Correct Answer - Option 4 :
\(\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)\)
\(\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2} \sin \theta drd\theta d\phi \)
= \( \mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{\pi }{4}} \mathop \smallint \limits_0^1 \left( {{r^2} \times dr} \right) \times \sin \theta d\theta d\phi \)
= \(\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{\pi }{4}} \left[ {\frac{{{r^3}}}{3}} \right]_0^1 \times \sin \theta d\theta d\phi \)
= \( \frac{1}{3}\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{\pi }{4}} \sin \theta \times d\theta \times d\phi \)
= \(\frac{1}{3}\mathop \smallint \limits_0^{2\pi } - \left[ {\cos \theta } \right]_0^{\frac{\pi }{4}} \times d\phi \)
= \( \frac{1}{3}\mathop \smallint \limits_0^{2\pi } - \left[ {\frac{1}{{\sqrt 2 }} - 1} \right] \times d\phi \)
= \(\frac{1}{{3\sqrt 2 }}\left[ {\sqrt 2 \phi - \phi } \right]_0^{2\pi }\)
= \(\frac{{2\pi }}{{3\sqrt 2 }}\left[ {\sqrt 2 - 1} \right]\)
= \(\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)\)