Question

计算 ∫∫D√(5-x^2-y^2)dxdy,D是由圆x^2+y^2=1,x^2+y^2=4及直线y=x,y=0所包围的

Answer 1

极坐标：

∫∫(D) √(5-x²-y²) dxdy

=∫∫(D) r√(5-r²) drdθ

=∫[0→π/4] dθ∫[1→2] r√(5-r²) dr

=(π/4)∫[1→2] r√(5-r²) dr

=(π/8)∫[1→2] √(5-r²) d(r²)

=-(π/8)(2/3)(5-r²)^(3/2) |[1→2]

=(π/12)(8-1)

=7π/12