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Suppose, we have a cylinder with a flat base area $A$ and height $H$.

The volume $V$ of the cylinder is obtained by multiplying the two quantities: $V=AH$.

But what happens, when the base surface is not flat? For instance, if we take a little 'bite' at the bottom of the cylinder and put the bitten slice on top, the process mustn't change volume. It sure doesn't change the height, only the base surface changes (increases) to, say $A'$.

Now, obviously, as $A'>A$, we know, that $A'H>AH$.

Hence, area-height multiplication can't be a good way to calculate volume. So, I'm asking you: what is?

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One model for your solid is to take a plane region $D$ of area $A$, to take a real-valued (and, say, continuous) function $f$ with domain $D$, and to let the cylinder be the set of points $(x, y, z)$ such that $$ (x, y) \in D,\qquad f(x, y) \leq z \leq f(x, y) + H. $$

The volume of the cylinder is $AH$, independently of $f$, as you've observed.

One natural geometric interpretation of this formula is the height over an arbitrary point of $D$ multiplied by the area of the projection of the cylinder on the $(x, y)$-plane.

This type of formula holds in greater generality. If $f$ is a continuous function of $(n - 1)$ variables, then the "pointwise shearing" homeomorphism $$ (x_{1}, \dots, x_{i}, \dots, x_{n}) \mapsto (x_{1}, \dots, x_{i} + f(x_{1}, \dots, x_{i-1}, x_{i+1}, \dots, x_{n}), \dots, x_{n}) \tag{1} $$ preserves $n$-dimensional volume. If you start with the closure $C$ of an arbitrary bounded open subset of $\mathbf{R}^{n}$ and repeatedly apply transformations of the type (1), the resulting image has the same volume as $C$.

Your cylinder with non-flat base is the image of a "standard" cylinder $C = D \times [0, H]$ under $$ (x, y) \mapsto \bigl(x, y, z + f(x, y)\bigr). $$

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If a volume is cut out from base area as per an arbitrary function and shifted up by same height distance H there is no net volume change, no matter what new area is created/shifted. This is because each single differential shell is pushed up retaining its volume as

$$ \int dV = H \int dA = H \int 2 \pi r\, dr .$$

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  • $\begingroup$ This is exactly, what I'm implying. Actually, a friend of mine asked me this question in a form of a childish paradox: "If we cut a little cone from the bottom of the cylinder and put it on top of it, the base surface changes. Is the volume still [base area]x[height]? If so, how come, it increased (as the height is clearly the same)? If not, then, how can you calculate it?" $\endgroup$ – Chanto Jul 4 '15 at 18:14
  • $\begingroup$ Please see the above. What you lose at base you gain at the top, no net gain or loss. Have you started your integral calculus classes? $\endgroup$ – Narasimham Jul 4 '15 at 18:21
  • $\begingroup$ Yes, I've had my share of integral calculus. Your answer is not exactly, what I had in mind, but it sure is appreciated. My mistake was pulling $dH$ out of the integral and not considering, that $dH dA$ is not the same as $dV$, when there's angle between them. You have to take the projection of $dA$ on the plane, which has $dH$ as a normal vector and the angle depends on whatever the surface function is. I'd refine your answer as: $ \int dV = \int dH dA sin(t) $, where $t$ is the angle between $dH$ and $dA$. $\endgroup$ – Chanto Jul 4 '15 at 18:37

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