MvCalc59

From Example Problems
Jump to: navigation, search

Taking r,theta and z as cylindrical coordinates,the equation of the cone may be written as {\frac  {z}{r}}={\frac  {h}{a}}\,

Hence,its volume is given by V=\int _{{0}}^{{2\pi }}\int _{0}^{a}\int _{{z={\frac  {rh}{a}}}}^{{h}}rdrd\theta dz\,

=\int _{{0}}^{{2\pi }}\int _{0}^{a}[h-{\frac  {rh}{a}}]rdrd\theta \,

=h\int _{{0}}^{{2\pi }}[{\frac  {r^{2}}{2}}-{\frac  {r^{3}}{3a}}]_{0}^{a}d\theta \,

=h(2\pi )[{\frac  {a^{2}}{2}}-{\frac  {a^{2}}{3}}]\,

=2\pi h{\frac  {1}{6}}a^{2}\,

={\frac  {1}{3}}\pi a^{2}h\,

Main Page