# Volume of a partially full fuel tank

1. Aug 28, 2010

### Tregg Smith

1. The problem statement, all variables and given/known data Finding the volume of a partially full cylindrical tank that lies on it's length. L=6', D=4"

2. Relevant equations pi r sq times L = cu ft when full. Convert to gallons 1 gal. =
.1336 cu. ft. (I think)

3. The attempt at a solution (pi r sq times L)/2 = when half full

How about when the fuel is one foot off the bottom? Is this simply (pi r sq times L)/4 ?

How about when the fuel is 10" off the bottom?

The volume of this tank is about 560 gal. which is way too much for the application. Diesel fuel goes bad after a year or two. Only 50 gallons are used each year. If you fill it then you throw out 500 gal. each year. The intake tube from the tank to the generator motor is 6" off the bottom of the tank. I want to know how many gallons to put in the tank that would allow 75 gallons or so to be available-so you would calculate how many gallons are there at 6" then add 75 gallons. That's why I ask is this just a straight line problem where one foot off the bottom equals one fourth of the tanks total capacity etc.? If so then 6" being 1/8 total would be... 560gal/8=70gallons + another 75 gallons=145gal. To put it another way does the area of a circle decrease linearly as a line touching two points on the circle gets shorter?

Last edited: Aug 28, 2010
2. Aug 28, 2010

### HallsofIvy

No, you cannot use the formula for volume of a cylinder for part of a cylinder. In fact, this problem is not really "pre-calculus", it is a standard problem in integral calculus.

Imagine the fuel in the cylinder divided into "layers" of different heights. Each layer is a rectangle. The area of the rectangle is "length times width". We can take the length of each rectangle to be the same as the length of the cylinder: h. The width, however, varies with height. If we take x as the height and take x= 0 at the center of the circular end we can write the equation of the circle to be $x^2+ y^2= r^2$ so that $y= \pm\sqrt{r^2- x^2}$. The width of the rectangular layer is from "circle to circle"- twice y= $2\sqrt{r^2- x^2}$.

The area of each rectangle, then, is "length times width"= $2h\sqrt{r^2- x^2}$. If we take the thickness of the layer to be the infinitesmal dx, the volume is given by $2h\sqrt{r^2- x^2}dx$. The entire volume is the integral from the bottom, at x= -r, to the height of the fuel:
$$V= 2h\int_{-r}^{2r} \sqrt{r^2- x^2}dx$$

Last edited by a moderator: Aug 29, 2010
3. Aug 29, 2010

### Tregg Smith

Thx for your help. I don't quite see all of what you're saying-it's been 30 years since I had calculus. In the mean time can you give me the answer to how many gallons are at the six inch level off the tank bottom(before it starts sucking water into the engine-it's dangerously close) then I'll add 75 gal. run time to that.

4. Aug 29, 2010

### HallsofIvy

There was an error in what I wrote before: I had $2x\sqrt{r^2- x^2}$ when it should have been $2h\sqrt{r^2- x^2}$- the "h" is the length of the cylinder.

You posted this in the homework section and it is the policy of this forum to not give answers to homework. Are you saying it is not homework?

With h= 6' and D= 4' (you had 4" but surely that's not right? Not if you are asking about the volume when the fuel level is at 6"!), the radius is 2' and 6" above the bottom is 1/2 '. The volume of fuel 6 inches above the bottom is
$$2h\int_{-2}^{-3/2}\sqrt{4- x^2}dx$$

The substitution x= 2sin(t) should simplify that.

5. Aug 30, 2010

### Tregg Smith

No this isn't homework-I'm 62. I think it says homework or homework like questions. I don't understand the symbols. I would just like the answer as to how many gallons are at the 6" level. The diameter is four feet, length six feet.
You posted this in the homework section and it is the policy of this forum to not give answers to homework. Are you saying it is not homework?

With h= 6' and D= 4' (you had 4" but surely that's not right? Not if you are asking about the volume when the fuel level is at 6"!), the radius is 2' and 6" above the bottom is 1/2 '. The volume of fuel 6 inches above the bottom is
$$2h\int_{-2}^{-3/2}\sqrt{4- x^2}dx$$

The substitution x= 2sin(t) should simplify that.[/QUOTE]

6. Aug 31, 2010

### HallsofIvy

If you let x= 2 sin(t) the $4- 4 sin^2(t)= 4(1- sin^2(t))= 4cos^2(t)$ so that $\sqrt{4- x^2}= \sqrt{4cos^2(t)}= 2 cos(t)$. Also dx= 2 cos(t) dt so that $\sqrt{4- x^2} dx= 4 cos^2(t) dt$

When x= -2= 2sin(t), sin(t)= -1 and $t= -\pi/2$. When x= 3/2= 2 sin(t), sin(t)= 3/4 so that $t= arcsin(3/4)$
$$\int_{-2}^{3/2}\sqrt{4- x^2}dx= 4\int_{-\pi/2}^{arcsin(3/4)} cos^2(t)dt$$
$$= 2\int_{-\pi/2}^{arcsin(3/4)} 1+ cos(t) dt= 2\left(arcsin(3/4)+ \pi/4- 3/4+ \sqrt{2}/2\right)$$