Question

Find the volume of the solid that is enclosed by the cone \({\rm{z = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \)and the sphere \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 2}}\).

Short answer

Respectfully integrate \(\frac{{{\rm{1688\pi }}}}{{{\rm{15}}}}.\)

Step-by-step solution

Step1:The spherical coordinates triple integration formula.

\(_E\mathop f\limits_d (x,y,z) = {\text{ }}_{ca}^\alpha f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi ){\left( \rho \right)^2}\sin \phi d\rho d\theta d\phi \)

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

Define the integration region in the following manner using the supplied integral.

\(\begin{aligned}\int_{\rm{0}}^{\rm{\pi }} {\int_{\rm{0}}^{{\rm{2\pi }}} {\;\;\;} } {\rm{3(\rho sin}}\phi {\rm{cos\theta }}{{\rm{)}}^{\rm{2}}}{\rm{ + (\rho sin}}\phi {\rm{sin\theta }}{{\rm{)}}^{\rm{2}}}{{\rm{\rho }}^{\rm{2}}}{\rm{sin}}\phi {\rm{d\rho d\theta d}}\phi \rm &= \rho _{\rm{0}}^{\rm{2}}{\rm{si}}{{\rm{n}}^{\rm{2}}}\phi {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta + si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}\;\;\;{{\rm{\rho }}^{\rm{2}}}{\rm{sin}}\phi {\rm{d\rho d\theta d}}\phi \\&\begin{array}{*{20}{r}}{\rm{0}}&{\rm{0}}&{\rm{2}}\\{\rm{\pi }}&{{\rm{2\pi }}}&{\rm{3}}\end{array}\\\rm &= _{\rm{0}}{\rm{0}}_{{\rm{2\pi }}}^{\rm{2}}{{\rm{\rho }}^{\rm{2}}}{\rm{si}}{{\rm{n}}^{\rm{2}}}\phi {\rm{ * }}{{\rm{\rho }}^{\rm{2}}}{\rm{sin}}\phi {\rm{d\rho d\theta d}}\phi \\\rm &= {\int_{\rm{0}}^{\rm{\pi }} {\;\;\;} _{\rm{2}}}\;\;\;{{\rm{\rho }}^{\rm{4}}}{\rm{si}}{{\rm{n}}^{\rm{3}}}\phi {\rm{d\rho d\theta d}}\phi {\rm{.}}\end{aligned}\)

Step2:Can now assess the integral.

\(\begin{aligned}&= \int_0^\pi {_0^{2\pi }} \frac{{{3^5}}}{5} - \frac{{{2^5}}}{5}{\sin ^3}fd\theta df \\&= \frac{{211}}{5}\theta _0^{2\pi }{}_0^\pi {\sin ^3}fdf \\ &= \frac{{422\pi }}{5}\int_0^\pi {{{\sin }^3}} fdf. \\\end{aligned} \)

Taking Advantage of Trigonometric Identities

\(\frac{{{\rm{422\pi }}}}{{\rm{5}}}\int_{\rm{0}}^{\rm{\pi }} {{\rm{si}}{{\rm{n}}^{\rm{3}}}} \phi {\rm{d}}\phi {\rm{ = }}\frac{{{\rm{422\pi }}}}{{\rm{5}}}\int_{\rm{0}}^{\rm{\pi }} {{\rm{sin}}} \phi {\rm{1 - co}}{{\rm{s}}^{\rm{2}}}\phi \;\;{\rm{d}}\phi {\rm{.}}\)

\(\begin{aligned}\frac{{{\rm{422\pi }}}}{{\rm{5}}}_{\rm{1}}^{{\rm{ - 1}}}{\rm{ - 1 - }}{{\rm{k}}^{\rm{2}}}\;\;\;\rm dk &= \frac{{{\rm{422\pi }}}}{{\rm{5}}}\;\;\;{\rm{1 - }}{{\rm{k}}^{\rm{2}}}{\rm{dk}}\\\rm &= \frac{{{\rm{422\pi }}}}{{\rm{5}}}\;\;\;{\rm{k - }}\frac{{{{\rm{k}}^{\rm{3}}}}}{{\rm{3}}}\;\;\;\frac{{\rm{1}}}{{{\rm{ - 1}}}}\\\rm &= \frac{{{\rm{422\pi }}}}{{\rm{5}}}\;\;\;{\rm{1 - ( - 1) - }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ - }}\;\;\;{\rm{ - }}\frac{{\rm{1}}}{{\rm{3}}}\\\rm &= \frac{{{\rm{422\pi }}}}{{\rm{5}}}\;\;\;{\rm{2 - }}\frac{{\rm{2}}}{{\rm{3}}}\\\rm &= \frac{{{\rm{1688\pi }}}}{{{\rm{15}}}}\end{aligned}\)

Consequently, the integral becomes \(\frac{{{\rm{1688\pi }}}}{{{\rm{15}}}}.\)\(\)