The value of $\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2}\sin \theta dr\;d\theta \;d\phi $ will be:

Question

The value of $\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2}\sin \theta dr\;d\theta \;d\phi $ will be:

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 + \sqrt 3 } \right)$

TuteeHUB · Answer

$\frac{{{2\pi } }}{3}\left( {\sqrt 3 -1 } \right)$

TuteeHUB · Answer

$\frac{{ {2\pi } }}{3}\left( {\sqrt 2 - \sqrt 3 } \right)$

TuteeHUB · Answer

$\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)$

1.	The value of \(\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2}\sin \theta dr\;d\theta \;d\phi \) will be:
A.	\(\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 + \sqrt 3 } \right)\)
B.	\(\frac{{{2\pi } }}{3}\left( {\sqrt 3 -1 } \right)\)
C.	\(\frac{{ {2\pi } }}{3}\left( {\sqrt 2 - \sqrt 3 } \right)\)
D.	\(\frac{{\sqrt 2{\pi } }}{3}\left( {\sqrt 2 -1 } \right)\)
Answer» E.

The value of \(\mathop \smallint \nolimits_0^{2\pi } \mathop \smallint \nolimits_0^{\pi /4} \mathop \smallint \nolimits_0^1 {r^2}\sin \theta dr\;d\theta \;d\phi \) will be:

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