Answer to Question #143576 in Calculus for Promise Omiponle

Question

Use spherical coordinates to evaluate the triple integral
∭E (e^(−(x^2+y^2+z^2)))/√(x^2+y^2+z^2) dV,
where E is the region bounded by the spheres x^2+y^2+z^2=4 and x^2+y^2+z^2=9.

Answer 1

let's move on to spherical coordinates:

$x = rcos\varphi sin\theta ,y = rsin\varphi sin\theta ,z = rcos\theta$

$2< r < 3,\,\,0 < \varphi < 2\pi ,\,\,0 < \theta < \pi$

Then

${x^2} + {y^2} + {z^2} = {r^2}$

$\begin{array}{l} \int {\int {\int\limits_E {\frac{{{e^{ - \left( {{x^2} + {y^2} + {z^2}} \right)}}}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} } } dV = \int\limits_0^{2\pi } {d\varphi } \int\limits_2^3 {\frac{{{e^{ - {r^2}}}}}{r}} dr\int\limits_0^\pi {{r^2}\sin \theta d\theta } = \\ = \int\limits_0^{2\pi } {d\varphi } \int\limits_2^3 {r{e^{ - {r^2}}}} dr\int\limits_0^\pi {\sin \theta d\theta } = - \frac{1}{2}\int\limits_0^{2\pi } {d\varphi } \int\limits_2^3 {{e^{ - {r^2}}}} d( - {r^2})\int\limits_0^\pi {\sin \theta d\theta } = \\ = \frac{1}{2}\left. \varphi \right|_0^{2\pi } \cdot \left. {{e^{ - {r^2}}}} \right|_2^3 \cdot \cos \left. \theta \right|_0^\pi = \frac{1}{2}\left( {2\pi - 0} \right)\left( {{e^{ - 9}} - {e^{ - 4}}} \right)\left( {\cos \pi - \cos 0} \right) = \pi \left( {{e^{ - 9}} - {e^{ - 4}}} \right)( - 1 - 1) = \\ = 2\pi \left( {{e^{ - 4}} - {e^{ - 9}}} \right) \end{array}$

Answer: $2\pi \left( {{e^{ - 4}} - {e^{ - 9}}} \right)$