求不定积分应用题 10
1个回答
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r=a(1+cosθ),r'=-asinθ
利用对称性
长度=2∫(0,π)√r^2+r'^2dθ
=2∫(0,π)√a^2(2+2cosθ)dθ
=2a∫(0,π)√4cos^2(θ/2)dθ
=4a∫(0,π)cos(θ/2)dθ
=8a∫(0,π)cos(θ/2)dθ/2
=8asin(θ/2)|(0,π)
=8a
面积=2*1/2∫(0,π)r^2dθ
=∫(0,π)a^2(1+cosθ)^2dθ
=4a^2∫(0,π)cos^4(θ/2)dθ
=8a^2∫(0,π)cos^4(θ/2)dθ/2 (令θ/2=t)
=8a^2∫(0,π/2)cos^4tdt
=8a^2*3/4*1/2*π/2
=3/2*πa^2
利用对称性
长度=2∫(0,π)√r^2+r'^2dθ
=2∫(0,π)√a^2(2+2cosθ)dθ
=2a∫(0,π)√4cos^2(θ/2)dθ
=4a∫(0,π)cos(θ/2)dθ
=8a∫(0,π)cos(θ/2)dθ/2
=8asin(θ/2)|(0,π)
=8a
面积=2*1/2∫(0,π)r^2dθ
=∫(0,π)a^2(1+cosθ)^2dθ
=4a^2∫(0,π)cos^4(θ/2)dθ
=8a^2∫(0,π)cos^4(θ/2)dθ/2 (令θ/2=t)
=8a^2∫(0,π/2)cos^4tdt
=8a^2*3/4*1/2*π/2
=3/2*πa^2
追问
您好, 你能在纸上写下来,让我看看嘛?谢谢
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