SOLUTION: A cylindrical glass of (inside) diameter 6 cm and height 11 cm is filled and then tilted to a 45* angle so that some water overflows. How much water is left in it?

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Question 1178324: A cylindrical glass of (inside) diameter 6 cm and height 11 cm is filled and then tilted to a 45* angle so that some water overflows. How much water is left in it?
Found 3 solutions by mananth, greenestamps, ikleyn:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!


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Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The response from the other tutor is absurd from the beginning ("sine of 45 degrees = 1/2").

When the cylinder is tipped at 45 degrees, the water on one edge of the cylinder is the full 11 cm from the bottom; on the opposite edge of the cylinder it is 11-6=5 cm from the bottom.

Because of the symmetry of the cylinder, the average "depth" of water in the glass is (11+5)/2 = 8cm.

The volume of water left in the cylinder is then

(pi)(r^2)(h) = (pi)(3^2)(8) = 72pi.

ANSWER: 72pi cm^3
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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            The formulas and the plot in the post by @Mananth may shock anyone who has brain cells in his  (or her)  head.

            Therefore,  I came to present the solution in the normal form as it  SHOLUD  be done. 


Consider the cylinder of the diameter of 6 cm and the height of 6 cm.


The volume of the water which went out was HALF of the volume of this cylinder


     =  =  = 84.78 cm^3.


The volume of the water left in the ORIGINAL cylinder (remained in the original cylinder) is


     =  - 84.78 =  = 226.08 cm^3.      ANSWER

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Solved and normally presented.



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