Work Done Problem (20 分)

With reference to the following diagram:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Feb08/Crazyint1-1.jpg

From the diagram, by similar triangle, we have:
r = 0.4h
Then, the volume of the disc-shaped water with thickness dh is:
dV = πr2dh (in ft3)
with weight equal to:
dW = 62.4πr2dh (in lbs)
= 62.4π(0.4h)2dh
= 9.984πh2dh
= 4.538πh2dh (in kg)
And when bringing this disc-shaped water from height h to the top of the tank, work done is equal to:
4.538πh2dh × g × (10 - h)
= 45.38π(10 - h)h2dh (Taking g = 10)
Hence, the total work done can be found by integrating with respect to h from h = 0 to h = 6:
Work done = ∫(h = 0 → 6)45.38π(10 - h)h2dh
= 45.38π∫(h = 0 → 6)(10h2 - h3)dh
= 45.38π [10h3/3 - h4/4](h = 0 → 6)
= 45.38π (720 - 324)
= 17970π (in kg m s-2 ft)
To convert back the unit to J, we use the relation 1 ft = 0.3048 m which gives the work done equal to:
17970π × 0.3048 = 5477π J