# Work Applications of Calculus

• Apr 27th 2013, 07:28 PM
jamesknox
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!
• Apr 27th 2013, 09:47 PM
chiro
Re: Work Applications of Calculus
Hey jamesknox.

What is the formula for work? Is it the application of force of some path?
• Apr 28th 2013, 06:01 AM
jamesknox
Re: Work Applications of Calculus
Work=force•distance
• Apr 28th 2013, 07:22 AM
HallsofIvy
Re: Work Applications of Calculus
Quote:

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.

Quote:

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.

Quote:

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]