Question

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

F(x, y, z) = z2i + 2xyj + 5y2k.

Find the work done.

Answer 1

Answer:

the work done which is the lie integral = 48

Step-by-step explanation:

From the given question the normal to the plane formed  by the point is estimated as follows:

n = (2,0,0) (0,5,1) = (0,-2,10)

On the plane line, the equation is : -2y + 10z =0

10 z = 2y

$z = \dfrac{2y}{10}$

$z = \dfrac{y}{5}$

Let take an integral look at this function; Vg(x,y,z) = V(-2y+10z) = (0,-2,10)

$|V_g|(x,y,z)|= 2 \sqrt{26}$

$dS = \sqrt{1+z_x^2+xz_y^2} dA$

dS = $\dfrac{\sqrt{26} }{5} \ dA$

By definition; The curl of the vector field is:

$Curl \ F= (\dfrac{\partial R}{\partial y}- \dfrac{\partial Q}{\partial z})i + ( \dfrac{\partial P }{\partial z} - \dfrac{\partial R}{\partial x}) j + ( \dfrac{\partial Q }{\partial x}- \dfrac{\partial P}{\partial y}) k$

Given that F: z² i + 2xy j + 5y² k

$\bigtriangledown \times F = \begin{pmatrix} i \ \ \ \ j \ \ \ \ k \\ \\ \dfrac{\partial}{\partial x} \ \ \dfrac{\partial}{\partial y } \ \ \dfrac{\partial}{\partial z} \\ \\ z^2 \ \ 2xy \ \ 5y^2 \end{pmatrix} = \langle (10y -0), -(0-2z),(2y-0) \rangle$

$\bigtriangledown \times F = \langle 10y,2z,2y \rangle$

Now; by applying stokes theorem, we have :

$\int_c F \times dr = \int \int_s (\bigtriangledown \times F ) \times dS$

$\int_c F \times dr = \int \int_s \langle 10 y,2z,2y \rangle \times \dfrac{1}{2 \sqrt{26}} \langle 0,-2,10 \rangle\dfrac{\sqrt{26}}{5}dA$

$\int_c F \times dr =\dfrac{1}{10} \int \int_s \langle 10 y,2z,2y \rangle \times \langle 0,-2,10 \rangle \ dA$

$\int_c F \times dr =\dfrac{1}{10} \int \int_s \langle -4z+20y \rangle \ dA$

$\int_c F \times dr =\dfrac{1}{10} \int \int_s \langle -4 (\dfrac{1}{5}y)+20y \rangle \ dA$

$\int_c F \times dr =\dfrac{1}{10} * \dfrac{96}{5}\int \limits ^2_{x=0}\int \limits ^5_{y=0} \ ydydx$

$\int_c F \times dr =\dfrac{48}{25} * (2) * \dfrac{25}{2}$

$\int_c F \times dr =48$

the work done which is the lie integral = 48