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[itex] \int_{\vartheta=0}^{2\pi} \int_{

\varphi=0}^{\pi/2}\int_{\rho=0}^{R} 5Kr^2 (r^2 sin(\phi) d\rho d\phi d\theta [/itex]

which will then yield [itex] 2K\pi R^5 [/itex]

Here is the surface integral.

[itex] \oint\oint E * da [/itex]

where da will yield (4\pi R^2)

multiplying this times KR^3 gives us [itex] 4K\pi R^5 [/itex]

Any help on what to do here will be greatly appreciated.

THank you