# Hydrostatic Force and Integration

A homework problem states:

An aquarium 5 ft long, 2ft wide, and 3ft deep is full of water. Part C: find the hydrostatic force on one end of the aquarium.

Questions (I've completed part a and b. B asked for the force on the bottom which was 1875 lbs.:

1. What exactly does one end of the aquarium mean?

2. Is there a standard to applying long and wide in mathematics problems? It may not matter in calculating area, but in other instances it might. For example, calculating the area of a vertical standing slice would either have to be 2dx or 5dx.

3. My solutions manual shows that part C requires setting up an integral with a range from 0 to 3. This means that the "one end" of the aquarium is the top side? How else could the range be 0 to 3?

Here's my drawing of the aquarium:

-
Btw, the solutions manual solution to C is: Integral density*depth*2dx from 0 to 3. Which is Integral 62.5x * 2dx from 0 to 3. Why is the depth variable? I keep thinking "one end" means one side of the tank. Apparently, not.... – ShrimpCrackers Nov 21 '10 at 2:35
1. Objects are assumed to be longer than they are wide (unless a front and back are explicitly specified). – trutheality Nov 21 '10 at 2:36
2. You integrate over depth to get the force onto one side because at each horizontal cros-section the pressure is different (since it's a function of depth). – trutheality Nov 21 '10 at 2:38
trutheality. Thanks. Makes sense now. Add a formal reply and I can accept your answer for points. – ShrimpCrackers Nov 21 '10 at 2:46