Question

求二重积分根号下x^2+y^2,具体看图，麻烦了

Answer 1

Answer 2

解：因为
D={(x，y)|x²+y²≤4，0≤x≤2，0≤y≤2}
={(θ，r)|0≤θ≤π/4，0≤r≤2}
所以
∫∫D√(x²+y²)dσ
=∫[0,π/4]dθ∫[0,2]r·rdr
=(π/4)·(r³/3)|[0,2]
=(π/4)·(8/3)
=2π/3.

追问

你好，圆是大于等于4，积分区域错了

追答

抱歉！正确解法如下：
因为
D={(x，y)|x²+y²≤4，0≤x≤2，0≤y≤2}
={(θ，r)|0≤θ≤π/4，0≤r≤2/cosθ}+{(θ，r)|π/4≤θ≤π/2，0≤r≤2/sinθ}－{(θ，r)|0≤θ≤π/2，0≤r≤2}
所以
∫∫D√(x²+y²)dσ
=∫[0,π/4]dθ∫[0,2/cosθ]r·rdr＋∫[π/4,π/2]dθ∫[0,2/sinθ]r·rdr－∫[0,π/2]dθ∫[0,2]r·rdr
=∫[0,π/4](8/3)sec³θdθ＋∫[π/4,π/2](8/3)csc³θdθ－∫[0,π/2](8/3)dθ
=(4/3)[secθtanθ+ln(secθ+tanθ)]|[0,π/4]＋(4/3)[-cscθcotθ+ln(cscθ-cotθ)]|[π/4,π/2]－(π/2)(8/3)
=(4/3)[√2+ln(√2 +1)]+(4/3)[√2-ln(√2 -1)]-4π/3
=(8/3)[√2+ln(√2 +1)]-4π/3

第一个“≤”改为“≥”