Question

［急！］一道关于曲面积分的高数题

求I1-I2,其中I1=∫∫(S1)(x^2+y^2+z^2)dS
I2=∫∫(S2)(x^2+y^2+z^2)dS
S1为球面x^2+y^2+z^2=a^2(a>0)

S2为内接于S1的八面体的边界：|x|+|y|+|z|=a

Answer 1

∫∫(S1) (x² + y² + z²) dS - ∫∫(S2) (x² + y² + z²) dS
= a²∫∫(S1) dS - 3∫∫(S2) x² dS
= a² * 4πa² - 3 * 8∫∫D x²√3 dxdy
= 4πa⁴ - 24√3∫(0→a) x² dx ∫(0→a - x) dy
= 4πa⁴ - 24√3∫(0→a) x²(a - x) dx
= 4πa⁴ - 24√3∫(0→a) (ax² -x³) dx
= 4πa⁴ - 24√3 * [ (1/3)ax³ - (1/4)x⁴ ] |(0→a)
= 4πa⁴ - 24√3 * [ (1/3)a⁴ - (1/4)a⁴ ]
= 2(2π - √3)a⁴

Answer 2

l1为3\4*兀*a^5