
(i) the surface area of the can that is In contact with
the water when the sphere is inside the can,
(ii) the depth of water in the can before the sphere
was placed inside the can.

Answer 1

Hi Amy, this is a really good question. It takes into account the volume of a cylinder, volume of a sphere, and the surface are of a cylinder.

 

Given: the base radius of the can as 3.4 cm. Then the diameter of the can is 6.8 or twice the radius.

 

Volume of a can = area of the base times the height or π * r² * h

 

Volume of a sphere = 4 / 3 * π * r³

 

Surface area of cylinder = π * d * h, however, the area of the bottom of the can is also in contact with the sphere and needs to be added. The area of the bottom of the can = π * r².

 

When the sphere is placed into the cylinder, the water will cover the sphere to a height of the diameter of the sphere. From this information, we can answer part one of the question, what is the surface area of the can in contact with the water when the sphere is inside of the can.

 

Surface area in contact with the water = surface area of the cylinder in contact with the water plus the area of the base of the cylinder.

 

Surface area in contact with the water = π * 6.8 * 6.8 + π * 3.4²

 

                                                        = 145.27 + 36.32

 

                                                        = 181.59 sq. cm.

 

Part 2:

 

To find the height of the water without the sphere, we need to subtract the volume of the sphere from the volume of the cylinder if the sphere was not there.

 

Volume of the sphere = 4 / 3 * π * r³, were r = 3.4

 

                                = 164.64 cu. cm.

 

Volume of the can without the sphere = π * r² * h, where r = 3.4 and h = 6.8

 

                                                       = 247.00 cu. cm.

 

Subtract volume of sphere from volume of can = 247.00 - 164.64, or 82.36 cu. cm.

 

Last, to find the height of the water without the sphere, we need to find h in the volume of a cylinder formula:

 

Volume of water in can = 82.36, which is  = π * r² * h, or

 

82.36 = π * 11.56 * h

 

Divide both sides by π * 11.56:

 

h = 2.27 cm.

 

All values are rounded to two decimal places.

 

Questions?

Answer 2

Hi Amy!

 

By "the sphere fits exactly inside the can," I take it to mean the diameter of the sphere and the diameter of the can are equal, and thus their radii are equal.

 

Therefore, we have a sphere with a radius of 3.4 cm, and thus a diameter of 6.8 cm.  So, if the water in the can is just covering the sphere, the water must rise to a height of 6.8 cm.  It is thus in contact with the wall up to a height of 6.8 cm, and in contact with the entire lower surface of the cylinder.  I will also assume the total area of the contact points of the sphere with the cylinder are negligible, so that water covers this entire surface of the cylinder.

 

This means I need to calculate the vertical surface area of a cylinder of radius 3.4 cm and height 6.8 cm, and add in the total area of one endcap.  This means:

 

A_contact = vertical wall surface area + one endcap surface area = (2πrh) + πr² = [2π(3.4 cm)(6.8 cm)]+[π(3.4 cm)²] ≅ 181.6 cm²

 

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If there were water filling the entire inside of the cylinder in the situation above, the volume of water would be:

 

V_cyl = πr²h = π(3.4 cm)²(6.8 cm) = 78.6π cm³

 

However, a good portion of this volume is taken up by the sphere.  So the actual volume of the water in the above scenario is:

 

V_water = V_cyl - V_sph  

 

V_sph = (4/3)πr³ = (4/3)π(3.4 cm)³ = 52.4π cm³

 

The actual volume of water with the sphere in is then:

 

V_water =(78.6π - 52.4π) cm³ = 26.2π cm³

 

Once the sphere is removed, this volume of water collapses down into a smaller cylinder.  There still must be the same volume of water, though, so this new shorter cylinder must have a volume (V_{short cyl}) of 26.2π cm³.

 

The cylinder will still have the radius of the can, so r = 3.4 cm.  Thus, we have:

 

V_{short cyl} = πr²h_short = π(3.4 cm)²(h_short) = 26.2π cm³

 

Now, we can solve for h_short, the depth of the water in the can without the sphere in.  With some algebra, I get:

 

h_short = 2.3 cm

 

I cannot absolutely guarantee the algebra here, but the procedure should be correct.  And the result is reasonable, in that it should be significantly shorter than the diameter of the sphere.  [EDIT:  My answers do match Don's, so that should provide some support... or else we are both wrong in the same way! :D ]

 

If you have any questions about any of these steps, please let me know.  We can always go into more detail about any of them.  I hope this helps!