Re: [其他] 台大102學年度轉學考 微積分B參考答案
: 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
到這裡 我們證明了 F is conservative
而 line integral of any conservative vector field is independent of path
所以說原 integral 是 independent of path
先積f_x:
f(x, y) = x^2*sin(y) + g(y)
再積f_y:
f(x, y) = x^2*sin(y) - y^3 + h(x)
所以說 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) 根據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)
有兩種解法:
-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
所以說:
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
--
推
08/02 23:10, , 1F
08/02 23:10, 1F
推
08/03 00:21, , 2F
08/03 00:21, 2F
對 一定要這一步
※ 編輯: hsnuyi (118.160.160.135 臺灣), 09/03/2019 20:47:28
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