Compute the derivative $ \frac{d}{dR}\iiint_{\{(x,y,z)\in\textbf{R}^3: \sqrt{x^2+y^2+z^2} \leq R\}}f(x,y,z)\,dx\,dy\,dz. $ Let the function f and its first-order partial derivatives be continuous in $\textbf{R}^3$.  Suppose that 
$$ 
\iiint_{\textbf{R}^3}|f(x,y,z)|\,dx\,dy\,dz < \infty.
$$
Compute the derivative 
$$
\frac{d}{dR}\iiint_{\{(x,y,z)\in\textbf{R}^3: 
\sqrt{x^2+y^2+z^2} \leq R\}}f(x,y,z)\,dx\,dy\,dz.
$$
with the derivative given in terms of a surface integral.
Converting to spherical coordinates I found that
$$
\frac{d}{dR}\iiint_D f(x,y,z)\,dx\,dy\,dz =
\frac{d}{dR} \int_0^{2\pi} \int_0^\pi \int_0^R f(r,\theta,\phi)\;r^2 \sin(\phi)\,dr\,d\phi\,d\theta.
$$
My next step was
$$
\frac{d}{dR} \int_0^{2\pi} \int_0^\pi \int_0^R f(r,\theta,\phi)\;r^2 \sin(\phi)\,dr\,d\phi\,d\theta=
 \int_0^{2\pi} \int_0^\pi \frac{d}{dR}\int_0^R f(r,\theta,\phi)\;r^2 \sin(\phi)\,dr\,d\phi\,d\theta.
$$
In general, are you allowed to move the derivative through the  outside integrals?  Why?  My final step was
$$\int_0^{2\pi} \int_0^\pi \frac{d}{dR}\int_0^R f(r,\theta,\phi)\;r^2 \sin(\phi)\,dr\,d\phi\,d\theta=
\int_0^{2\pi} \int_0^\pi f(R,\theta,\phi)R^2 \sin(\phi)\,d\phi\,d\theta=
\iint_{S(R)}f(R,\theta,\phi)R^2\;dS \;\;\;\text{ where }\;\;\; dS=\sin(\phi)\,d\phi\,d\theta,\;\; S(R) \text{ is the sphere of radius R.}
$$
Is this correct?  Is there a better way of doing this problem?

In general moving the derivative in or outside an integral is not mathematically valid, but in your case you can. It makes no difference since the integrals with respect to $\phi$ and $\theta$ are not dependant on $R$, thus differentiating the results of these integrals with respect to $R$ will yield $0$.
