
微积分求解答
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∫D∫e^[-(x^2+y^2)]dσ
=4∫(0,π/2)dθ∫(2,3)e^(-r^2)rdr
=-2∫(0,π/2)dθ∫(2,3)e^(-r^2)d(-r^2)
=-2∫(0,π/2)[e^(-r^2)](2,3)dθ
=-2[e^(-3^2)-e^(-2^2)]∫(0,π/2)dθ
=-2(1/e^9-1/e^4)[θ](0,π/2)
=2(e^9-e^4)/e^13(π/2-0)
=π(e^9-e^4)/e^13
=4∫(0,π/2)dθ∫(2,3)e^(-r^2)rdr
=-2∫(0,π/2)dθ∫(2,3)e^(-r^2)d(-r^2)
=-2∫(0,π/2)[e^(-r^2)](2,3)dθ
=-2[e^(-3^2)-e^(-2^2)]∫(0,π/2)dθ
=-2(1/e^9-1/e^4)[θ](0,π/2)
=2(e^9-e^4)/e^13(π/2-0)
=π(e^9-e^4)/e^13
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