# calculus

a large vertical rectangular plate of glass is to be inserted in the wall of an aquarium underwater so visitors can see into th tank of fish. the glass is 10 feet high, 25 feet long and the top of the glass is 3 feet below the top of the water. the pressure exerted by water at depth b feet below the surface is .433b pounds/square inch. your task is to set up the integral to compute the TOTAL pressure of the water in pouds against the glass

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1. What they appear to be asking for is the total FORCE, not the total pressure. Pressure is not measured in pounds, but force is. When you multiply or integrate froce times area, you get force.

There are two ways to do this. The easiest way is to multiply the average pressure on the glass, 0.433 lb/in^2 ft*8 ft = 3.464 lb/in^2, by the window area in square inches (36,000 in^2). I multiplied by 8 ft because that is the average depth of the window beneath the surface. The window depth runs from 3 to 13 ft.

Since they want you to get the answer by integration, write

F = 300 in*12 in/ft*
INTEGRAL 0.433 b (lb/(in^2) db(ft)
b from 3 to 13 ft
= 3600*0.433 INTEGRAL b db
b from 3 to 13 ft
= [(13^2)/2- (3^2)/2]*1558.8 lb

Note that [(13^2)/2- (3^2)/2] is
(window height)*(avg. depth)
= (13-3)*(13+3)/2 in square feet
The b db integral leads to the b^2/2 term, which gets evaluated at b = 13 and 3 feet.

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