For my question, I came up with this very simple analogy for the original question I have in mind.

Case 1 let's say there is a rectangle with the length $X$ and width $Y$, so the area will be $XY$. If I have $2$ of these rectangles The combined area would be $2XY$.

Case 2 Let's say I have a cuboid with length $X$, width $Y$ and height $2$, so the volume will be $2XY$.

Now my question is $2XY$ here as $2$ interpretations, so how am I supposed to distinguish that multiplying with $2$ gives me a $2-$dimensional answer or takes me into the 3rd dimension.

I came up with this problem when I was studying the integration by cylindrical shells method. That why do we get the thickness of the shell when we multiply it with delta $x$.

  • $\begingroup$ Using units of length might help with a dimensional analysis $\endgroup$
    – Henry
    Commented Sep 21, 2020 at 18:57

2 Answers 2


@user's answer is correct.

In the first case, the total area is $XY$. In the second case, the volume is $2XY [L]$, where $[L]$ is the unit of length you are using, e.g. meter, foot etc. This is assuming the $X$ and $Y$ include units along with the numerical value.

If $X$ and $Y$ are just numbers, the units are even more clear: $2XY\ [L]^2$, $2XY\ [L]^3$


In the first case the factor $2$ is a dimensionless number obtained by addition of two surfaces

$$XY+XY=2XY \quad [L^2]+[L^2]=[L^2]$$

while in the second case $2$ is a length times a surface, that is

$$2 \cdot X \cdot Y = 2XY \quad [L]\cdot[L^2]=[L^3]$$

Without specifying the dimensions involved is not possible establish a distinction between these expressions.

  • $\begingroup$ does this not allow for higher than 3-dimensional objects? $\endgroup$ Commented Sep 21, 2020 at 19:04
  • $\begingroup$ @blackmamba Yes we can extend to higher dimensions of course. When we are dealing with physical quantities indeed we need always to specify units. $\endgroup$
    – user
    Commented Sep 21, 2020 at 19:08
  • $\begingroup$ Okay then why does multiplying with delta x with 2pi.r.h gives us the thickness of the cylinder and not a 4-dimensional object. $\endgroup$ Commented Sep 21, 2020 at 19:12
  • $\begingroup$ @blackmamba $2\pi r$ is the circumference, not area. Therefore, if you add along all $\Delta x$ (or integrate), then you get the volume of the 3d shape. $\endgroup$
    – KingLogic
    Commented Sep 21, 2020 at 19:13
  • 1
    $\begingroup$ In this case dimensions are $2\pi \cdot r\cdot h \cdot dx = [L]\cdot [L] \cdot [L]=[L^3]$. $\endgroup$
    – user
    Commented Sep 21, 2020 at 19:14

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