Question

Evaluate \(\iiint_{\text{B}}{{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}} \right)}^{\text{2}}}}\)dV, Where B is the ball with center the origin and radius 5

Short answer

The value of the given triple integral is \(\frac{{{\rm{312500\pi }}}}{{\rm{7}}}\).

Step-by-step solution

Explanation of solution

The formula for triple integration in spherical coordinates:

\(\iiint_{E}{f}(x,y,z)=\int_{c}^{d}{\int_{\alpha }^{\beta }{\int_{a}^{b}{f}}}(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi ){{\rho }^{2}}\sin \phi d\rho d\theta d\phi \).

E is a spherical wedge indicated by

\(E = \{ (\rho ,\theta ,\phi )\mid a \le \rho \le b,\alpha \le \theta \le \beta ,c \le \phi \le d\} \).

The region of integration can be defined in the following way using the given integral:

\({\rm{B = \{ (\rho ,}}\theta ,\phi )\mid 0 \le \rho \le {\rm{5,0}} \le \theta \le {\rm{2\pi ,0}} \le \phi \le {\rm{\pi \} }}\)

Using \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = }}{{\rm{\rho }}^{\rm{2}}}\)

\(\int_0^\pi {\int_0^{2\pi } {\int_0^5 {{{\left( {{\rho ^2}} \right)}^2}} } } {\rho ^2}\sin \phi d\rho d\theta d\phi = \int_0^\pi {\int_0^{2\pi } {\int_0^5 {{\rho ^6}} } } \sin \phi d\rho d\theta d\phi \).

Calculation

\(\begin{aligned}\int_0^\pi {\int_0^{2\pi } {\int_0^5 {{\rho ^6}} } } \sin \phi d\rho d\theta d\phi &= \left. {\int_0^\pi {\int_0^{2\pi } {\left( {\frac{{{\rho ^7}}}{7}\sin \phi } \right)} } } \right|_0^5d\theta d\phi \\ &= \int_0^\pi {\int_0^{2\pi } {\frac{{{5^7}}}{7}} } \sin \phi d\theta d\phi \end{aligned}\)

\(\begin{aligned}\int_0^\pi {\int_0^{2\pi } {\int_0^5 {{\rho ^6}} } } \sin \phi d\rho d\theta d\phi \rm &= \left. {\frac{{{\rm{78125}}}}{{\rm{7}}}{\rm{\theta }}} \right|_{\rm{0}}^{{\rm{2\pi }}}\int_{\rm{0}}^{\rm{\pi }} {{\rm{sin}}} \phi {\rm{d}}\phi \\ \rm &= \frac{{{\rm{156250\pi }}}}{{\rm{7}}}\int_{\rm{0}}^{\rm{\pi }} {{\rm{sin}}} \phi {\rm{d}}\phi \\ \rm &= \left. {\frac{{{\rm{156250\pi }}}}{{\rm{7}}}{\rm{( - cos}}\phi {\rm{)}}} \right|_{\rm{0}}^{\rm{\pi }}\\ \rm &= \frac{{{\rm{156250\pi }}}}{{\rm{7}}}{\rm{( - cos\pi + cos0)}}\end{aligned}\)

\(\begin{aligned}\int_0^\pi {\int_0^{2\pi } {\int_0^5 {{\rho ^6}} } } \sin \phi d\rho d\theta d\phi \rm &= \frac{{{\rm{156250\pi }}}}{{\rm{7}}}{\rm{*2}}\\ \rm &= \frac{{{\rm{312500\pi }}}}{{\rm{7}}}\end{aligned}\)

Hence, the value of the given triple integral is \(\frac{{{\rm{312500\pi }}}}{{\rm{7}}}\).