2. We need to find the dimensions of the given cylinder without the new conical top so that we can find the radius of our cone top.

Input := 


solR = Solve[ Pi R^2 12 == 354,R]

Output =


              59                 59

{{R -> -Sqrt[----]}, {R -> Sqrt[----]}}

             2 Pi               2 Pi

Thus the radius R of our tin can must be 3.06433 cm.

Therefore we want to construct a cone top that will have a base radius

of 3.06433 cm.

Recall the relationship between the outer perimeter of the cut out circle and the perimeter of the base of the cone from equation 1 above. We reeneter this formula with our R in it.

We solve for t, the angle of the cut, for this specific radius, R, of the base of the cone.

And substitute the radius of the tin can r = 4 cm into this expression to find that our angle for the cut must be 1.46974 radians or 84.21 degrees.

Then find the additional volume from the cone atop the can is 25.2812 cm^3

And the total volume of the can cone combination is 628.467 cm^3.

Input := 


TotalVol = coneVol + Pi 4^2 12//N

Output =


628.467