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A particle moves along line segments from the origin to the points (2, 0, 0), (2, 5, 1), (0, 5, 1), and back to the origin under the influence of the force field
[tex]F(x,y,z) = z^2i+5xyj+2y^2k[/tex]
Use Stokes Theorem to find the work done:

Sagot :

okpalawalter8

A particle moves along line segments from the origin to the points, work done is mathematically given as

W=184.5units

What is the solution of an equation that matches the model.?

Considering line segments from (0,0,0)origin to (2,0,0) , (2,5,1) and (0,5,1) under the infulence of force

Generally,  the equation for force is mathematically given as

F = z2i + 5xyj + 2y2k

Therefore, Considering u*v

[tex]u * v = (u_1j+u_2j+u_3k) * (v_1i+v_2j+v_3k)[/tex]

[tex]u* v = u_1v_1(i * i) + u_1v_2(i * j)+u_1v_3(i * k) + u_2v_1(j * i) + u_2v_2(j * j)+u_2_3(j* k) + u_3v_1(k * i) + u_3v_2(k * j)+u_3v_3(k * k)[/tex]

Where

[tex]i * i = j *j=k * k=0[/tex]

Hence

[tex]u* v = u_1v_2k-u_1v_3j-u_2v_1k+u_2v_3i +u_3v_1j - u_3v_2i[/tex]

[tex]u = (u_1,u_2,u_3) = (0,5,1)\\\\v = (v_1,v_2,v_3) = (-3,0,0)[/tex]

The normal equation formed

-2y + 15z = 0

z= (1/5)y

Considering the level surface and differential surface area

h(x,y,z) = -y + 5z =0

[tex]dS = |grad(h)| dA[/tex]

In terms of the x and y coordinates of (2,0,0) and (2,5,1) and (0,5,1), we can state that the ranges are 0 to 3 and 5 respectively we have

[tex]0 \leq x \leq 2 \ and \ 0 \leq y \leq 5[/tex]

Using strokes theorem to evaluate

[tex]F = z2i + 5xyj + 2y2k[/tex]

[tex]curl \ F = 4yi+2zj+5yk = (4y,2z,5y)[/tex]

[tex]curl \ F * nds= \frac{1}{5}(-2z + 25y)\ dy \ dx[/tex]

In conclusion, The work done is

[tex]W=\int _CF *dr[/tex]

[tex]\int _C F * dr = \int \int curl \ F * n ds[/tex]

[tex]\int _C F *dr = \frac{123}{2}\int_0^3 \ dx[/tex]

[tex]\int _C F * dr = \frac{123*3}{2}[/tex]

W= 184.5units

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