let's move on to spherical coordinates:
x = r c o s φ s i n θ , y = r s i n φ s i n θ , z = r c o s θ x = rcos\varphi sin\theta ,y = rsin\varphi sin\theta ,z = rcos\theta x = rcos φ s in θ , y = rs in φ s in θ , z = rcos θ
2 < r < 3 , 0 < φ < 2 π , 0 < θ < π 2< r < 3,\,\,0 < \varphi < 2\pi ,\,\,0 < \theta < \pi 2 < r < 3 , 0 < φ < 2 π , 0 < θ < π
Then
x 2 + y 2 + z 2 = r 2 {x^2} + {y^2} + {z^2} = {r^2} x 2 + y 2 + z 2 = r 2
∫ ∫ ∫ E e − ( x 2 + y 2 + z 2 ) x 2 + y 2 + z 2 d V = ∫ 0 2 π d φ ∫ 2 3 e − r 2 r d r ∫ 0 π r 2 sin θ d θ = = ∫ 0 2 π d φ ∫ 2 3 r e − r 2 d r ∫ 0 π sin θ d θ = − 1 2 ∫ 0 2 π d φ ∫ 2 3 e − r 2 d ( − r 2 ) ∫ 0 π sin θ d θ = = 1 2 φ ∣ 0 2 π ⋅ e − r 2 ∣ 2 3 ⋅ cos θ ∣ 0 π = 1 2 ( 2 π − 0 ) ( e − 9 − e − 4 ) ( cos π − cos 0 ) = π ( e − 9 − e − 4 ) ( − 1 − 1 ) = = 2 π ( e − 4 − e − 9 ) \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} ∫ ∫ E ∫ x 2 + y 2 + z 2 e − ( x 2 + y 2 + z 2 ) d V = 0 ∫ 2 π d φ 2 ∫ 3 r e − r 2 d r 0 ∫ π r 2 sin θ d θ = = 0 ∫ 2 π d φ 2 ∫ 3 r e − r 2 d r 0 ∫ π sin θ d θ = − 2 1 0 ∫ 2 π d φ 2 ∫ 3 e − r 2 d ( − r 2 ) 0 ∫ π sin θ d θ = = 2 1 φ ∣ 0 2 π ⋅ e − r 2 ∣ ∣ 2 3 ⋅ cos θ ∣ 0 π = 2 1 ( 2 π − 0 ) ( e − 9 − e − 4 ) ( cos π − cos 0 ) = π ( e − 9 − e − 4 ) ( − 1 − 1 ) = = 2 π ( e − 4 − e − 9 )
Answer: 2 π ( e − 4 − e − 9 ) 2\pi \left( {{e^{ - 4}} - {e^{ - 9}}} \right) 2 π ( e − 4 − e − 9 )
#340153 on Dec 2023