Stokes's theorem F=(x-z)i+(y-

Question

use Stokes's theorem to evalute ∫c F‧Tds

where C is the boundary of the part of the plane x+4y+z=12

lying in the first octant, and F=(x-z)i+(y-x)j+(z-y)k

Answer 1

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∫c F‧Tds = ∫∫(▽xF)‧Ndσ (over region R)

i　　　j　　　k

∂/∂x　∂/∂y　∂/∂z

x-z　　y-x　　z-y

this is a 3x3 matrix denoted it by M

take det(M)

then ▽xF = curl(F) = det(M) = (-1, -1, -1) is a vector

∫∫(▽xF)‧Ndσ= ∫∫(-1, -1, -1)‧(1, 4, 1)*[1/(1, 4, 1)‧(0, 0, 1)]*dxdy

for N = (1, 4, 1) an upward-pointing normal vector

=∫∫(-1 - 4 -1)*1/1*dxdy over region R

the surface is the portion of the graph of z = 12-x-4y corresponding to (x, y) in the region R in the xy-plane bounded by the coordinate axes and the line x + 4y = 12

= ∫∫(-6)dxdy where the area of the region R = 1/2*12*3 = 18

= (-6)*18 = -108........Ans

Source(s): me