An open tank has the shape of a right circular cone. The tank is 8 feet across the top and 18 feet high. Water is pumped in through the bottom of the tank. (Assume water weighs 62.4 lb per cubic foot.) (a) How much work is done to fill the tank to a depth of 6 feet? foot-pounds (b) How much work is done to fill the tank from a depth of 12 feet to a depth of 18 feet? foot-pounds I am having so much trouble with this one -_- Help is appreciated GREATLY.

I'm not sure but you should try integrating the volume formula/weight of water of a cone by using 0 and 6 in the first example and 12 and 18 in the second.

That's all I'm given... I am guessing the radius would be 4 if its 8ft across... formula is probably w(y)= density(volume)(distance) just that I'm not very good at doing all that junk : )

No, the radius will be 4ft at the top but in a cone, the radius drops down as you go down. Wait, I'll try again.

You could try integrating it with this 0 / 6 (that's the integration sign with the starting value 0 and the ending value 6) 62.4(density) * Pie . r ^2. h. dh Then find the radius of the cone in terms of H and solve.

Wow, I will see what I can do... I gotta head out to work but will come back in 6-7 hrs and see how it comes... thanks so muchh : )

Thank god school is out for life!!!! Integration.....it was on my exam 2 months ago but I literally cannot remember what to do

Any current calculus book should have a section on how to do this. I just grabbed Stewart, 3rd edition and it's in section 6.5 Applications to Physics and Engineering. The example they give is similar to this one: a tank has the shape of an inverted circular cone with height 10 m and base radius 4m. It is filled with water to a height of 8m. Find the work required to empty the tank by pumping all of the water to the top of the tank. It provides the density of water so that you can figure out the amount of work required. The original poster should have a textbook with a pretty close example that can be modified for this problem. If not, a trip to the school library or math department should turn up a textbook with a suitable example.

Is the cone right side up? As in the base is on the ground. If so here is the solution. By using similar triangles, you find the radius x = 2/9 (18-y) Disk of Water weight = 62.4 * pi * x^2 * (delta)y Distance water has to travel = y (delta)w= Water weight * distance so integrate (delta)w a) from 0 to 6 b) from 12 to 18 answers a)34502.2 ft lbs b)9409.7 ft lbs I am 99.99% sure this is correct.