补线L1：y = 0.dy = 0.逆时针方向,x由- 3变到3.封闭区域运用格林公式
∮(L+L1) ( 3y - x² ) dx + ( 7x + √(y⁴ + 1) ) dy
= ∫∫D [ ∂/∂x ( 7x + √(y⁴ + 1) ) - ∂/∂y ( 3y - x² ) ] dxdy
= ∫∫D [ 7 - 3 ] dxdy
= 4∫∫D dxdy
= 4 * 1/2 * π * 3²
= 18π
∫L1 ( 3y - x² ) dx + ( 7x + √(y⁴ + 1) ) dy
= ∫(- 3,3) ( - x² ) dx
= - 2 * x³/3：(0,3)
= - 2 * 27/3
= - 18
即∫L + ∫L1 = ∮(L+L1) = 18π
∫L - 18π = 18π
得∫L ( 3y - x² ) dx + ( 7x + √(y⁴ + 1) ) dy = 18(π + 1)