Integration Work

by darthxepher
Tags: integration, work
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 Sounds to me like a trough with triangular ends. Are we to presume that the vertices of the triangles are up or down? And are we to presume that the triangles are isosceles? Assuming the triangles are isosceles with vertices down, start by drawing a picture of a triangular end. Draw a horizontal line across the triangle at height x. That represents one "layer" of water to be lifted at height 3- x to the top of the trough. Work= force times distance so the work done to lift that "layer" is its weight times 3- x. Weight is density times volume. Density is a constant but the volume depends upon x. Each "layer" a rectangle with length 6 m and width equal to the length of that horizontal line (call that w for now). We can take the thickness of the "layer" to be dx. Now the volume of that "layer" is 6wdx and its weight is $6\delta w dx$ where $\delta$ is the density of water. The work done in lifting that "layer" to the top of the trough is $6\delta w (3- x)dx$. The work done in lifting all of the water in the trough out of the trough is [tex]6\delta\int_0^2 (3-x)w dx[/itex] Now, to find w as a function of x, use "similar triangles". The triangle of water has height x and base width w and is the same shape as the entire end of the trough which has height 3 and base width 4. w/x= 4/3.