Applications of Integrals

[Caldus]

I am trying to solve a problem dealing with applying integrals to physics. Here's one that I am having trouble with:

A trough is 4 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of x^6 from x = -1 to x =1. The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.

Here's my attempt at it:

I drew an x-axis which starts at the top (and at the center) of the trough.

Volume of a small slice (horizontal slice):
Length * Width
4 * 2(-x+1)^(-1/2)dx

Weight (Force) of a small slice (horizontal slice):
Volume * Density
(4 * 2(-x+1)^(-1/2)dx)*62

Work required for small slice:
Force * Distance
((4 * 2(-x+1)^(-1/2)dx)*62)x

And then I took the integral of that between 0 and 1 and it did not work. Any ideas?

 

[Tide]

Where did the square roots come from?

 

[M98Ranger]

First of all, great job on attempting to apply integrals to a physics problem! Integrals are very useful in solving problems related to physics, as they help us find the total amount of a quantity (such as work) by adding up smaller parts (such as the work done by each small slice of water).

Now, let's take a closer look at the problem and your approach. The key to solving this problem is to understand that the trough is being emptied by pumping the water over the top. This means that the work done is actually the energy required to lift the water to a certain height, rather than just the force required to move it horizontally.

To calculate the work done, we need to consider the change in potential energy of the water as it is lifted from the bottom of the trough to the top. This can be represented by the integral:

Work = ∫ (force * distance) dx

Where the force is the weight of the water (62 pounds per cubic foot) and the distance is the height at which the water is lifted.

Now, let's break down the problem into smaller parts. We can divide the trough into infinitesimally thin slices, each with a width of dx. The height of each slice can be represented by the function x^6, as given in the problem.

The distance that each slice is lifted can be calculated by subtracting the height of the slice from the total height of the trough (1 foot). This can be represented by (1-x^6).

So, the work done by each small slice can be written as:

dW = (62 * x^6 * dx) * (1-x^6)

Note that we have multiplied by dx to account for the infinitesimally small width of each slice.

To find the total work done, we need to integrate this expression from x = -1 to x = 1, as given in the problem.

Therefore, the total work done can be calculated as:

Work = ∫ (62 * x^6 * dx) * (1-x^6) from -1 to 1

Solving this integral will give you the required amount of work in foot-pounds to empty the trough by pumping the water over the top.

I hope this helps you understand the problem better and gives you a starting point to solve it. Keep practicing and applying integrals to different physics problems, and you will become