高等数学,求大神解析,谢谢!
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用柱坐标法:
I = ∫<0, π/2> dt ∫<0, 1> r^2sintcost rdr ∫<0, √(1-r^2)> zdz
= ∫<0, π/2> sintcost dt ∫<0, 1> (1/2)r^3(1-r^2)dr
= (1/2) ∫<0, π/2> sintcost dt [r^4/4-r^6/6]<0, 1>
= (1/24) ∫<0, π/2> sint dsint = 1/48
用球坐标法:
I = ∫<0, π/2> dΦ ∫<0, π/2> dθ ∫<0, 1> r^3cosΦ(sinΦ)^2sinθcosθ r^2sinΦdr
= ∫<0, π/2> cosΦ(sinΦ)^3dΦ ∫<0, π/2> sinθcosθdθ ∫<0, 1> r^5dr
= (1/6)∫<0, π/2> (sinΦ)^3dsinΦ ∫<0, π/2> sinθdsinθ
=(1/6)(1/4)(1/2) = 1/48
I = ∫<0, π/2> dt ∫<0, 1> r^2sintcost rdr ∫<0, √(1-r^2)> zdz
= ∫<0, π/2> sintcost dt ∫<0, 1> (1/2)r^3(1-r^2)dr
= (1/2) ∫<0, π/2> sintcost dt [r^4/4-r^6/6]<0, 1>
= (1/24) ∫<0, π/2> sint dsint = 1/48
用球坐标法:
I = ∫<0, π/2> dΦ ∫<0, π/2> dθ ∫<0, 1> r^3cosΦ(sinΦ)^2sinθcosθ r^2sinΦdr
= ∫<0, π/2> cosΦ(sinΦ)^3dΦ ∫<0, π/2> sinθcosθdθ ∫<0, 1> r^5dr
= (1/6)∫<0, π/2> (sinΦ)^3dsinΦ ∫<0, π/2> sinθdsinθ
=(1/6)(1/4)(1/2) = 1/48
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