Deriving the volume of a right circular cylinder using triple integral
Just for fun, I was wondering how one could derive the formula for the volume of a right circular cylinder using triple integrals in cylindrical coordinates.
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$\begingroup$Let $r$ be the radius and $h$ be the height. The ranges for each variable are given by $\rho \in [0,r], z \in [0,h], \phi \in ]-\pi,\pi]$. The volume van be evaluated using a triple integral $$\begin{align*} \iiint_{\mathrm{Cyl}} dV &= \int_0^r \int_0^h \int_{-\pi}^\pi \rho \,d\phi\,dz\,d\rho \\ &= \int_0^r \int_0^h 2\pi \rho\,dz\,d\rho \\ &= \int_0^r 2\pi \rho h \,d\rho \\ &= \left. \pi \rho^2 h \right|_0^r = \pi r^2 h \end{align*}$$
$\endgroup$ $\begingroup$Let $C = \{ (x,y,z) \in \Bbb R^3 \mid x^2+y^2 \leq R^2, ~0\leq z \leq h \}$. If you put $f(r, \theta, z) = (r\cos\theta, r\sin\theta, z)$, then by the change of variables theorem we'll have: $$\begin{align} {\rm Vol}(C) &= \int_C 1 = \int_{f([0,R]\times[0,2\pi]\times[0,h])}1 \\ &= \int_{[0,R]\times[0,2\pi]\times[0,h]}|\det Df(r,\theta,z)|\,{\rm d}r\,{\rm d} \theta\,{\rm d}z \\ &= \int_0^h\int_0^{2\pi}\int_0^R r \,{\rm d}r\,{\rm d}\theta\,{\rm d}z \\ &=2\pi h \frac{r^2}{2}\bigg|_0^R = \pi R^2h. \end{align}$$
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