2个回答
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(1)利用球坐标求解
原式=∫(0,2π)dθ∫(0,2π)dφ∫(2,3)r^4*sin^3φ*sin^2θdr
=(211/5)*∫(0,2π)dθ∫(0,2π)sin^3φ*sin^2θdφ
=0
(2)利用柱坐标求解
原式=∫(0,2π)dθ∫(0,1)r^3dr∫(r,√(1-r^2))dz
=2π*∫(0,1)r^3*[√(1-r^2)-r]dr
=2π*[∫(0,1)r^3*√(1-r^2)dr-∫(0,1)r^4dr]
=2π*∫(0,π/2)sin^3x*cos^2xdx-2π/5
=2π*[(-1/3)*cos^3x+(1/5)*cos^5x]|(0,π/2)-2π/5
=4π/15-2π/5
=-2π/15
原式=∫(0,2π)dθ∫(0,2π)dφ∫(2,3)r^4*sin^3φ*sin^2θdr
=(211/5)*∫(0,2π)dθ∫(0,2π)sin^3φ*sin^2θdφ
=0
(2)利用柱坐标求解
原式=∫(0,2π)dθ∫(0,1)r^3dr∫(r,√(1-r^2))dz
=2π*∫(0,1)r^3*[√(1-r^2)-r]dr
=2π*[∫(0,1)r^3*√(1-r^2)dr-∫(0,1)r^4dr]
=2π*∫(0,π/2)sin^3x*cos^2xdx-2π/5
=2π*[(-1/3)*cos^3x+(1/5)*cos^5x]|(0,π/2)-2π/5
=4π/15-2π/5
=-2π/15
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