# Thread: Find height given volume

1. ## Find height given volume

So I heard an interesting problem on the radio, can't crack it. A trucker's gas gauge is broken, he knows that the tank diameter is 20 inches, if he dips the dipstick and gas is at half the length obviously it's half-full. What height does he need to mark off on the dipstick in order to measure when the tank is 1/4 full?

I figure let's forget about the unnecessary third dimension and solve this for a unit circle. I tried thinking of this as a circle where I'm solving for the top 1/4 area and I don't know the length from the "fluid level" to the top of the tank; nor the length from the $x = 0$ line to the point where the fluid level meets the edge of the tank. So I try setting up my function as $x^{2} + (y - b)^{2} = 1$ where $b$ is the necessary number to set the fluid-level at the x-axis. I then need to integrate from 0 to $a$ where $a$ is the length of the fluid-level line from the $x = 0$ line to the edge of the tank. Integrating $\displaystyle \int_{0}^{a} \Big( b + \sqrt{1 - x^{2}} \Big) dx$ and setting it equal to $\pi / 8$ is a bit laborious and gives an answer that doesn't really seem to help.

I've heard that there is a purely trigonometric solution, but I don't know any trigonometry that deals with area except the basic formula $A = \pi r^{2}$.

2. Originally Posted by ragnar
So I heard an interesting problem on the radio, can't crack it. A trucker's gas gauge is broken, he knows that the tank diameter is 20 inches, if he dips the dipstick and gas is at half the length obviously it's half-full. What height does he need to mark off on the dipstick in order to measure when the tank is 1/4 full?

I figure let's forget about the unnecessary third dimension and solve this for a unit circle. I tried thinking of this as a circle where I'm solving for the top 1/4 area and I don't know the length from the "fluid level" to the top of the tank; nor the length from the $x = 0$ line to the point where the fluid level meets the edge of the tank. So I try setting up my function as $x^{2} + (y - b)^{2} = 1$ where $b$ is the necessary number to set the fluid-level at the x-axis. I then need to integrate from 0 to $a$ where $a$ is the length of the fluid-level line from the $x = 0$ line to the edge of the tank. Integrating $\displaystyle \int_{0}^{a} \Big( b + \sqrt{1 - x^{2}} \Big) dx$ and setting it equal to $\pi / 8$ is a bit laborious and gives an answer that doesn't really seem to help.

I've heard that there is a purely trigonometric solution, but I don't know any trigonometry that deals with area except the basic formula $A = \pi r^{2}$.
From the Scham Mathematical handbook:

3. However, just saying that the gas tank "has a diameter of 20 inches" doesn't tell us the shape or orientation. Captain Black assumed a cylindrical tank "on its side", which is probably most likely, but a spherical tank or a cylindrical tank with vertical axis are also possible.