# Problem involving work

• September 30th 2008, 04:53 PM
PensFan10
Problem involving work
The cylinder stands vertically. Radius = 4. Height = 20. The water level is 12. Weight of water = 62.4
The question is to determine how much work is need to drain the water from the cylinder.

My integral is:
Int(between 0 and 8). 62.4(pi(4)^2)(20-x)
When I solved this out i got 127795.2 pi

I just wanted to know if I did this right.
• September 30th 2008, 06:03 PM
skeeter
if all the water is being pumped to the top of the tank, limits of integration should be from 0 to 12 ...

$W = 62.4 \cdot 16\pi \int_0^{12} (20-y) \, dy$
• September 30th 2008, 06:16 PM
PensFan10
doh
i was trying to follow some problems i found. i accidentally used the height on a diff problem. thank you