What does the area under a volume vs time graph represent?

[orangeraindrops]

Homework Statement

I have a function showing the volume of water in a bay at different times in the day, and I want to know what the area under this curve would represent (if it represents anything meaningful). I know how to integrate, so that isn't a problem.

Homework Equations

I am familiar with physics graphs; I know that the area under an acceleration-time graph gives the change in velocity and the area under a velocity-time graph gives the total distance. However, I am unsure how this would work with volume vs time.

The Attempt at a Solution

My first thought was that the area represents the total volume of water over the full time period, but that doesn't seem to fit with the pattern since the area under a velocity-time graph, for example, gives the total distance and not the change in velocity.

 

[Mark44]

Homework Statement

I have a function showing the volume of water in a bay at different times in the day, and I want to know what the area under this curve would represent (if it represents anything meaningful). I know how to integrate, so that isn't a problem.

Homework Equations

I am familiar with physics graphs; I know that the area under an acceleration-time graph gives the change in velocity and the area under a velocity-time graph gives the total distance. However, I am unsure how this would work with volume vs time.

The Attempt at a Solution

My first thought was that the area represents the total volume of water over the full time period, but that doesn't seem to fit with the pattern since the area under a velocity-time graph, for example, gives the total distance and not the change in velocity.

I don't think the integral represents anything meaningful. If the ordinate on the graph (i.e., the y-value) represents the water volume, then integrating the graph would result in a number with units of volume x time.

 

[fresh_42]

How about the amount of water filling a bathtub?

 

[Mark44]

How about the amount of water filling a bathtub?

The amount of water (volume) would just be the ordinate on the graph at some particular time. Once the graph touches the horizontal line y = V_tub, the tub is full. I still don't think the area under the graph is meaningful, as it would be in units of, say, $\text{m}^3\cdot \text{sec}$.

If the ordinate represented the rate of change of volume relative to time, then the area under the graph would be meaningful.

 

[fresh_42]

I had some fun searching for such diagrams and actually found some applications. Of course this doesn't imply that the integral has a meaning, but I have the feeling, that whenever there is a integrable curve, someone does it and finds an interpretation. The translations are mainly done by Google, so please forgive me strange sentences.

On a Wiki page about construction sites:

A line construction site has - unlike a point construction site - a large longitudinal extent with minimum transverse extent. This is the case when creating networks for supply and disposal (e.g. power grid, water supply, communication, pipelines, road or rail transport).
...
Difference: line construction site - point construction site:

• dimensions
• smaller number of units
• the precursor process determines the working direction of the subsequent process
• are controlled by line diagrams (and in the first instance not with the bar chart)

Types of line chart:
Volume-time diagram (gives a general overview of the project)
...

So at least the graph is used somewhere. So maybe the integral could be viewed as an "overall achievement from start to current time versus expectations".

In a medical book about breathing:

The volume-time diagram shows the time course of the inspiratory applied and exspiratory exhaled volume. ... During volume-controlled incorporation, the volume increases continuously during the inspiratory flow phase, while during the pause or no-flow phase, the applied volume remains constant, as no breathing gas flows into the lungs. The maximum volume value is a measure of the delivered breath volume. During exhalation, the curve shows a downward slope due to the passive breather of the tidal volume.

Another one about breath:

The volume-time curve of the bell-type spirometer shows the static lung volumes in an easily comprehensible manner.
Unfortunately the bell spirometers are technically obsolete, so we have to be content with figures in tabular form.

... yet a breather:
https://vitalograph.de/pdf_library/...rac/Vitalograph_Spirotrac_V_Muster-Report.pdf

Seems as lung doctors use these graphs.

 

[Ray Vickson]

Homework Statement

I have a function showing the volume of water in a bay at different times in the day, and I want to know what the area under this curve would represent (if it represents anything meaningful). I know how to integrate, so that isn't a problem.

Homework Equations

I am familiar with physics graphs; I know that the area under an acceleration-time graph gives the change in velocity and the area under a velocity-time graph gives the total distance. However, I am unsure how this would work with volume vs time.

The Attempt at a Solution

My first thought was that the area represents the total volume of water over the full time period, but that doesn't seem to fit with the pattern since the area under a velocity-time graph, for example, gives the total distance and not the change in velocity.

Over a given time interval $[0,T]$ your integral $I(T) = \int_0^T V(t) \, dt$ may not represent much, but $A = I(T)/T$ represents average volume (that is, a time-average), and that quantity may, indeed, have some meaning and/or importance.

 

[orangeraindrops]

The amount of water (volume) would just be the ordinate on the graph at some particular time. Once the graph touches the horizontal line y = V_tub, the tub is full. I still don't think the area under the graph is meaningful, as it would be in units of, say, $\text{m}^3\cdot \text{sec}$.

If the ordinate represented the rate of change of volume relative to time, then the area under the graph would be meaningful.

I actually do have the equation for the derivative (dv/dt). For some reason it didn't occur to me to use that graph instead. The equation I have models the change in water volume caused by the tides (increasing water height), so I am assuming the area under the derivative would give me the total water volume moved into the bay in a given time period?

Thanks for your help.

 

[orangeraindrops]

Over a given time interval $[0,T]$ your integral $I(T) = \int_0^T V(t) \, dt$ may not represent much, but $A = I(T)/T$ represents average volume (that is, a time-average), and that quantity may, indeed, have some meaning and/or importance.

I don't think there is a way I can apply average volume in the project I'm doing, but it's interesting to know that the integral could be useful. Thanks for your reply.

 

[Mark44]

I actually do have the equation for the derivative (dv/dt). For some reason it didn't occur to me to use that graph instead. The equation I have models the change in water volume caused by the tides (increasing water height), so I am assuming the area under the derivative would give me the total water volume moved into the bay in a given time period?

Yes