Question

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z) = z2i + 3xyj + 3y2k.Find the work done.
C


F · dr =

Answer 1

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 $\int_{C}^{}F.dx$.

According to the question the force $F=z^{2}i+5xyj+4y^2k$ and all four points are in the plane $z=\frac{1}{4}y$.

therefore, if S is the at surface with boundary C, so that S is the portion of the plane $z=\frac{1}{4}y$ over the rectangle D=[0,2]\times [0,4].

Now,

$curlF=\begin{vmatrix}i &j &k \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ z^{2}&5xy &4y^{2} \end{vmatrix}=8yi+2zj+5yk$

By the Stoke's theorem:

$\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA$

$=\iint_{D}^{}\left ( \frac{39}{8}y \right )dA\, \, \, \, \, \, \, \, \, Since\, \, z=\frac{1}{4}y\\\\=\int_{0}^{2}\int_{0}^{4}\frac{39}{8}y\, \, \, dy\, dx$

$\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \, dx\\\\=39\times 2\\\\=78 \, N$