Compito

(1)

\begin{align} $\int_{\pi \over 4}^{5 \pi \over 4} \int_{0 }^{2} \rho^2 \cos \theta (\rho \sin \theta -1) d \rho d \theta =$ $={1 \over 2 }\int_{\pi \over 4}^{5 \pi \over 4} \sin(2 \theta) \, 4 \, d \theta - {8 \over 3} \int_{\pi \over 4}^{5 \pi \over 4} \cos ( \theta) 4 d \theta = $ $=[- \cos 2 \theta] _{5{ \pi} \over 4}^{\pi \over 4} - {8 \over 3 }[- \sin \theta]_ {5{ \pi} \over 4}^{\pi \over 4} $ $=- {8 \over 3} (\sin {5 \over 4 \pi }- \sin {\pi \over 4}) =$ $=( - {\sqrt 2 \over 2 }- {\sqrt 2 \over 2 }) (- {8 \over 3})$ \end{align}

page revision: 1, last edited: 03 Sep 2009 13:23