Triple Integrals in Spherical Coordinates
Solved Problems
Example 3.
Evaluate the integral \[\iiint\limits_U {xyzdxdydz} ,\] where the region \(U\) is a portion of the ball \[{x^2} + {y^2} + {z^2} \le {R^2},\] lying in the first octant \(x \ge 0, y \ge 0, z \ge 0.\)
Solution.
We convert the integral to spherical coordinates. Change the variables:
\[x = \rho \cos \varphi \sin \theta ,\;\; y = \rho \sin \varphi \sin \theta ,\;\; z = \rho \cos \theta ,\;\; dxdydz = {\rho ^2}\sin \theta d\rho d\varphi d\theta .\]
The new variables range within the limits:
\[0 \le \rho \le R,\;\; 0 \le \varphi \le \frac{\pi }{2},\;\; 0 \le \theta \le \frac{\pi }{2}.\]
Then the integral in spherical coordinates becomes
\[
I = \iiint\limits_U {xyzdxdydz}
= \iiint\limits_{U'} {\left[ {\rho \cos \varphi \sin \theta \cdot \rho \sin \varphi \sin \theta \cdot \rho \cos \theta \cdot {\rho ^2}\sin \theta d\rho d\varphi d\theta } \right]}
= \int\limits_0^{\frac{\pi }{2}} {\cos \varphi \sin \varphi d\varphi } \int\limits_0^R {{\rho ^5}d\rho } \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}\theta \cos \theta d\theta }
= \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{1}{2}\sin 2\varphi d\varphi } \right)} \int\limits_0^R {{\rho ^5}d\rho } \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}\theta \cos \theta d\theta }
= \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2\varphi d\varphi } \int\limits_0^R {{\rho ^5}d\rho } \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}\theta d\left( {\sin \theta } \right)}
= \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2\varphi d\varphi } \int\limits_0^R {{\rho ^5}d\rho } \cdot \left[ {\left. {\left( {\frac{{{{\sin }^4}\theta }}{4}} \right)} \right|_{\theta = 0}^{\theta = \frac{\pi }{2}}} \right]
= \frac{1}{8}\int\limits_0^{\frac{\pi }{2}} {\sin 2\varphi d\varphi } \int\limits_0^R {{\rho ^5}d\rho } \cdot \left( {{{\sin }^4}\frac{\pi }{2} - {{\sin }^4}0} \right)
= \frac{1}{8}\int\limits_0^{\frac{\pi }{2}} {\sin 2\varphi d\varphi } \int\limits_0^R {{\rho ^5}d\rho } \cdot 1
= \frac{1}{8}\int\limits_0^{\frac{\pi }{2}} {\sin 2\varphi d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^6}}}{6}} \right)} \right|_0^R} \right]
= \frac{{{R^6}}}{{48}}\left[ {\left. {\left( { - \frac{{\cos 2\varphi }}{2}} \right)} \right|_0^{\frac{\pi }{2}}} \right]
= \frac{{{R^6}}}{{96}}\left( { - \cos \pi + \cos 0} \right)
= \frac{{{R^6}}}{{96}} \cdot 2
= \frac{{{R^6}}}{{48}}.\]
Example 4.
Find the triple integral
\[\iiint\limits_U {\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}}} \right)dxdydz},\]
where the region \(U\) is bounded by the ellipsoid \[{\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}}} = 1.\]
Solution.
To calculate the integral we use generalized spherical coordinates by making the following change of variables:
\[x = a\rho \cos \varphi \sin \theta ,\;\; y = b\rho \sin \varphi \sin \theta ,\;\; z = c\rho \cos \theta .\]
The absolute value of the Jacobian of the transformation is \(\left| I \right| = abc{\rho ^2}\sin \theta .\) Therefore, the following relation is valid for the differentials:
\[dxdydz = abc{\rho ^2}\sin \theta d\rho d\varphi d\theta .\]
The integral in the new coordinates becomes
\[
I = \iiint\limits_U {\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}}} \right)dxdydz}
= \iiint\limits_{U'} {\left[ {\frac{{{{\left( {a\rho \cos \varphi \sin \theta } \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {b\rho \sin \varphi \sin \theta } \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {c\rho \cos \theta } \right)}^2}}}{{{c^2}}}} \right]abc{\rho ^2}\sin \theta d\rho d\varphi d\theta }
= \iiint\limits_{U'} {\left[ {{\rho ^2}{{\cos }^2}\varphi \,{{\sin }^2}\theta + {\rho ^2}{\sin^2}\varphi \,{{\sin }^2}\theta + {\rho ^2}{{\cos }^2}\theta } \right]abc{\rho ^2}\sin \theta d\rho d\varphi d\theta }
= \iiint\limits_{U'} {\Big[ {{\rho ^2}{{\sin }^2}\theta \underbrace {\left( {{{\cos }^2}\varphi + {\sin^2}\varphi } \right)}_1 + {\rho ^2}{{\cos }^2}\theta } \Big] abc{\rho ^2}\sin \theta d\rho d\varphi d\theta }
= \iiint\limits_{U'} {{\rho ^2}\underbrace {\left( {{\sin^2}\theta + {{\cos }^2}\theta } \right)}_1 }\cdot { abc{\rho ^2}\sin \theta d\rho d\varphi d\theta }
= abc\iiint\limits_{U'} {{\rho ^4}\sin \theta d\rho d\varphi d\theta } .\]
The region of integration \(U'\) in spherical coordinates is a rectangular parallelepiped and defined by the inequalities
\[0 \le \rho \le 1,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .\]
Then the triple integral can be written as
\[I = abc\iiint\limits_{U'} {{\rho ^4}\sin \theta d\rho d\varphi d\theta } = abc\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}d\rho } \int\limits_0^\pi {\sin \theta d\theta } = abc\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}d\rho } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] = abc\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}d\rho } \cdot \left( { - \cos \pi + \cos 0} \right) = 2abc\int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}d\rho } = 2abc\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^5}}}{5}} \right)} \right|_0^1} \right] = \frac{{2abc}}{5}\int\limits_0^{2\pi } {d\varphi } = \frac{{2abc}}{5} \cdot \left[ {\left. \varphi \right|_0^{2\pi }} \right] = \frac{{2abc}}{5} \cdot 2\pi = \frac{{4abc\pi }}{5}.\]
Example 5.
Evaluate the integral
\[\int\limits_0^1 {dx} \int\limits_0^{\sqrt {1 - {x^2}} } {dy} \int\limits_0^{\sqrt {1 - {x^2} - {y^2}} } {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}dz},\]
using spherical coordinates.
Solution.
The region of integration is a portion of the ball lying in the first octant (Figures \(2,3\)) and, hence, it is bounded by the inequalities
\[0 \le \rho \le 1,\;\; 0 \le \varphi \le \frac{\pi }{2},\;\; 0 \le \theta \le \frac{\pi }{2}.\]
Taking into account that the integrand is
\[{\left( {{x^2} + {y^2} + {z^2}} \right)^2} = {\left[ {{{\left( {\rho \cos \varphi \sin \theta } \right)}^2} + {{\left( {\rho \sin \varphi \sin \theta } \right)}^2} + {{\left( {\rho \cos \theta } \right)}^2}} \right]^2} = {\left[ {{\rho ^2}{{\sin }^2}\theta \underbrace {\left( {{{\cos }^2}\varphi + {{\sin }^2}\varphi } \right)}_1 + {\rho ^2}{{\cos }^2}\theta } \right]^2} = {\left[ {{\rho ^2}{{\sin }^2}\theta + {\rho ^2}{{\cos }^2}\theta } \right]^2} = {\left[ {{\rho ^2}\underbrace {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}_1} \right]^2} = {\rho ^4},\]
and the differential is expressed by the formula
\[dxdydz = {\rho ^2}\sin \theta d\rho d\varphi d\theta ,\]
we have
\[I = \int\limits_0^1 {dx} \int\limits_0^{\sqrt {1 - {x^2}} } {dy} \int\limits_0^{\sqrt {1 - {x^2} - {y^2}} } {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^2}dz} = \int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {\left( {{\rho ^4} \cdot {\rho ^2}d\rho } \right)} \int\limits_0^{\frac{\pi }{2}} {\sin \theta d\theta } = \int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^6}d\rho } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^{\frac{\pi }{2}}} \right] = \int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^6}d\rho } \cdot \left( { - \cos \frac{\pi }{2} + \cos 0} \right) = \int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^1 {{\rho ^6}d\rho } \cdot 1 = \int\limits_0^{\frac{\pi }{2}} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^7}}}{7}} \right)} \right|_0^1} \right] = \frac{1}{7}\int\limits_0^{\frac{\pi }{2}} {d\varphi } = \frac{1}{7} \cdot \left[ {\left. \varphi \right|_0^{\frac{\pi }{2}}} \right] = \frac{1}{7} \cdot \frac{\pi }{2} = \frac{\pi }{{14}}.\]