Question

二重积分：∫∫√(R^2-X^2-Y^2)dxdy, 其中D是由圆域X^2 Y^2=Rx 求过程和结果 急！

Answer 1

D:  x^2+(y-R/2)^2=(R/2)^2

∫∫[D]√(R^2-X^2-Y^2)dxdy

=∫[0,π]dθ∫[0,Rsinθ]ρ√(R^2-ρ^2)dρ

=∫[0,π][-1/3(R^2-ρ^2)^(3/2)|[0,Rsinθ]dθ

=∫[0,π][R^3/3-R^3/3cos^3(θ)]dθ

=πR^3/3-R^3/3∫[0,π](1-sin^2(θ)dsinθ

=πR^3/3-R^3/3sin(θ)|[0,π]+R^3/9sin^3(θ)|[0,π]

=πR^3/3