Advanced Calculus

Question

The cross section of a 3meter long water trough is a semicircle with a radius of 1 meter. If the water trough is full, how much force does the fluid exert on the end of the trough? How much work is done in pumping all the fluid out of the trough

Answer 1

For the first part, you have to calculate the integral of PdA where P is P_atm + ρgy where y is the depth into the trough. The length of the trough is irrelevant to the force on the end. The net pressure at any point at the end is ρgy (but the question implies the former expression). Density is 1000 kg/m³ and g is 9.8 m/s².

There are many approaches to the integral: single integral in y with x(y), double integral with x(y) limits, and double integral in polar coordinates with y as rsin(θ) and area element rdrdθ. The polar solution is the easiest integral.

Ex. integral from 0 to 1 of (P_atm + gρgy)(2sqrt(1-y²))dy where dA is 2x(y)dy

or integral for r = 0 to 1, θ = 0 to pi of (P_atm + ρgrsinθ) rdrdθ

The work is the integral of ρgydV for the cylinder. dV can be interpreted as 2x(y)Ldy or the cylindrical version which will be rLdrdθ where L is the length of the trough.

Good luck!