Question

A particle moves along line segments from the origin to the points (3, 0, 0), (3, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field
F(x, y, z) = z2i + 5xyj + 4y2k.

I have tried multiple times to solve this problem but the answer is still not right.

Answer 1

Here is some of the work so far that I have had helped with:

A (3, 0, 0), B (3, 3, 1), C (0, 3, 1)
the plane containing these points has normal vector
n = <3 -3, 0 -3, 0 -1> x <3 -0, 3 -3, 1 -1> = <0, -3, -1> x <3, 0, 0> = <0, -3, 9>
then curl F = del x F = <d/dx, d/dy, d/dz> x <z^2, xy, 4y^2> = <8y -0, -(0 -2z), 5y -0> = <8y, 2z, 5y>
by Stokes theorem, the line integral around the path on which the particle allegedly travels is the surface integral of the dot product of the curl F and the normal to that surface
the region of integration is the projection of the plane onto the xy-plane, a triangle of vertices
(3, 0), (3, 3), (0, 3)
which corresponds to the region bounded by the lines
y = 3
x = 0
y = -x +3
x = 3
n dot curl F = <0, -3, 9> dot <8y, 2z, 5y> = 0 -6z +45y
need to solve for z in terms of x, y, using the equation of the tangent plane containing each point
choosing point A (3, 0, 0)
n dot <x -x0, y -y0, z -z0> = 0
<0, -3, 9> dot <x -3, y, z> = 0 -3y +9z = 0
9z = 3y
z = y/3
evaluate the double integral of
-6(y/3) +45y dy dx from [3, -x +3] [3, 0]
= 43y^2 /2 dy dx from [3, -x +3] [3, 0]
= 43*9/2 -(43/2)(-x +3)(-x +3) dx [3, 0]
= 43*9/2 -(43/2)(x^2 -6x +9) dx [3, 0]
= 43*9/2 x -(43/2)(x^3 /3 -6x^2 /2 +9x) [3, 0]
= 43*9/2 *3 -(43/2)(27/3 -6*9/2 +27)
= 387

Answer 2

the origin is the point (0, 0, 0) not (3, 0, 0).

Answer 3

𝖜𝖍𝖆𝖙 𝖙𝖍𝖊 𝖋𝖚𝖈𝖐