Most of these tanks are cylinders, with their axes horizontal. You are to "stick the tank" by inserting a measuring rod through a hole in the center of the top until it touches the bottom, then pulling it out and reading off the liquid level showing on the stick.
Assuming that the cross sectional radius of the tank is R and its length is L, calibrate the stick for them. That is, provide a formula to convert height showing on the stick to volume of liquid. Do this by evaluating a definite integral that gives the filled volume in terms of the height, h, on the stick.
Thanks!
Assuming that the cross sectional radius of the tank is R and its length is L, calibrate the stick for them. That is, provide a formula to convert height showing on the stick to volume of liquid. Do this by evaluating a definite integral that gives the filled volume in terms of the height, h, on the stick.
Thanks!
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The problem is really only 2 dimensional. You need to find the cross sectional area of the fluid as a function of its height. Multiply this by L to get the volume.
You can set this up in multiple ways. I'll consider the cross section to be the circle of radius R centered at (0, R)---this is sitting on top of the x-axis so the axis of symmetry for the cylinder would come out of the plane and have equation y = R, x = 0.
The circle has equation in polar coordinates r = 2RsinΘ. If the height is the line y = h, then in polar this is
r = h cscΘ.
The circle and line intersect when
h cscΘ = 2R sinΘ ==> sin²Θ = h/2R ==> Θ = arcsin(√(h/2R)).
The area inside the circle and under the line y = h has an area that can be split into two parts. The area inside the circle for 0 ≤ Θ ≤ arcsin(√(h/2R)) (with its mirror image in the second quadrant) and the triangle formed by the line x = 0, Θ = arcsin(√(h/2R)), and the fluid line. (See the linked site for a figure.)
The circle area is
arcsin(√(h/2R))
∫ (2RsinΘ)² dΘ = 2R² arcsin(√(h/2R)) - R√(2Rh - h²).
0
The triangle has area
h√(2Rh - h²).
Add these to get the cross sectional area. The volume is therefore
V = [2R² arcsin(√(h/2R)) - (R - h)√(2Rh - h²)] L.
A diagram is here.
http://www.flickr.com/photos/54652214@N0…
To get the area of the triangle, just get half of the base by noting that if half of the base is Q then
Q/h = cot(arcsin(√(h/2R)) ==> Q = h √(2R/h - 1) = √(2Rh - h²).
You can set this up in multiple ways. I'll consider the cross section to be the circle of radius R centered at (0, R)---this is sitting on top of the x-axis so the axis of symmetry for the cylinder would come out of the plane and have equation y = R, x = 0.
The circle has equation in polar coordinates r = 2RsinΘ. If the height is the line y = h, then in polar this is
r = h cscΘ.
The circle and line intersect when
h cscΘ = 2R sinΘ ==> sin²Θ = h/2R ==> Θ = arcsin(√(h/2R)).
The area inside the circle and under the line y = h has an area that can be split into two parts. The area inside the circle for 0 ≤ Θ ≤ arcsin(√(h/2R)) (with its mirror image in the second quadrant) and the triangle formed by the line x = 0, Θ = arcsin(√(h/2R)), and the fluid line. (See the linked site for a figure.)
The circle area is
arcsin(√(h/2R))
∫ (2RsinΘ)² dΘ = 2R² arcsin(√(h/2R)) - R√(2Rh - h²).
0
The triangle has area
h√(2Rh - h²).
Add these to get the cross sectional area. The volume is therefore
V = [2R² arcsin(√(h/2R)) - (R - h)√(2Rh - h²)] L.
A diagram is here.
http://www.flickr.com/photos/54652214@N0…
To get the area of the triangle, just get half of the base by noting that if half of the base is Q then
Q/h = cot(arcsin(√(h/2R)) ==> Q = h √(2R/h - 1) = √(2Rh - h²).