Question

Evaluate \(\iint_{\text{E}}{\text{x}}\text{yzdV}\),where E lies between the spheres \({\rm{\rho = 2}}\) and \({\rm{\rho = 4}}\)and above the cone \(\phi {\rm{ = \pi /3}}\).

Short answer

The evaluation of \(\iint_{\text{E}}{\text{x}}\text{yzdV}\), is zero.

Step-by-step solution

Setting spherical coordinates.

\(\int_{\rm{0}}^{\frac{{\rm{\pi }}}{{\rm{3}}}} {\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{2}}^{\rm{4}} {{{\rm{\rho }}^{\rm{5}}}} } } {\rm{si}}{{\rm{n}}^{\rm{3}}}{\rm{fco}}{{\rm{s}}^{\rm{2}}}{\rm{\theta sin\theta d\rho d\theta d}}\phi \)

Splitting into individual integrals.

Notice that the entire expression is product so split into individual integrals

\(\int_{\rm{0}}^{\frac{{\rm{\pi }}}{{\rm{3}}}} {{\rm{si}}{{\rm{n}}^{\rm{3}}}} \phi {\rm{cos}}\phi {\rm{d}}\phi \int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{sin}}} {\rm{\theta cos\theta d\theta }}\int_{\rm{2}}^{\rm{4}} {{{\rm{\rho }}^{\rm{5}}}} {\rm{d\rho = 0}}\)

Therefore, the evaluation of \(\iint_{\text{E}}{\text{x}}\text{yzdV}\), is zero since \({\rm{sin\theta cos\theta = }}\frac{{{\rm{sin2\theta }}}}{{\rm{2}}}\).