求大神解答极坐标方程r=1+sinθ与r=1所围公共部分的面积
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解决时间 2021-11-27 20:38
- 提问者网友：疯子也有疯子的情调
- 2021-11-27 07:52
求大神解答极坐标方程r=1+sinθ与r=1所围公共部分的面积
最佳答案
- 五星知识达人网友：刀戟声无边
- 2021-11-27 08:49
θ= 0,r=1+sinθ与r=1重合,
θ= PI,r=1+sinθ与r=1重合.
所以,θ的积分范围是 0 -> PI.
0 1 + sinθ.
面积 = (1/2)S_{0->PI}dθ S_{1 -> 1 + sinθ}r^2dr
= (1/6)S_{0->PI} [(1 + sinθ)^3 - 1]dθ
= (1/6)S_{0->PI} [(sinθ)^3 + 3(sinθ)^2 + 3sinθ]dθ
= (1/6)S_{0->PI} [(sinθ)^3 + 3sinθ]dθ + (1/6)S_{0->PI} [3(sinθ)^2]dθ
S_{0->PI} [(sinθ)^3 + 3sinθ]dθ = S_{0->PI} [4 - (cosθ)^2]sinθdθ
= 8 - 2/3 = 22/3
S_{0->PI} [(sinθ)^2]dθ = (1/2)S_{0->PI} [1 - cos(2θ)]dθ
= (1/2)[PI]
面积 = (1/6)S_{0->PI} [(sinθ)^3 + 3sinθ]dθ + (1/6)S_{0->PI} [3(sinθ)^2]dθ
= (1/6)(22/3) + (1/6)*3*PI/2
= 11/9 + PI/4追问不对啊，答案是(5π/4)-2
θ= PI,r=1+sinθ与r=1重合.
所以,θ的积分范围是 0 -> PI.
0 1 + sinθ.
面积 = (1/2)S_{0->PI}dθ S_{1 -> 1 + sinθ}r^2dr
= (1/6)S_{0->PI} [(1 + sinθ)^3 - 1]dθ
= (1/6)S_{0->PI} [(sinθ)^3 + 3(sinθ)^2 + 3sinθ]dθ
= (1/6)S_{0->PI} [(sinθ)^3 + 3sinθ]dθ + (1/6)S_{0->PI} [3(sinθ)^2]dθ
S_{0->PI} [(sinθ)^3 + 3sinθ]dθ = S_{0->PI} [4 - (cosθ)^2]sinθdθ
= 8 - 2/3 = 22/3
S_{0->PI} [(sinθ)^2]dθ = (1/2)S_{0->PI} [1 - cos(2θ)]dθ
= (1/2)[PI]
面积 = (1/6)S_{0->PI} [(sinθ)^3 + 3sinθ]dθ + (1/6)S_{0->PI} [3(sinθ)^2]dθ
= (1/6)(22/3) + (1/6)*3*PI/2
= 11/9 + PI/4追问不对啊，答案是(5π/4)-2
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