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# Calculate.$\int_0^{\pi / 2} \int_0^{3 \sin \theta} r^2 d r d \theta$

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$\begin{gathered} \int_0^{\pi /2} {\int_0^{3sin\theta } {{r^2}} drd\theta } \\ \textcolor{#4257b2}{ {\text{Integrating with respect to }}r} \\ \int_0^{\pi /2} {\left[ {\frac{{{r^3}}}{3}} \right]_0^{3\sin \theta }d} \theta \\ \textcolor{#4257b2}{{\text{Evaluate}}{\text{, use The Fundamental Theorem of Integral Calculus}}} \\ \int_0^{\pi /2} {\left[ {\frac{{{{\left( {3\sin \theta } \right)}^3}}}{3} - \frac{{{{\left( 0 \right)}^2}}}{2}} \right]d\theta } \\ \int_0^{\pi /2} {9{{\sin }^3}\theta d\theta } \\ 9\int_0^{\pi /2} {{{\sin }^2}\theta \sin \theta d\theta } \\ \textcolor{#4257b2}{ {\text{Apply the pythagorean identity }}{\cos ^2}x + {\sin ^2}x = 1} \\ 9\int_0^{\pi /2} {\left( {1 - {{\cos }^2}\theta } \right)\sin \theta d\theta } \\ 9\int_0^{\pi /2} {\left( {\sin \theta - {{\cos }^2}\theta \sin \theta } \right)d\theta } \\ \textcolor{#4257b2}{{\text{Integrate and evaluate}}} \\ 9\int_0^{\pi /2} {\left( {\sin \theta - {{\cos }^2}\theta \sin \theta } \right)d\theta } = 9\left[ { - \cos \theta + \frac{{{{\cos }^3}\theta }}{3}} \right]_0^{\pi /2} \\ 9\left[ {\theta + \frac{{{{\cos }^3}\theta }}{2}} \right]_0^{\pi /2} = 9\left[ { - \cos \left( {\frac{\pi }{2}} \right) + \frac{{{{\cos }^3}\left( {\pi /2} \right)}}{3}} \right] - 9\left[ { - \cos \left( 0 \right) + \frac{{{{\cos }^3}0}}{3}} \right] \\ \textcolor{#4257b2}{{\text{Simplify}}} \\ 9\left[ {\theta + \frac{{{{\cos }^3}\theta }}{2}} \right]_0^{\pi /2} = 9\left[ {0 + \frac{0}{2}} \right] - 9\left[ { - 1 + \frac{1}{3}} \right] \\ 9\left[ {\theta + \frac{{{{\cos }^3}\theta }}{2}} \right]_0^{\pi /2} = 6 \\ \end{gathered}$

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