曲线积分与曲面积分问题？

这道题你题目没给错的话是可以直接用高斯公式的呀，不需要补平面啊，它给的就是闭曲面的外侧啊，所以积分值直接就是棱锥的体积的二倍。

那两个半圆平面要算啊，
【s是X=0,Y=0以及x^2+y^2+z^2=a^2(x>=0,y>=0)所围成的闭曲面】
【闭曲面啊，要连起来，封闭啊。】
在曲面x^2+y^2+z^2=a^2(x>=0,y>=0)上，dS = a*db*a*dc. b:0->π/2, c:-π/2->π/2.
∫∫（x^2+y^2+z^2)dS = ∫(0->π/2)db∫(-π/2->π/2)a^2 *a^2*dc
= a^4*(π/2)*π = a^4π^2/2
在x=0[（a^2-z^2)^(1/2) >= y>=0]上,dS = dydz, z:-|a|->|a|. y:0->（a^2-z^2)^(1/2).
∫∫（x^2+y^2+z^2)dS = ∫(-|a|->|a|)dz∫(0->（a^2-z^2)^(1/2))(y^2+z^2)dy
= 2∫(0->|a|){[（a^2-z^2)^(1/2)]^3/3 + z^2（a^2-z^2)^(1/2)}dz
z = |a|sint, dz = |a|costdt,t:0->π/2.
∫∫（x^2+y^2+z^2)dS = 2∫(0->|a|){[（a^2-z^2)^(1/2)]^3/3 + z^2（a^2-z^2)^(1/2)}dz
= 2∫(0->π/2){（|a|cost)^3/3 + a^2(sint)^2|a|cost}|a|costdt
= 2a^4∫(0->π/2){（cost)^4/3 + (sint)^2(cost)^2}dt
= 2a^4∫(0->π/2){-2（cost)^4/3 + (cost)^2}dt
∫(0->π/2)（cost)^4dt = (1/4)∫(0->π/2)[cos(2t) + 1]^2dt = (1/4)∫(0->π/2){[cos(2t)]^2 + 2cos(2t) + 1}dt = (1/4)∫(0->π/2){[cos(4t) + 1]/2 + 2cos(2t) + 1}dt = (1/4)∫(0->π/2){cos(4t)/2 + 3/2 + 2cos(2t)}dt = (1/4){3/2*π/2} = 3π/16.
∫(0->π/2)(cost)^2dt = (1/2)∫(0->π/2)[cos(2t)+1]dt = (1/2)(π/2) = π/4.
∫∫（x^2+y^2+z^2)dS = 2a^4∫(0->π/2){-2（cost)^4/3 + (cost)^2}dt
= 2a^4{-2/3*3π/16 + π/4} = a^4π/8
由对称性，
在y=0[（a^2-z^2)^(1/2) >= x>=0]上,
∫∫（x^2+y^2+z^2)dS = a^4π/8.
在整个封闭曲面上，
∫∫（x^2+y^2+z^2)dS = a^4π^2/2 + a^4π/8 + a^4π/8 = a^4π^2/2 + a^4π/4
【哦，和答案也不一样啊。哈，奇怪~~~】
2，
还要算上Z=1那个面上的积分。那个1/2π就是这个积分的结果吧。
3，
柱面坐标, x = rcost, y = rsint, z:0->2arcost-r^2, t:-π/2->π/2. r:a->2a