把积分化为极坐标形式
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/05 02:58:43
把积分化为极坐标形式
![](http://img.wesiedu.com/upload/9/c4/9c4b3b4670a9285a9f1721d1dfba58b0.jpg)
![](http://img.wesiedu.com/upload/9/c4/9c4b3b4670a9285a9f1721d1dfba58b0.jpg)
![把积分化为极坐标形式](/uploads/image/z/8615062-46-2.jpg?t=%E6%8A%8A%E7%A7%AF%E5%88%86%E5%8C%96%E4%B8%BA%E6%9E%81%E5%9D%90%E6%A0%87%E5%BD%A2%E5%BC%8F)
积分域D:由直线y=x,x=a,及x轴所围得平面域;将此平面域换成极坐标形式,则是:
0≦r≦a/cosθ,0≦θ≦π/4;
故原式=【0,π/4】∫dθ【0,a/cosθ】∫r²dr=【0,π/4】∫dθ[r³/3]【0,a/cosθ】
=【0,π/4】(a³/3)∫(1/cos³θ)dθ=【0,π/4】(a³/3)∫sec³dθ=【0,π/4】(a³/3)∫secθd(tanθ)
=【0,π/4】(a³/3)[secθtanθ-∫tanθd(secθ)]=【0,π/4】(a³/3)[secθtanθ-∫secθtan²θdθ]
=【0,π/4】(a³/3)[secθtanθ-∫secθ(sec²θ-1)dθ=【0,π/4】(a³/3)[secθtanθ-∫sec³θdθ+∫secθdθ]
【移项,得:】
【0,π/4】(2a³/3)∫sec³dθ=【0,π/4】(a³/3)[secθtanθ+∫secθdθ]
=(a³/3)[secθtanθ+ln(secθ+tanθ)]【0,π/4】=(a³/3)[√2+ln(√2+1)]
故原式=【0,π/4】(a³/3)∫sec³dθ=(a³/6)[√2+ln(√2+1)]
0≦r≦a/cosθ,0≦θ≦π/4;
故原式=【0,π/4】∫dθ【0,a/cosθ】∫r²dr=【0,π/4】∫dθ[r³/3]【0,a/cosθ】
=【0,π/4】(a³/3)∫(1/cos³θ)dθ=【0,π/4】(a³/3)∫sec³dθ=【0,π/4】(a³/3)∫secθd(tanθ)
=【0,π/4】(a³/3)[secθtanθ-∫tanθd(secθ)]=【0,π/4】(a³/3)[secθtanθ-∫secθtan²θdθ]
=【0,π/4】(a³/3)[secθtanθ-∫secθ(sec²θ-1)dθ=【0,π/4】(a³/3)[secθtanθ-∫sec³θdθ+∫secθdθ]
【移项,得:】
【0,π/4】(2a³/3)∫sec³dθ=【0,π/4】(a³/3)[secθtanθ+∫secθdθ]
=(a³/3)[secθtanθ+ln(secθ+tanθ)]【0,π/4】=(a³/3)[√2+ln(√2+1)]
故原式=【0,π/4】(a³/3)∫sec³dθ=(a³/6)[√2+ln(√2+1)]