# Calculating Volume of a Gas Tank/Gas within

1. Dec 6, 2009

### marcus768

1. The problem statement, all variables and given/known data
I am taking a calc 2 class and I'm studying for finals. I've got this problem i need to figure out and I am having a tough time with it. if someone could give me some guidance in solving this I would greatly appreciate it. I am not looking for answers to this assignment - I would really like to understand how this problem is working, since similar problems will presumably be on the final. If anyone could help guide me through this I would really appreciate it.

A cylindrical gas tank with radius 4ft. and length 22ft is buried on its side.
-Develop an integral that could be used to find the volume of gas in the tank and number of gallons of gas in the tank, given the depth of the gasoline level h
- Test the integral by using it to find the amount of gas in the tank (by vol in cubic ft and gallons) for h=8 ft (full) h=4ft h=0ft h=6 ft

2. Relevant equations
I am trying to see if I can set up my integral in a way where it is not dependent on height. V=(pi)(r)^2L is what I have been trying. I am trying to set up my integral in disc format. V=(integral sign) pi[f(x)]^2dx

3. The attempt at a solution
So what I've been trying to do is first create a function with the V=(pi)(r)^2L formula. Since the radius is given, and the length is given also, where is my variable? Is my radius (4-x)? I guess I am having trouble understanding how to set up the integral. When it comes to evaluating the integral, I am fine. i really need to understand this concept, I am not looking for answers.

any help is appreciated very much, thank you

2. Dec 6, 2009

### LCKurtz

Suppose you are looking directly at the end of the tank, which is a circle of radius 4. Suppose the water is to depth h. Can you set up an integral to calculate the wet area of the end of the tank? The answer, of course, will depend on h. And the volume of water in the tank is just L times the wet area, no?

3. Dec 6, 2009

### marcus768

Ok, so by using the formula for volume of a cylinder V=pi*(r^2)h
I am coming up with V=pi((4-x)^2)22
I have been trying to use the integral format V=(integral)A(h) dh
I am getting V= (integral evaulated from 0 to 8) (pi(4-x)^2)22
After I evaluate the integral, I am getting approx 2949. This is way off to what it should be, when I use the formula for vol. of a cylinder I get
V= pi(4^2)22 = approx. 1105.
I do not get where I am going wrong!!

4. Dec 6, 2009

### LCKurtz

Did you even read my post? You don't need to integrate the cross section area along the length of the tank. All the cross sections are the same. You need to calculate the area of the wet portion of the end for a depth h and multiply it by the length of the tank to get the volume of water.

Can you set up the integral for the area of the end covered when the water has depth h? That is what you need to do.