求大神解答,利用极坐标计算二重积分∫∫(x+y)^2dσ (σ)={(x,y)|(x^2+y^2)
求大神解答,利用极坐标计算二重积分∫∫(x+y)^2dσ(σ)={(x,y)|(x^2+y^2)^2<=2a(x^2-y^2),(a>0)}正确有好评...
求大神解答,利用极坐标计算二重积分∫∫(x+y)^2dσ (σ)={(x,y)|(x^2+y^2)^2<=2a(x^2-y^2),(a>0)} 正确有好评
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2014-05-05
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那你就给硬好评了!
(x² + y²)² = 2a(x² - y²)
r⁴ = 2ar²(cos²θ - sin²θ)
r² = 2acos2θ
r = √(2acos2θ),双纽线
∫∫D (x + y)² dxdy
= ∫∫D (x² + 2xy + y²) dxdy
= ∫∫D (x² + y²) dxdy
= 4∫∫D1 r³ drdθ
= 4∫(0,π/4) dθ ∫(0,√(2acos2θ)) r³ dr
= 4∫(0,π/4) a²cos²(2θ) dθ
= 4a²∫(0,π/4) (1 + cos4θ)/2 dθ
= 2a²(θ + 1/4 * sin4θ) |(0,π/4)
= 2a² * π/4
= πa²/2
(x² + y²)² = 2a(x² - y²)
r⁴ = 2ar²(cos²θ - sin²θ)
r² = 2acos2θ
r = √(2acos2θ),双纽线
∫∫D (x + y)² dxdy
= ∫∫D (x² + 2xy + y²) dxdy
= ∫∫D (x² + y²) dxdy
= 4∫∫D1 r³ drdθ
= 4∫(0,π/4) dθ ∫(0,√(2acos2θ)) r³ dr
= 4∫(0,π/4) a²cos²(2θ) dθ
= 4a²∫(0,π/4) (1 + cos4θ)/2 dθ
= 2a²(θ + 1/4 * sin4θ) |(0,π/4)
= 2a² * π/4
= πa²/2
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