Calculate the line integral of F⃗ F→ along the curve CC.?
Suppose
F(x,y)=9sin(x/2)sin(y/2)i ,−9cos(x/2)cos(y/2)j
and C is the curve from P to Q in the figure. Calculate the line integral of F along the curve C.
The labeled points are P=(−3π/2,3π/2), Q=(−3π/2,−3π/2), R=(3π/2,3π/2), and S=(3π/2,−3π/2). The curves PR and SQ are trigonometric functions of period 2π and amplitude 1.
∫C[F]⋅dr = ?
The C is supposed to be down in the bottom right corner of the integral.
I know that I only have to worry about going from P to Q, but I m confused what I have to do the given F(x,y) equation to get the answer. Thanks!
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For F = Pi + Qj and parametric curve C: r(t) = x(t)i + y(t)j, then
∫c F⋅dr = ∫c Pdx + Qdy
Now it doesn't matter what the curves PR and SQ are. They are irrelevant.
We are only interested in the curve from P to Q, which you say nothing about.
However, since value of integral is independent of the path, we can use the simplest path from P to Q ----> a vertical line
C:
Straight line from P(−3π/2, 3π/2) to Q(−3π/2, −3π/2)
x = −3π/2 ----> dx = 0 dt
y = −t ----> dy = −1 dt
−3π/2 ≤ t ≤ 3π/2
F = < 9sin(x/2)sin(y/2) ,−9cos(x/2)cos(y/2) >
F = < 9sin(−3π/4)sin(−t/2) ,−9cos(−3π/4)cos(−t/2) >
F = < 9(−√2/2)(−sin(t/2)) ,−9(−√2/2)cos(t/2) >
F = < 9√2/2 sin(t/2) , 9√2/2 cos(t/2) >
∫c F⋅dr
= ∫c Pdx + Qdy
= ∫[−3π/2 to 3π/2] (9√2/2 sin(t/2)) (0 dt) + (9√2/2 cos(t/2)) (−1 dt)
= ∫[−3π/2 to 3π/2] (−9√2/2) cos(t/2) dt
= −9√2 sin(t/2) | [−3π/2 to 3π/2]
= −9√2 [sin(3π/4) − sin(−3π/4)]
= −9√2 [√2/2 + √2/2]
= −18