这张图片中的面积用定积分怎么求,求教~
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联立ρ=2^0.5sin(θ)
ρ^2=cos(2θ)
解得θ=π/6
θ=5π/6
两阴影面积一样
只需计算右侧阴影
S1=1/2∫2sin^2(θ))dθ
(a=0
b=π/6)
=1/2∫2sin^2(θ))dθ
(a=0
b=π/6)
=1/2∫(1-cos(2θ))dθ
(a=0
b=π/6)
=1/4(2θ-sin(2θ))
(a=0
b=π/6)
=1/4(π/3-sin(π/3)
S2=1/2∫cos(2θ)
dθ
(a=π/6
b=π/4)
=1/4
sin(2θ)
(a=π/6
b=π/4)
=1/4(1-sin(π/3))
S=2(S1+S2)
=(π/3+1-2sin(π/3))/2
=(π/3+1-3^0.5)/2
ρ^2=cos(2θ)
解得θ=π/6
θ=5π/6
两阴影面积一样
只需计算右侧阴影
S1=1/2∫2sin^2(θ))dθ
(a=0
b=π/6)
=1/2∫2sin^2(θ))dθ
(a=0
b=π/6)
=1/2∫(1-cos(2θ))dθ
(a=0
b=π/6)
=1/4(2θ-sin(2θ))
(a=0
b=π/6)
=1/4(π/3-sin(π/3)
S2=1/2∫cos(2θ)
dθ
(a=π/6
b=π/4)
=1/4
sin(2θ)
(a=π/6
b=π/4)
=1/4(1-sin(π/3))
S=2(S1+S2)
=(π/3+1-2sin(π/3))/2
=(π/3+1-3^0.5)/2
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