Answer :
The Triple Intergral of the given expression is [tex]4 \pi \times ln (\frac{a}{b})[/tex]
[tex]\int\limits \int\limits \int\limits_w \frac{dx dy dz}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}[/tex]
[tex]\text{Where } x^2 + y^2 + z^2 = a^2 \text{ and } x^2 + y^2 + z^2 = b^2, 0 < b < a[/tex]
Changing to Spherical Coordinates
[tex]b \leq \rho \leq a, 0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi[/tex]
[tex]\therefore x^2 + y^2 + z^2 = \rho^2; dx dy dz = \rho^2 sin \phi .d\phi .d\theta. d\rho[/tex]
[tex]\implies \int\limits^a_b \int\limits^{2\pi}_{0} \int\limits^{\pi}_{0} \frac{1}{\rho^3} \rho^2 sin \phi .d\phi .d\theta. d\rho[/tex]
[tex]=\int\limits^a_b {\frac{1}{\rho}} \, d\rho \int\limits^0_{2\pi} d\theta \int\limits^{\pi}_0 \sin{\phi} \, d\phi[/tex]
[tex]= [ln \rho]^{a}_{b} [\theta]^{2\pi}_{0} [-\cos \phi]^{\pi}_{0}[/tex]
[tex]= [\ln a - \ln b] \times [2\pi] \times [- \cos \pi - (-\cos 0)][/tex]
[tex]= 2\pi \times 2 \times (\ln a - \ln b)[/tex]
[tex]= 4 \pi \, \ln{\frac{a}{b}}[/tex]
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