自己畫圖：積分區(qū)域關(guān)于y軸對(duì)稱,而被積函數(shù)關(guān)于x是偶函數(shù),因此
∫∫（1-y²）^1/2 dσ積分區(qū)域?yàn)镈
=2∫∫（1-y²）^1/2 dσ 積分區(qū)域?yàn)镈的第一象限部分
用極坐標(biāo),（1-y²）^1/2=（1-sin²θ）^1/2=|cosθ|
=2∫∫ rcosθ drdθ 由于第一象限余弦為正,絕對(duì)值可去掉
=2∫[π/4---->π/2]cosθdθ ∫[0--->1] r dr
=2sinθ |[π/4---->π/2] * (1/2)r² |[0--->1]
=(1-√2/2)(1-0)
=1-√2/2可是答案是2-√2我做錯(cuò)了，（1-y²）^1/2=（1-sin²θ）^1/2這個(gè)是不對(duì)的，應(yīng)該是（1-y²）^1/2=（1-r²sin²θ）^1/2∫（1-y²）^1/2 dσ積分區(qū)域?yàn)镈 =2∫∫（1-y²）^1/2 dσ 積分區(qū)域?yàn)镈的第一象限部分 用極坐標(biāo)，（1-y²）^1/2=（1-r²sin²θ）^1/2 =2∫∫ r（1-r²sin²θ）^1/2 drdθ =2∫[π/4---->π/2]dθ ∫[0--->1] r（1-r²sin²θ）^1/2dr =∫[π/4---->π/2]dθ ∫[0--->1] （1-r²sin²θ）^1/2d(r²) =(2/3)∫[π/4---->π/2] (-1/sin²θ)(1-r²sin²θ)^(3/2) |[0--->1] dθ =(2/3)∫[π/4---->π/2] (1/sin²θ)(1-cos³θ) dθ =(2/3)∫[π/4---->π/2] csc²θ dθ-(2/3)∫[π/4---->π/2] cos³θ/sin²θ dθ =(2/3)(-cotx)-(2/3)∫[π/4---->π/2] cos²θ/sin²θ d(sinθ) =(2/3)(-cotx)-(2/3)∫[π/4---->π/2] (1-sin²θ)/sin²θ d(sinθ) =(2/3)(-cotx)-(2/3)∫[π/4---->π/2] (1/sin²θ-1) d(sinθ) =(2/3)(-cotx)+(2/3)(1/sinθ)+(2/3)sinθ|[π/4---->π/2] =0+2/3+2/3+2/3-(2/3)√2-(2/3)(√2/2) =2-√2