# 高等微積分...急

Given any nonvertical plane P parallel to the x-axis, let C be the curve of intersection of P with the cylinder X^2+y^2=a^2 . Show that

∫_c▒〖[(yz-y)dx+(xz+x)dy] =2 π a^2

Let F(x,y,z)=(x^2+y^2+z^2)(xi+yj+zk) and let S be the sphere of radius a about the origin . compute ∬_s▒Fn both directly and by the divergence theorem.

Let x=(x,y,z) and g(x)= |x|^-1 =(x^2+y^2+z^2)^-1/2

a.compute ∇g (x) for x ≠0

b.Show by direct calculation that ∬▒∂g/∂n dA =-4π if S is any sphere centered at the origin

### 1 個解答

• 最佳解答

Given any nonvertical plane P parallel to the x-axis, let C be the curve of intersection of P with the cylinder x²+y²=a² . Show that

∫_c〖[(yz-y)dx+(xz+x)dy] =2 π a²

向量F=(yz-y, zx+x, 0), curl(F)=(-x,y,2)

設平面P平行x軸,設P的法向量與z軸夾θ角(θ不是90度)

則P的方程式為ysinθ+zcosθ=k,

A為C所圍成的平面區域(橢圓內部)

B為A在xy平面上的投影區域(圓 x²+y²=a²)

由Stokes定理:

∫_c [(yz-y)dx+(zx+x)dy]=∫∫_A (-x,y,2)‧nda

=∫∫_B (-x,y,2)‧(0,sinθ,cosθ)/cosθ dxdy

=∫∫_B (ysinθ/cosθ+2)dydx

=∫∫_B (ytanθ)dydx+∫∫_B 2 dydx

=0+ 2πa² (註:因對稱性,故第一個積分=0)

Let F(x,y,z)=(x²+y²+z²)(xi+yj+zk) and let S be the sphere of radius a about the origin . compute ∬_s F‧ndA both directly and by the divergence theorem.

(散度定理)

F=(x(x²+y²+z²), y(x²+y²+z²),z(x²+y²+z²)),

div(F)=(3x²+y^2+z²)+(3y²+x²+z²)+(3z²+x²+y²)=5(x²+y²+z²)

∫∫_s F‧ndA=∫∫∫_V 5(x²+y²+z²)dv (V為球體內部)

=∫[0~2π]∫[0~π]∫[0~a] 5(r²)(r²sinφ)dr dφ dθ

=∫[0~2π]∫[0~π] a^5 sinφ dφ dθ

=∫[0~2π] 2a^5 dθ= 4πa^5

(直接積分)

∫∫_s F‧ndA=∫∫_s a²(x,y,z)‧(x,y,z)/a dA

=∫∫_s a³ dA=a³ (4πa²)=4πa^5 (註:球表面積=4πa²)

Let x=(x,y,z) and g(x)= 1/|x| =1/(x²+y²+z²)^(1/2)

a.compute ∇g (x) for x ≠0

b.Show by direct calculation that ∬_s ∂g/∂n dA =-4π if S is any sphere centered at the origin

∂g/∂x= -x/(x²+y²+z²)^(3/2),故