Verify pulling out the partial derivative.

[yungman]

For spherical coordinates, $u(r,\theta,\phi)$ is function of $r,\theta,\phi$. $a$ is constant and is the radius of the spherical region. Is:

$$\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\partial\;u(r,\theta,\phi)}{\partial {r}}a^2\sin\theta d\theta d\phi=\frac{\partial}{\partial {r}}\left[\int_{0}^{2\pi}\int_{0}^{\pi}u(r,\theta,\phi)a^2\sin\theta d\theta d\phi\right]$$

Thanks

 

[fzero]

This is true as long as $\partial u/\partial r$ exists and is continuous on the appropriate domain. This is known as the Leibniz Integral Rule and a proof is given at the link.