Answer: Work done by the particle is 78 N.
Step-by-step explanation:
If C is the path the particle follows, then work done is [tex]\int_{C}^{}F.dx[/tex].
According to the question the force [tex]F=z^{2}i+5xyj+4y^2k[/tex] and all four points are in the plane [tex]z=\frac{1}{4}y[/tex].
therefore, if S is the at surface with boundary C, so that S is the portion of the plane [tex]z=\frac{1}{4}y[/tex] over the rectangle D=[0,2]\times [0,4].
Now,
[tex]curlF=\begin{vmatrix}i &j &k \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ z^{2}&5xy &4y^{2} \end{vmatrix}=8yi+2zj+5yk[/tex]
By the Stoke's theorem:
[tex]\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA[/tex]
[tex]=\iint_{D}^{}\left ( \frac{39}{8}y \right )dA\, \, \, \, \, \, \, \, \, Since\, \, z=\frac{1}{4}y\\\\=\int_{0}^{2}\int_{0}^{4}\frac{39}{8}y\, \, \, dy\, dx\\\\[/tex]
[tex]\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \, dx\\\\=39\times 2\\\\=78 \, N[/tex]
