L由y = √(a² - x²) 和 y = x 和 y = - x围成参数化：t：- π/4 → π/4x = acost,y = asintdx = - asintdt,dy = acostdtds = adt∫L (x + y)e^(x² + y²) ds= ∫(- π/4,π/4) (acost + asint)e...答案是1\√2[e^(a^2)-1]+√2a^2e^(a^2)算漏了两条线- -
y = x，dy = dx
∫L (x + y)e^(x² + y²) ds
= ∫(0，a/√2) 2xe^(2x²)√(1 + 1) dx
= 1/√2 * (e^a² - 1)
y = - x，dy = - dx
∫L (x + y)e^(x² + y²) ds
= ∫L (x - x)e^(2x²)√(1 + 1) dx
= 0
都加起就好了