Determine el valor promedio de la función sobre la región solida

Veamos... El volumen de V se calcula con la fórmula de la esfera:

$$\begin{align}&V=\frac{4\pi}{3}(3^3-2^3)=\frac{76\pi}{3}\end{align}$$

Luego utilizamos las coordenadas esféricas

$$\begin{align}&x=\rho\sin\phi\cos \theta\\&y=\rho\sin\phi\sin \theta\\&z=\rho\cos\phi\\&\\&\text{Donde: }\theta\in[0,2\pi),\phi\in[0,\pi],\rho\in[2,3]\end{align}$$

Así

$$\begin{align}&\iiint_{Q}(x^2+y^2+z^2)^{-3/2}dV=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{2}^{3}\dfrac{1}{\rho^3}\cdot \rho^2\sin\phi~d\rho~d\phi~ d\theta\\&\\&\iiint_{Q}(x^2+y^2+z^2)^{-3/2}dV=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{2}^{3}\dfrac{1}{\rho}\sin\phi~d\rho~d\phi~ d\theta\\&\\&\iiint_{Q}(x^2+y^2+z^2)^{-3/2}dV=\int_{0}^{2\pi} d\theta\int_{0}^{\pi}\sin\phi~d\phi\int_{2}^{3}\dfrac{1}{\rho}~d\rho\\&\\&\iiint_{Q}(x^2+y^2+z^2)^{-3/2}dV=2\pi\int_{0}^{\pi}\sin\phi~d\phi\int_{2}^{3}\dfrac{1}{\rho}~d\rho\\&\\&\iiint_{Q}(x^2+y^2+z^2)^{-3/2}dV=4\pi\int_{2}^{3}\dfrac{1}{\rho}~d\rho\\&\\&\iiint_{Q}(x^2+y^2+z^2)^{-3/2}dV=4\pi\ln\dfrac{3}{2}\end{align}$$

Finalmente...

$$\begin{align}&\text{Prom }[f]=\dfrac{4\pi\ln\dfrac{3}{2}}{\dfrac{76\pi}{3}}\\&\\&\\&\boxed{\text{Prom }[f]=\dfrac{3}{19}\ln\dfrac{3}{2}}\end{align}$$