: 15. 6 + (π/6)^3 令 F = <2xsin(y), x^2*cos(y) - 3y^2> f_x = 2xsin(y), f_y = x^2*cos(y) - 3y^2 f_xy = 2xcos(y), f_yx = 2xcos(y) f_xy = f_yx 到这里 我们证明了 F is conservative 而 line integral of any conservative vector field is independent of path 所以说原 integral 是 independent of path 先积f_x: f(x, y) = x^2*sin(y) + g(y) 再积f_y: f(x, y) = x^2*sin(y) - y^3 + h(x) 所以说 f(x, y) = x^2*sin(y) - y^3 + c ∫ 2xsin(y)dx + [x^2*cos(y) - 3y^2]dy = ∫ <2xsin(y), x^2*cos(y) - 3y^2>·<dx, dy> C C |(0, -2) = ∫ F·dr = [x^2*sin(y) - y^3]| = 6 + (π/6)^3 C |(2, π/6) : 16. (a) 0 : (b) 2π (a) 根据Green's Theorem: x^2 + y^2 - 2x^2 -x^2 - y^2 + 2y^2 原式 = ∫∫ [------------------ - -------------------]dxdy = 0 R (x^2 + y^2)^2 (x^2 + y^2)^2 R is the region enclosed by C (b) 有两种解法: -sin(θ) cos(θ) 原式 + lim ∫ [----------*ε*-sin(θ)dθ + ---------*ε*cos(θ)]dθ = 0 ε->0 C' ε ε C': x = εcos(θ), y = εsin(θ) 且C'是clockwise 所以说: 原式 + ∫ dθ = 原式 - 2π = 0 C' 原式 = 2π 另解: -sin(θ) cos(θ) 原式 = ∫ [----------*a*-sin(θ)dθ + ---------*a*cos(θ)]dθ = ∫ dθ = 2π C a a C -- 对 一定要这一步 ※ 编辑: hsnuyi (118.160.160.135 台湾), 09/03/2019 20:47:28