∫∫D|1-x²-y²|dxdy,其中D={(x,y)|x²+y²≤x,y≥0}

题目

∫∫D|1-x²-y²|dxdy,其中D={(x,y)|x²+y²≤x,y≥0}

答案

∵在区域D={(x,y)|x²+y²≤x,y≥0}中,1-x²-y²≥0
∴∫∫|1-x²-y²|dxdy=∫∫(1-x²-y²)dxdy
=∫dθ∫(1-r²)rdr (作极坐标变换)
=∫[cos²x(2-cos²x)/2]dθ
=(1/32)∫[5+4cos(2θ)-cos(4θ)]dθ (应用倍角公式)
=(1/32)[5(π/2)+2sin(2(π/2))-sin(4(π/2))/4]
=(1/32)(5π/2)
=5π/64.