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) + πr^{2} = [2π(3.4 cm)(6.8 cm)]+[π(3.4 cm)^{2}] ≅ **181.6 cm**^{2}

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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} = πr^{2}h = π(3.4 cm)^{2}(6.8 cm) = 78.6π cm^{3}_{}

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)πr^{3} = (4/3)π(3.4 cm)^{3} = 52.4π cm^{3}

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

V_{water} =(78.6π - 52.4π) cm^{3} = 26.2π cm^{3}

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 cm^{3}.

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

V_{short cyl} = πr^{2}h_{short} = π(3.4 cm)^{2}(h_{short}) = 26.2π cm^{3}

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!

Amy S.

08/18/16