设积分区域D为x2+y2≤R2,则∫∫De-x2-y2dσ=____．

设积分区域D为x2+y2≤R2,则∫∫De-x2-y2dσ=____．
【正确答案】：π(1-e-R2)。解析：∫∫De-x2-y2dσ=∫02πdθ∫0Re-r2•rdr=2π•[-(1/2)]∫0Re-r2d(-r2)=2π•[-(1/2)]•e-r2∣0R=π(1-e-R2)