Answer:
0
Step-by-step explanation:
We want to compute the curve integral (or line integral)
[tex] \bf \int_{C}F[/tex]
where the force field F is defined by
F(x,y) = (y, -x)
and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).
We can write
C = [tex] \bf C_1+C_2[/tex]
where
[tex] \bf C_1[/tex] = line segment from (1, 5) to (0, 0)
[tex] \bf C_2[/tex] = line segment from (0, 0) to (0, 9)
so,
[tex] \bf \int_{C}F=\int_{C_1}F+\int_{C_2}F[/tex]
Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with
r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1
Hence [tex] \bf C_1[/tex] can be parameterized as
[tex] \bf r_1(t) = (1-t, 5-5t)[/tex] for 0 ≤ t ≤ 1
and [tex] \bf C_2[/tex] can be parameterized as
[tex] \bf r_2(t) = (0, 9t)[/tex] for 0 ≤ t ≤ 1
The derivatives are
[tex] \bf r_1'(t) = (-1, -5)[/tex]
[tex] \bf r_2'(t) = (0, 9)[/tex]
and
[tex] \bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0[/tex]
[tex] \bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0[/tex]
In consequence,
[tex] \bf \int_{C}F=0[/tex]
