Evaluate this Surface Integral?

Question

Evaluate ∫∫ (curl F) X dS where S is the portion of the surface of the sphere x^2 +y^2 + z^2 = 1 above the plane x + y + z = 1 and F = (xi + yj + zk) X (i + j + k). Note: Here, X means cross product.

Answer 1

F = (xi + yj + zk) X (i + j + k)=

i|y z| + j|z x| + k|x y|

.|1 1| . . |1 1| . . |1 1|

F=(y-z)i+(z-x)j+(x-y)k

. . . . . |. i . . . j . . k .|

curlF= |∂/∂x ∂/∂y ∂/∂z| = -2i -2j -2k

. . . . . |y-z . z-x . x-y|

∫∫[S] (curl F)•dS = -2∫∫[S] (i+j+k)•(idydz+jdzdx+kdxdy)=

-2∫∫[S] (dydz+dzdx+dxdy)=-2(∫∫[A3] dydz+∫∫[A2] dzdx+∫∫[A1] dxdy)=

-2(A1+A2+A3)=-6*A1=J, where A1, A2 and A3 are the orthogonal projections

of S on xOy, xOz and yOz coordinate planes respectively.

x^2+y^2+z^2=1...(1) and x+y+z=1...(2) intersect in a circle, which orthogonal projections on the 3 coordinate planes are congruent ellipses. The equation of the ellipse on the xOy plane is obtained by substituting z=1-x-y from (2) in (1) ==>

2x²+2xy+2y²-2x-2y=0

x² +xy+y²-x-y=0 ==> it passes through the coordinate origin.

A1=πab, where a and b are the semimajor and semiminor axis respectively.

The minor axis is on the first quadrant bisector y=x ==>

Its ends are the roots of the equation x² +xx+x²-x-x=0 ==>

The end points of the minor axis are (0,0) and (2/3, 2/3) ==>

The center of the ellipse is at (1/3, 1/3) ==> b=√2/3

The major axis is on the line y-1/3=-(x-1/3) or

y=2/3-x, which intersects the ellipse at the points where

x²+x(2/3-x)+(2/3-x)²-x-(2/3-x)=0

9x²-6x-2=0

x1=(1+√3)/3, y1=(1-√3)/3

x2=(1-√3)/3, y2=(1+√3)/3

a=√[(x1-1/3)²+(y1-1/3)²]

a=√[(√3/3)²+(-√3/3)²]=√(2/3)

J=-6*A1=-6π√(2/3)√2/3

J=-4π/√3=-4π√3/3