Klotets massa ges av integralen

M = \iiint_ {K} \rho (x, y, z) d x d y d z = \iiint_ {K} \frac {\rho_ {0} c _ {0} z}{\left(1 + c _ {0} ^ {2} \left(x ^ {2} + y ^ {2} + z ^ {2}\right)\right) ^ {3 / 2}} d x d y d z.

För att beräkna denna byter vi till rymdpolära koordinater. Då är $z = r\cos \theta$ , $x^{2} + y^{2} + z^{2} = r^{2}$ , $0\leq \phi \leq 2\pi$ , $0\leq \theta \leq \pi /2$ , $0\leq r\leq R$ , och Jacobianen är $r^2\sin \theta$ . Vi får

\begin{array}{l} M = \int_ {0} ^ {2 \pi} \left(\int_ {0} ^ {\pi / 2} \left(\int_ {0} ^ {R} \frac {\rho_ {0} c _ {0} r \cos \theta}{\left(1 + c _ {0} ^ {2} r ^ {2}\right) ^ {3 / 2}} r ^ {2} \sin \theta d r\right) d \theta\right) d \phi \\ = 2 \pi \rho_ {0} c _ {0} \left(\int_ {0} ^ {\pi / 2} \cos \theta \sin \theta d \theta\right) \left(\int_ {0} ^ {R} \frac {r ^ {3}}{\left(1 + c _ {0} ^ {2} r ^ {2}\right) ^ {3 / 2}} d r\right). \\ \end{array}

Här är

\int_ {0} ^ {\pi / 2} \cos \theta \sin \theta d \theta = \frac {1}{2} \int_ {0} ^ {\pi / 2} \sin 2 \theta d \theta = \frac {1}{2} \left[ - \frac {1}{2} \cos 2 \theta \right] _ {0} ^ {\pi / 2} = \frac {1}{2}.

Variabelbytet $1 + c_0^2 r^2 = t$ ger $r^2 = c_0^{-2}(t - 1)$ och $rdr = 2^{-1}c_0^{-2}dt$ , så

\begin{array}{l} \int_ {0} ^ {R} \frac {r ^ {3}}{\left(1 + c _ {0} ^ {2} r ^ {2}\right) ^ {3 / 2}} d r = \int_ {1} ^ {1 + c _ {0} ^ {2} R ^ {2}} \frac {c _ {0} ^ {- 2} (t - 1)}{t ^ {3 / 2}} 2 ^ {- 1} c _ {0} ^ {- 2} d t \\ = \frac {1}{2 c _ {0} ^ {4}} \int_ {1} ^ {1 + c _ {0} ^ {2} R ^ {2}} \left(\frac {1}{t ^ {1 / 2}} - \frac {1}{t ^ {3 / 2}}\right) d t \\ = \frac {1}{2 c _ {0} ^ {4}} \left[ \frac {t ^ {1 / 2}}{1 / 2} - \frac {t ^ {- 1 / 2}}{- 1 / 2} \right] _ {1} ^ {1 + c _ {0} ^ {2} R ^ {2}} \\ = \frac {1}{c _ {0} ^ {4}} \left[ t ^ {1 / 2} + t ^ {- 1 / 2} \right] _ {1} ^ {1 + c _ {0} ^ {2} R ^ {2}} \\ = \frac {1}{c _ {0} ^ {4}} \left(\sqrt {1 + c _ {0} ^ {2} R ^ {2}} + \frac {1}{\sqrt {1 + c _ {0} ^ {2} R ^ {2}}} - 2\right). \\ \end{array}

Tillsammans har vi

\begin{array}{l} M = 2 \pi \rho_ {0} c _ {0} \cdot \frac {1}{2} \cdot \frac {1}{c _ {0} ^ {4}} \left(\sqrt {1 + c _ {0} ^ {2} R ^ {2}} + \frac {1}{\sqrt {1 + c _ {0} ^ {2} R ^ {2}}} - 2\right) \\ = \frac {\pi \rho_ {0}}{c _ {0} ^ {3}} \left(\sqrt {1 + c _ {0} ^ {2} R ^ {2}} + \frac {1}{\sqrt {1 + c _ {0} ^ {2} R ^ {2}}} - 2\right). \\ \end{array}

Svaret kan skrivas om till

M = \pi \rho_ {0} c _ {0} \left(\frac {2}{c _ {0} ^ {4}} \left(\sqrt {1 + c _ {0} ^ {2} R ^ {2}} - 1\right) - \frac {R ^ {2}}{c _ {0} ^ {2} \sqrt {1 + c _ {0} ^ {2} R ^ {2}}}\right) = \frac {\pi \rho_ {0} R ^ {2}}{c _ {0} \sqrt {1 + c _ {0} ^ {2} R ^ {2}}} \cdot \frac {\sqrt {1 + c _ {0} ^ {2} R ^ {2}} - 1}{\sqrt {1 + c _ {0} ^ {2} R ^ {2}} + 1}.