曲线积分与曲面积分的问题,求助

Question

计算第一类去面积分
1. ∫∫（x^2+y^2+z^2)dS,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，这题我算的是2πa^4答案是2/3πa^4，那两个半圆平面算不算啊？求详解！！
2.∫∫（x^2+y^2)dS,s为立体（x^2+y^2）开方<=z<=1的边界曲面。
答案是1/2π+根号2/2π，我不知道它的那个1/2π是怎么算出来的，我晕，求助！

第二类曲线积分
∫∫z^2/(x^2+y^2)dxdy,S为上半球面Z=根号(2ax-x^2-y^2)（a>0)在圆柱面x^2+y^2=a^2的外面部分的上侧

我算的跟他的答案完全不着边，着急啊

Answer 1

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

Answer 2

面积是正数，积分可以是负数。结合图像再看看吧！