I have a problem that I just don't get.
I am supposed to evaluate the line integral for the work done by the two-dimensional force F = (y, 2x) going from the origin O to the point P = (1,1). The path goes from O to Q = (1,0) along the x-axis, and from Q straight up to P = (1,1) along the y-axis. The textbook define the integrals as follow (this is an example in the book):
$\int_{0}^{1} F_x(x,0)$ $dx$ + $\int_{0}^{1} F_y(1,y)$ $dy$
= 0 + 2 $\int_{0}^{1} dy$ = 2
I understand how they get $F_x(x,0)$ and $F_y(1,y)$ (I think). What I don't understand is how this becomes 0 and 2.
Best Answer
Notation can be very confusing. Your force is given by the function: $F(x,y) = \langle y, 2x \rangle$.
Therefore, the components of the force are given by the functions:
$F_x(x,y)= y$, and $F_y(x,y)= 2x$
On the first path, $x$ is varying but y is constantly 0, so:
$F_x(x,y)= F_x(x,0)= 0$
On the second path, $y$ is varying but x is constantly 1, so:
$F_y(x,y)= F_y(1,y)= 2(1)=2$