Consider the [[Integration in Cartesian coordinates|triple integral]] $\int \int \int_{\Omega} f \, dV$ where $\Omega \subseteq R^{3}$ is the volume under a [[Algebraic Surface|surface]]in $\mathbb{R}^{3}$. Suppose $\Omega$ has rotational symmetry about the $z$ axis, then instead of using *cartesian coordinates*, we might consider using [[Cylindrical Coordinates of Element of Real 3-Space|cylindrical coordinates]] $(r,\theta,z)$. Note: $dV=r \,dz \,dr \,d\theta$. # Examples > [!Example] Example (Volume of cylinder) > Find the volume of a cylinder with base radius $a$, height $h$. > >**Solution** $\begin{align} V &= \int_{0}^{2\pi} \int_{0}^{a} \int_{0}^{h} r \, dz \, dr \, d\theta \\ & = \int_{0}^{2\pi} \, d\theta \int_{0}^{a} r \, dr \int_{0}^h \, dz \\ & = 2\pi \cdot \frac{a^{2}}{h} \cdot h \\ & = \pi a^{2} h \end{align}$ > [!Example] Example (Volume of cone) > Find the volume of a cone with base radius $a$, height $h$. > >**Solution** $ \begin{align} V &= \int_{0}^{2\pi} \int_{0}^{a} \int_{\frac{hr}{a}}^{h} r \, dz \, dr \, d\theta \\ & = 2\pi \int_{0}^{a} r\left( h-\frac{hr}{a} \right) \, dr \\ & = 2\pi h \left[ \frac{r^{2}}{2} - \frac{r^{3}}{3a} \right]_{0}^{a} \\ & = \frac{\pi}{3} a^{2}h \end{align}$ > [!Example] Example (Centroid of paraboloid) > Find the coordinates of the [[Centre of mass (Centroid)|centre of mass]] of the solid bounded by the [[Paraboloid|paraboloid]] $z=1-x^{2}-y^{2}$ and the $x$-$y$ plane. Assume that the density $\rho$ is uniform. > > **Solution**. In cylindrical coordinates, the paraboloid is given by $z=1-r^{2}$. >Let $(\bar{x},\bar{y},\bar{z})$ denote the cartesian coordinates of its centre of mass. By symmetry, $\bar{x}=\bar{y}=0$. $\begin{align} \text{Volume} &= \int \int \int \, dV = \int_{0}^{2\pi} \int_{r=0}^{1} \int_{z=0}^{1-r^{2}} r \, dz \, dr \, d\theta \\ & = 2\pi \int_{0}^{1} r(1-r^{2} ) \, dr \\ & = 2\pi \left[ \frac{r^{2}}{4} -\frac{r^{4}}{4} \right]_{0}^{1} = \frac{\pi}{2} \end{align}$Now $\begin{align} \int \int \int z \, dV &= \int_{0}^{2\pi} \int_{r=0}^{1} \int_{z=0}^{1-r^{2}} rz \, dz \, dr \, d\theta \\ & = \frac{\pi}{6} \\ \implies z = \frac{\pi}{6} / \frac{\pi}{2} = \frac{1}{3} \end{align}$ >Hence centre of mass is at $\left( 0,0, \frac{1}{3} \right)$. ^bcbd53