将二次积分化为极坐标形式的二次积分
∫(0→2)dx∫(0→x)f(√(x^2+y^2))dy答案是∫(π/4→π/3)dθ∫(0→2secθ)f(ρ)ρdρ为什么是π/4→π/3而不是0→π/4...
∫(0→2)dx∫(0→x)f(√(x^2+y^2))dy
答案是∫(π/4→π/3)dθ∫(0→2secθ)f(ρ)ρdρ
为什么是π/4→π/3而不是0→π/4 展开
答案是∫(π/4→π/3)dθ∫(0→2secθ)f(ρ)ρdρ
为什么是π/4→π/3而不是0→π/4 展开
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积分区域是半圆,化成极坐标为:r=2acosθ,(0≤θ≤π)
原式=∫[0,π/2]dθ ∫[0,2acosθ ] (r^2*r)dr
=∫[0,π/2]dθ [0,2acosθ [ r^4/4
=(1/4)∫[0,π/2]dθ [0,2acosθ ] (cosθ )^4
=(16a^4/4)∫[0,π/2]dθ [1+cos2θ)^2/4
=a^4∫[0,π/2]dθ [1+2cos2θ+(cos2θ)^2]
=a^4[θ+sin2θ+θ/2+(sin4θ)/8][0,π/2]
=a^4(3/2*π/2+0+0)
=3πa^4/4.
原式=∫[0,π/2]dθ ∫[0,2acosθ ] (r^2*r)dr
=∫[0,π/2]dθ [0,2acosθ [ r^4/4
=(1/4)∫[0,π/2]dθ [0,2acosθ ] (cosθ )^4
=(16a^4/4)∫[0,π/2]dθ [1+cos2θ)^2/4
=a^4∫[0,π/2]dθ [1+2cos2θ+(cos2θ)^2]
=a^4[θ+sin2θ+θ/2+(sin4θ)/8][0,π/2]
=a^4(3/2*π/2+0+0)
=3πa^4/4.
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