Refer to the diagram shown below.
In spherical coordinates,
r² = x² + y² + z²
x = r sinφ cos θ
y = r sinφ sin θ
z = r cosφ
The element of volume is
dV = r² sinφ dr dθ dφ
Therefore, for the sphere with radius = 3,
[tex]\int (x^{2}+y^{2}+z^{2})^{2} dV = \int_{0}^{ \pi } d\phi \int_{0}^{2 \pi }d\theta \int_{0}^{3} (r^{4})r^{2} sin\phi \,dr [/tex]
The integration yields
[tex]2 \pi \int_{0}^{ \pi } sin \phi \, d\phi \, [ \frac{r^{7}}{7} ]_{0}^{3} = 2 \pi [-cos\phi]_{0 }^{ \pi } (312.429) = 2 \pi (2)(312.429) = 1250 \pi [/tex]
Answer: 1250π
