MvCalc22

From Example Problems
Jump to: navigation, search

Changing into polars,we get r=a\, from \theta =0\, to \theta ={\frac  {\pi }{2}}\,,hence the given integral becomes

\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{0}^{a}r\cos \theta (r\sin \theta )r^{n}(rdr)d\theta \,

=\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{0}^{a}r^{{n+3}}\cos \theta \sin \theta d\theta \,

=\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin \theta \cos \theta d\theta \int _{0}^{a}r^{{n+3}}dr\,

=[{\frac  {\sin ^{2}\theta }{2}}]_{{0}}^{{{\frac  {\pi }{2}}}}[{\frac  {r^{{n+4}}}{n+4}}]_{0}^{a}\, [Since n+3>0]

={\frac  {1}{2}}{\frac  {a^{{n+4}}}{n+4}}={\frac  {a^{{n+4}}}{2(n+4)}}\,

Main Page