Finding the Mass of a Lamina
A lamina in the shape of the cone \(z=6-\sqrt{x^{2}+y^{2}}\) lies between the planes \(z=2\) and \(z=5\). If the mass density of the lamina is \(\rho (x,y,z)=\sqrt{x^{2}+y^{2}}\), find the mass \(M\) of the lamina.
First we find \(\mathbf{r}_{r}\times \mathbf{r}_{\theta }.\) \[ \begin{eqnarray*} \mathbf{r}_{r}\times \mathbf{r}_{\theta } &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos \theta & \sin \theta & -1 \\ -r\sin \theta & r\cos \theta & 0 \end{array}\right| =r\cos \theta \,\mathbf{i} + r\sin \theta \,\mathbf{j}+\left( r\cos ^{2}\theta +r\sin ^{2}\theta \right) \mathbf{k}\\ &=&r\cos \theta \,\mathbf{i}+r\sin \theta \,\mathbf{j}+r\,\mathbf{k} \\[4pt] \hspace{-2pt}\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert &=&\sqrt{r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta +r^{2}}=r\sqrt{\cos ^{2}\theta +\sin ^{2}\theta +1}=r\sqrt{1+1}=\sqrt{2}\,r \end{eqnarray*} \]
Since the mass density of the lamina is \(\rho (x,y,z)=\sqrt{x^{2}+y^{2}}=r\) and \(dS=\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert dr\,d\theta =\sqrt{2}r\,dr\,d\theta ,\) the mass \(M\) of the lamina is given by \[ \begin{eqnarray*} M &=&\iint\limits_{\kern-3ptS}\rho (x,y,z)\,dS=\iint\limits_{\kern-3ptR}\rho \left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert dr\,d\theta =\iint\limits_{\kern-3ptR}r(\sqrt{2}r) dr\,d\theta \\ &=&\sqrt{2}\int_{0}^{2\pi }\int_{1}^{4}r^{2}\,dr\,d\theta =\sqrt{2} \int_{0}^{2\pi }\left[ \dfrac{r^{3}}{3}\right] _{1}^{4}\,d\theta =21\sqrt{2} \int_{0}^{2\pi }d\theta =42\sqrt{2}\,\pi \end{eqnarray*} \]