The drain of a bathtub is in the xy-plane with center at the origin and radius 1 cm. Find ∫c F·dr where C is the edge of the drain, oriented clockwise when viewed from above and where
F = 1/2.
F = 1/2
-
First of all, your C is rotated in the wrong direction, so simply replace your solution for its negative. ∫C = -∫-C
The counter-clockwise curve -C can be traced out by the position vector r = < cos t, sin t, 0 >, as t goes from 0 to 2π.
dr/dt = <-sin t, cos t, 0>
dr = <-sin t, cos t, 0> dt
Since we are looking for ∫-C F ⋅ dr we can substitute
∫-C F ⋅ <-sin t, cos t, 0> dt
And all x= r cos t, y= r sin t, z=0, and r=1
F = 1/2 turns into
F = 1/2
∫-C F ⋅ <-sin t, cos t, 0> dt
∫-C 1/2 ⋅ <-sin t, cos t, 0> dt
∫-C 1/2 (-sin² t - cos² t + 0) dt
∫-C -1/2 dt
-t/2 ]]-C
-π
But your original question was for the clockwise curve C. So we negative the answer.
π
The counter-clockwise curve -C can be traced out by the position vector r = < cos t, sin t, 0 >, as t goes from 0 to 2π.
dr/dt = <-sin t, cos t, 0>
dr = <-sin t, cos t, 0> dt
Since we are looking for ∫-C F ⋅ dr we can substitute
∫-C F ⋅ <-sin t, cos t, 0> dt
And all x= r cos t, y= r sin t, z=0, and r=1
F = 1/2
F = 1/2
∫-C F ⋅ <-sin t, cos t, 0> dt
∫-C 1/2
∫-C 1/2 (-sin² t - cos² t + 0) dt
∫-C -1/2 dt
-t/2 ]]-C
-π
But your original question was for the clockwise curve C. So we negative the answer.
π