Prove that $\frac{d}{dr}\frac{1}{4 \pi r^2} \iint_{S_r} u\ dS = \frac{1}{4 \pi r^2} \iint_{S_r} \nabla u \cdot d\vec{S}$

Question

How can I show that if $u$ is a $C^1$ function and $S_r = \{(x, y, z) : x^2 + y^2 + z^2 = r\}$, then for all $r$:

$$\frac{d}{dr}\frac{1}{4 \pi r^2} \iint_{S_r} u\ dS = \frac{1}{4 \pi r^2} \iint_{S_r} \nabla u \cdot d\vec{S}$$

Answer 1

Get the "$r$" out of the limits of integration. With the substitution $rz = y$ you have $r^2 dS(u) = dS(y)$ so that $$\frac 1{4\pi r^2} \iint_{S_r} u(y) \, dS(y) = \frac 1{4\pi} \iint_{S_1} u(rz) \, dS(z) = \frac 1{4\pi} \iint_{S_1} u(rz_1,rz_2) \, dS(z).$$ The derivative is calculated using the chain rule: $$\frac{d}{dr} u(rz_1,rz_2) = \frac{\partial u}{\partial z_1}u(rz_1,rz_2) z_1 + \frac{\partial u}{\partial z_2} u(rz_1,rz_2) z_2 = \nabla u(rz_1,rz_2) \cdot (z_1,z_2) = \nabla u(rz) \cdot z.$$ Re-apply the substitution $rz = y$ to obtain $$ \frac{d}{dr} \frac 1{4\pi} \iint_{S_1} u(rz_1,rz_2) \, dS(z) = \frac 1{4\pi} \iint_{S_1} \nabla u(rz) \cdot z \, dS(z) = \frac 1{4\pi r^2} \iint_{S_r} \nabla u(y) \cdot \frac yr \, dS(y)$$

Finally observe that the exterior unit normal vector to the sphere $S_r$ at a point $y \in S_r$ is just $\dfrac yr$.