Question

高数曲面积分：计算∫(x+y)e^(x^2+y^2)ds 其中L为圆弧y=√(a^2-x^)和直线y=x与y=-x围成的扇形边界

Answer 1

L由y = √(a² - x²) 和 y = x 和 y = - x围成
参数化：t：- π/4 → π/4
x = acost，y = asint
dx = - asintdt，dy = acostdt
ds = adt
∫L (x + y)e^(x² + y²) ds
= ∫(- π/4，π/4) (acost + asint)e^a² adt
= a²e^a²∫(- π/4，π/4) (sint + cost) dt
= a²e^a² * 2[sint] |(0，π/4)
= √2a²e^a²

追问

答案是1\√2[e^(a^2)-1]+√2a^2e^(a^2)

追答

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

Answer 2

L由y = √(a² - x²) 和 y = x 和 y = - x围成
参数化：t：- π/4 → π/4
x = acost，y = asint
dx = - asintdt，dy = acostdt
ds = adt
∫L (x + y)e^(x² + y²) ds
= ∫(- π/4，π/4)