Question

二重积分∫∫√(x^2+y^2-x)/(2x-x^2-y^2)

Answer 1

问：二重积分∫∫√(x^2+y^2-x)/(2x-x^2-y^2)

答：原式=∫dθ∫√(1-r²)rdr (做变换：x=1+rcosθ,y=rsinθ,则dxdy=rdθdr)=2π(-1/2)∫√(1-r²)d(1-r²)=-π(2/3)[(1-1²)^(3/2)-(1-0²)^(3/2)]=-π(2/3)(0-1)=2π/3.

答：∫∫(x^2+y^2-x)dxdy=∫[0,2]dy∫[y/2,y](x^2+y^2-x)dx=∫[0,2]∫(x^3/3+xy^2-x^2/2)[x=y/2,y]dy=13/6