Question Number 197874 by yaslm last updated on 01/Oct/23
Answered by aleks041103 last updated on 02/Oct/23
$${v}:\begin{cases}{\mathrm{0}\leqslant{z}\leqslant\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\\{\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} }\end{cases} \\ $$$${in}\:{cylimdrical}\:{coordinates} \\ $$$${v}:\begin{cases}{\mathrm{0}\leqslant{z}\leqslant{r}}\\{\mathrm{0}\leqslant{r}\leqslant{a}}\end{cases} \\ $$$$\Rightarrow{V}=\underset{{v}} {\int\int\int}{dxdydz}=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:{a}} \int_{\mathrm{0}} ^{\:{r}} {det}\left({J}\right){dzdrd}\theta \\ $$$${the}\:{metric}\:{is}: \\ $$$${dl}^{\mathrm{2}} ={r}^{\mathrm{2}} {d}\theta^{\mathrm{2}} +{dr}^{\mathrm{2}} +{dz}^{\mathrm{2}} \Rightarrow{g}={diag}\left(\mathrm{1},{r}^{\mathrm{2}} ,\mathrm{1}\right) \\ $$$$\Rightarrow{det}\left({J}\right)=\sqrt{{det}\left({g}\right)}=\sqrt{\mathrm{1}.\mathrm{1}.{r}^{\mathrm{2}} }={r} \\ $$$$\Rightarrow{V}=\left(\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {d}\theta\right)\left(\int_{\mathrm{0}} ^{\:{a}} \left(\int_{\mathrm{0}} ^{\:{r}} {dz}\right){rdr}\right)= \\ $$$$=\mathrm{2}\pi\int_{\mathrm{0}} ^{\:{a}} {r}^{\mathrm{2}} {dr}=\frac{\mathrm{2}}{\mathrm{3}}\pi{a}^{\mathrm{3}} \\ $$$$\Rightarrow{V}=\frac{\mathrm{2}}{\mathrm{3}}\pi{a}^{\mathrm{3}} \\ $$