For a cylinder, the volume formula is given by $V = \pi r^2 H $. To find the differential ( dV ), we differentiate ( V ) with respect to ( r ) and ( H ), yielding $\left( dV = 2 \pi r H dr + \pi r^2 dH \right)$. Upon integrating ( dV ) again, we obtain $( 2 \pi r^2 H )$. However, I'm confused about where the factor of two originates.
Additionally, in calculus, it's taught that to derive the formula for the volume of a cylinder, we perform a double integral of ( da ) over the cylinder's surface area ( A ) and ( dh ) over the height ( H ). This suggests that $( V = \iint 2 \pi r dr dh )$, or equivalently $( V = \int \pi r^2 dH )$, which only includes the second term of the earlier ( dV ) expression. Does this imply that the first term is zero?
Before getting into differentials, let's talk about that calculus formula, $V = \iint 2 \pi r\,\mathrm dr \,\mathrm dh$. How do we know that this gives the volume of a cylinder?
I think it is easier to justify the formula $$V = \iiint_S f(r,\theta,h)\, r\,\mathrm d\theta \,\mathrm dr \,\mathrm dh$$ where $f(r,\theta,h)=1$, which is the integral of the constant function $1$ over the interior of a cylinder $S$ in cylindrical coordinates and $r$ is the Jacobian of the coordinate transformation. (That is, the volume element is $\mathrm dV = r\,\mathrm d\theta \,\mathrm dr \,\mathrm dh$.) We choose the cylindrical coordinates such that the axis of the cylinder $S$ is the axis of the coordinate system, so the integral over $\theta$ can simply run from $0$ to $2\pi$ for every $r$ and $h$. Then by doing the integration over $\theta$ first we reduce the integral to $V = \iint 2 \pi r\,\mathrm dr \,\mathrm dh$.
The point here is that $r$ and $h$ in this integral are not the radius and height of the cylinder $S$; they're just cylindrical coordinates of points inside the cylinder over which we integrate.
Getting back to how we get the volume of a cylinder from the differential $2\pi r H\,\mathrm dr + \pi r^2\,\mathrm dH$, I think the first question to ask why we would ever think in the first place that we could use a differential to find the volume of a three-dimensional shape.
Since the differential $2\pi r H\,\mathrm dr + \pi r^2\,\mathrm dH$ is only a first-order differential, we should only do a single integral on it, not a double integral. Moreover, we don't want to wind up with an unknown constant of integration, so we want something more analogous to a definite integral of the form $$ \int_a^b f(t)\,\mathrm dt, $$ which gives us a value. Note that the value of a definite integral depends both on the starting point and on the ending point.
There's also a problem in the notation that needs to be addressed. I don't see how I can clearly explain the necessary integration if I have to write things like "integrate over $r$ from $0$ to $r$." I need to be able to distinguish when I'm talking about something inside an integral and when I'm talking about the radius $r$ and height $H$ of a particular cylinder whose dimensions I'm going to plug into the formula $V = 2\pi r^2 H$.
In another answer the notational problem is resolved by using $r$ and $H$ for the radius and height of the cylinder and using $\gamma_1(t)$ and $\gamma_2(t)$ for the "radius" and "height" variables inside the integral. I'll take a slightly different approach, which is to write $r$ when I mean the radius of a particular cylinder $S$ whose volume I want to know, and $r(t)$ when I mean a variable quantity that I can use inside the integral, which isn't always equal to the radius of the cylinder $S$. Likewise I'll write $H$ when I mean the height of the particular cylinder $S$, and $H(t)$ for the corresponding variable quantity I use inside the integral.
I will use the same variable $t$ in $r(t)$ and $H(t)$ because the integration needs to be a one-dimensional integral. We can make the integral one-dimensional by integrating over $t$ from a starting value $a$ to an ending value $b$. A common convention is to start at $a=0$ and end at $b=1$, but all that really matters is that the integration starts at $r(a)$ and $H(a)$ and ends at $r(b)$ and $H(b)$.
We can think of $r(t)$ and $H(t)$ as the radius and height of an inflatable cylinder that we can enlarge at will. The differential of the volume of this cylinder tells us how the volume of this inflatable cylinder increases as we increase its dimensions. We can write that differential as $$dV(t) = 2\pi r(t) H(t)\,\mathrm dr(t) + \pi (r(t))^2\,\mathrm dH(t).$$
Conveniently, a "cylinder" of radius $0$ and height $0$ is just a single point, which has volume $0$. So if we let an integration of $\mathrm dV(t)$ start at this point, $(r(a),H(a)) = (0,0)$, and stop at the point $(r(b),H(b)) = (r,H)$, where the inflatable cylinder has grown to exactly match the size and shape of the cylinder $S$, the integral (which literally represents how much the volume of the inflatable cylinder grew) will be equal to the volume of the cylinder $S$.
We can write $\mathrm dr(t) = r'(t)\,\mathrm dt$ and $\mathrm dH(t) = H'(t)\,\mathrm dt$, so another way to write the differential is
$$\mathrm dV(t) = (2\pi r(t) H(t) r'(t) + \pi (r(t))^2 H'(t))\,\mathrm dt.$$
Then the volume of the cylinder $S$ is $$V = \int_{t=a}^{t=b} \mathrm dV(t) = \int_a^b (2\pi r(t) H(t) r'(t) + \pi (r(t))^2 H'(t))\,\mathrm dt.$$
There are a few more details to settle before we can evaluate this integral. What are the values of $a$ and $b$ and how do the values of $r(t)$ and $H(t)$ increase from $0$ to $r$ and $H$ (respectively) as $t$ goes from $a$ to $b$?
We have a lot of freedom in how we answer those questions. In the other answer previously mentioned, the choices are $a=0$, $b=1$, $r(t)=tr$, and $H(t)=tH$. But really any differentiable functions $r(t)$ and $H(t)$ that have the desired values at $t=a$ and $t=b$ will do. And if you're willing to integrate a function that is only piecewise continuous and is undefined at the single point between each consecutive pair of pieces, you can drop the strict requirement that $r(t)$ and $H(t)$ are differentiable everywhere on the interval $(a,b)$ and only require that they are differentiable at all but a finite number of points in the interval $(a,b)$.
That opens up the following alternative integrations:
Alternative A
We can set $a=0$, $b=r+H$, \begin{align} r(t) &= \begin{cases} t & t < r, \\ r & t \geq r, \end{cases} \\ H(t) &= \begin{cases} 0 & t < r, \\ t - r & t \geq r. \end{cases} \end{align}
In other words, we start at $(r(0),H(0)) = (0,0)$, we first increase $r(t)$ from $0$ to $r$ while holding $H(t)$ constant at $0$, then we increase $H(t)$ from $0$ to $H$ while holding $r(t)$ constant at $r$. Then \begin{align} r'(t) &= \begin{cases} 1 & t < r, \\ 0 & t > r, \end{cases} \\ H'(t) &= \begin{cases} 0 & t < r, \\ 1 & t > r, \end{cases} \end{align} and therefore $$ \mathrm dV(t) = (2\pi r(t) H(t) r'(t) + \pi (r(t))^2 H'(t))\,\mathrm dt = \begin{cases} 0\,\mathrm dt & t < r, \\ \pi r^2 \,\mathrm dt & t > r. \end{cases} $$ Keep in mind that $r$ is a constant in the expression $\pi r^2\,\mathrm dt$ on the right-hand side of the last equation. Also note that the expression $2\pi r(t) H(t) r'(t)$ is always $0$ (because either $H(t)=0$ or $r'(t)=0$) except at $t=r$, where it is undefined.
We then find that \begin{align} V &= \int_{t=0}^{t=r+H} \mathrm dV(t) \\ &= \int_{t=0}^{t=r} \mathrm dV(t) + \int_{t=r}^{t=r+H} \mathrm dV(t) \\ &= \int_0^r 0\,\mathrm dt + \int_r^{r+H} \pi r^2 \,\mathrm dt \\ &= 0 + \int_0^H \pi r^2 \,\mathrm dt \\ &= \pi r^2 H. \end{align}
Alternative B
We can set $a=0$, $b=H+r$, \begin{align} r(t) &= \begin{cases} 0 & t < H, \\ t - H & t \geq H, \end{cases} \\ H(t) &= \begin{cases} t & t < H, \\ H & t \geq H. \end{cases} \end{align}
In other words, we start at $(r(0),H(0)) = (0,0)$, first increase $H(t)$ from $0$ to $H$ while holding $r(t)$ constant at $0$, then increase $r(t)$ from $0$ to $r$ while holding $H(t)$ constant at $H$. Then \begin{align} r'(t) &= \begin{cases} 0 & t < H, \\ 1 & t > H, \end{cases} \\ H'(t) &= \begin{cases} 1 & t < H, \\ 0 & t > H, \end{cases} \end{align} and therefore $$ \mathrm dV(t) = (2\pi r(t) H(t) r'(t) + \pi (r(t))^2 H'(t))\,\mathrm dt = \begin{cases} 0\,\mathrm dt & t < H, \\ 2\pi (t-H) H \,\mathrm dt & t > H. \end{cases} $$
Here $H$ is a constant in the expression $2\pi (t-H) H\,\mathrm dt$ on the right-hand side of the last equation and the expression $\pi (r(t))^2 H'(t)$ is always $0$ except at $t=H$, where it is undefined.
Then (leaving a couple of obvious steps implicit that were spelled out for Alternative A) \begin{align} V &= \int_0^r 0\,\mathrm dt + \int_H^{H+r} 2\pi (t-H) H \,\mathrm dt \\ &= 0 + \int_0^r 2\pi t H \,\mathrm dt \\ &= \pi r^2 H. \end{align}
Summary
You can only do a single integral on the first-order differential $\mathrm dV$, and since there are two variables (radius and height) you must define a path integral that starts at a point in a graph of radius and height where both radius and height are zero and ends at a point where the radius is $r$ and the height is $H$.
One of the ways you can do this (Alternative A) is to first increase the radius from $0$ to $r$, which doesn't add anything to the volume of the cylinder since the height is still zero, and then increase the height from $0$ to $H$, which (in effect) integrates the constant $\pi r^2$ from $0$ to $H$.
Another way you can do this (Alternative B) is to first increase the height from $0$ to $H$, which doesn't add anything to the volume since the radius is still zero, and then increase the radius from $0$ to $r$, which (in effect) integrates the function $2\pi t H$ over $t$ from $0$ to $r$.
Each of these methods legitimately integrates the differential $2\pi r(t) H(t)\,\mathrm dr(t) + \pi (r(t))^2\,\mathrm dH(t)$ over a path that starts at zero radius and zero height and ends at radius $r$ and height $H$. It just happens that the first part of the differential, $2\pi r(t) H(t)\,\mathrm dr(t)$, is always zero on one path, whereas the second part of the differential, $\pi (r(t))^2\,\mathrm dH(t)$ is always zero on the other path.
When you integrated the differential and got the result $2 \pi r^2 H$, you assumed that $H(t) = H$ in $2\pi r(t) H(t)\,\mathrm dr(t)$ and that $r(t) = r$ in $\pi (r(t))^2\,\mathrm dH(t)$. As Alternatives A and B showed, you can indeed do the integral in such a way that one of those assumptions is true. But you cannot satisfy both assumptions in the same integration. If $H(t)=H$ as $r(t)$ increases from $0$ to $r$, you can't have $r(t)=r$ while $H(t)$ increases from $0$ to $H$.
Indeed if you assume that $H(t) = H$ in $2\pi r(t) H(t)\,\mathrm dr(t)$, then $\pi (r(t))^2\,\mathrm dH(t)$ contributes nothing to the integral. But if you assume that $r(t) = r$ in $\pi (r(t))^2\,\mathrm dH(t)$ then $2\pi r(t) H(t)\,\mathrm dr(t)$ contributes nothing to the integral.
By assuming two incompatible assumptions, you ended up with an integral that was the same as if you integrated according to Alternative A and then repeated the integration according to Alternative B and added these two results together. Naturally, by integrating the entire volume twice, you ended up with twice the correct volume.
The general solution, as shown in the answer already mentioned, is to assume neither that $H(t)=H$ nor that $r(t)=r$ while evaluating any part of the differential at any time during the integration. Rather, you let both $r(t)$ and $H(t)$ grow concurrently. You therefore have to know the value of $r(t)$ at each value of $H(t)$ and vice versa in order to work out the integral.
Typically you will find that the left half of the differential $2\pi r(t) H(t)\,\mathrm dr(t) + \pi (r(t))^2\,\mathrm dH(t)$ will contribute less than $\pi r^2 H$ to the final integral and the right half also will contribute less than $\pi r^2 H$, but when you add up the contributions of each part you will get the total $\pi r^2 H$.
Aside
In one of the comments it is suggested to consider a potential function for a conservative force field.
The relevance to this question is that the volume of a cylinder should depend only on its radius $r$ and height $H$, not on the particular path you took while integrating the differential from $(0,0)$ to $(r,H)$. The existence of a potential function in a conservative force field likewise depends on the fact that it doesn't matter what path you take from one point to another; the integral of the force along that path is the same.
You can imagine that the radius $r$ and height $H$ of the cylinder are plotted as a point in a plane with the radius along one coordinate axis and the height along the other coordinate axis, and that there is a "potential function" defined on that plane such that the potential at coordinates $(r,H)$ is the volume of a cylinder of radius $r$ and height $H$. The differential $2\pi r(t) H(t)\,\mathrm dr(t) + \pi (r(t))^2\,\mathrm dH(t)$ then corresponds to the conservative force field that supports the potential function.