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Let S be the part of the plane 2x + ly + z = 2 which lies in the first octant, oriented upward. Use the Stokes theorem to find the flux of the vector field F = 1i + 1j + 2k across the surface S

User Ben Lesh
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We can use Stokes' theorem to relate the flux of the vector field F across the surface S to the circulation of the vector field around its boundary. The boundary of S is the curve C formed by the intersection of S with the xy-plane, which is given by 2x + ly = 2.

To compute the circulation of F around C, we first need to parameterize C. Solving for y in terms of x, we get:

y = (2 - 2x)/l

This gives us a parameterization of C in the form r(t) = ti + [(2 - 2t)/l]j, where 0 <= t <= 1. We can compute the circulation of F around C using the line integral:

int_C F . dr = int_0^1 F(r(t)) . r'(t) dt

where F(r) = 1i + 1j + 2k and r'(t) = i - (2/l)j. Substituting in, we get:

int_C F . dr = int_0^1 (i - (2/l)j) . (i + j + 2k) dt

= int_0^1 (1 - (2/l)) dt

= 1 - 2/l

Now, we can apply Stokes' theorem:

flux of F across S = int_S curl(F) . dS

= int_C F . dr

= 1 - 2/l

To determine the value of l that gives us the part of the plane in the first octant oriented upward, we need to find the intersection of the plane with the coordinate planes. The intersection with the xy-plane is given by 2x + ly = 2, which intersects the positive x- and y-axes at (1,0) and (0,2/l), respectively. The intersection with the xz-plane is given by 2x + z = 2, which intersects the positive x- and z-axes at (1,0,1) and (0,0,2), respectively. Therefore, the part of the plane in the first octant oriented upward is given by:

S = (x,y,z)

To compute the flux of F across S, we need to compute the curl of F:

curl(F) = (2i - j)

Now, we can apply the surface integral:

flux of F across S = int_S curl(F) . dS

= int_0^1 int_0^(2/l) int_0^(2-2x-ly) (2i-j) . (i j -k) dz dy dx

= int_0^1 int_0^(2/l) [z(2i-j)]_0^(2-2x-ly) dy dx

= int_0^1 int_0^(2/l) [(2-2x-ly)(2i-j)] dy dx

= int_0^1 [2y-yl-2y^2/l]_0^(2/l) dx

= int_0^1 (4/l - 2/l^2 - 8/l^3) dx

= 4/l - 1/l^2 - 4/l^3

Therefore, the flux of F across S is given by 4/l - 1/l^2 - 4/l^3.

User Darksaga
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