# Thread: Work Applications of Calculus

1. ## Work Applications of Calculus

I have a few questions in my Calculus class that I really need help understanding. I don't know much about these work problems; I never took physics. I appreciate any help anyone can give me.

1. Emptying a tank of oil: a spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubic foot. Find the work required to pump oil out through a hole in the top of the tank.

2. Work done by an expanding gas: a quantity of gas with an initial volume of 1 cubic foot and a pressure of 500 paid expands to a volume of 2 cubic feet. Find the work done by the gas (Assume that the pressure is inversely proportional to the volume).

Again, thanks!

2. ## Re: Work Applications of Calculus

Hey jamesknox.

What is the formula for work? Is it the application of force of some path?

3. ## Re: Work Applications of Calculus

Work=force•distance

4. ## Re: Work Applications of Calculus

Originally Posted by jamesknox
I have a few questions in my Calculus class that I really need help understanding. I don't know much about these work problems; I never took physics. I appreciate any help anyone can give me.
The only physics required is "work equals force times distance" which you say you know. Mainly this is an exercise in geometry.

1. Emptying a tank of oil: a spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubic foot. Find the work required to pump oil out through a hole in the top of the tank.
You can write the equation of the sphere as $x^2+ y^2+ z^2= 64$. (All distances in feet) If we imagine a single, thin, "layer of oil" at a given height z, it forms a circle with radius less than or equal to 8 feet. Looking at the sphere "from the side" we have the circle $x^2+ z^2= 64$. At a given z, x, the radius of the circular "layer of oil" is $\sqrt{64- z^2}$ so the surface area is $\pi r^2= \pi(64- z^2)$. Taking the "thickness" of this layer to be dz, the volume of the layer is $\pi(64- z^2)dz$. Multiply that by the density of the oil, 50 pounds per cubic foot, the weight of each such layer is $50\pi(64- z^2)dz$. Each layer has to be raised from z to 8, a distance of 8- z feet. Since "work= force * distance", the work done raising each layer of oil out of the tank is $50\pi(64- z^2)(8- z))dz$. Integrate that from z= -8 to 8.

2. Work done by an expanding gas: a quantity of gas with an initial volume of 1 cubic foot and a pressure of 500 paid expands to a volume of 2 cubic feet. Find the work done by the gas (Assume that the pressure is inversely proportional to the volume).
"Pressure is inversely proportional to volume" means P= k/V. Knowing that it has "initial volume of 1 cubic foot and a pressure of 500 paid" tells you that 500= k/1 so that k= 500. The pressure is P= 500/k.
Nothing is said about the "geometry", whether the gas is in a rectangular solid with one side moving, a cube with all six sides moving, or a sphere with r changing, so let try all three ways.

Suppose we have the gas in a rectangular solid with a side length a, b, and x, with x changing. The moving face has area ab and, since "pressure" is "force divided by area"
the force on that side is (500/V)(ab)= (500/abx)(ab)= 500/x and we can take the slight movement to be $dx$ so that the work done in pushing the face that far is $(500/x)dx$ and the total work is the integral of that.

If the gas is in a cube with side length x then the volume is $x^3$ and each face has area $x^2$ and so the force against it is $(500/x^3)(x^2)= 500/x$. Of course the movement is dx so you want to integrate (500/x)dx.

If the gas is in a sphere with radius r, then the volume is $(4/3)\pi r^3$ and the surface area is $4\pi r^2$ so the force against the surface is $(500/(4/3)\pi r^3)(4\pi r^2)= 1500/r$. Integrate (1500/r) dr.

Again, thanks![/QUOTE]