Answer:
The circulation of F is 0.
Step-by-step explanation:
Consider the provided vector field.
Some times F tends to clockwise and some time F tends to anti clock wise, Therefore the circulation of F is 0.
It is given that [tex]F=\left \langle y-x,x\right \rangle[/tex] and [tex]r(t)=\left \langle-9\text{cos} t , 9\text{sin}t\right \rangle[/tex]
[tex]F(t)=\left \langle 9\sin t+9\cos t,-9\cos t\right \rangle[/tex]
[tex]dr=\left \langle 9\sin t+9\cos t\right \rangle[/tex]
[tex]F(t)\cdot dr=\left \langle 9\sin t+9\cos t,-9\cos t,\right \rangle \cdot \left \langle 9\sin t+9\cos t\right \rangle[/tex]
[tex]F(t)\cdot dr=81\sin^2t+81\sin t\cos t-81\cos^2t[/tex]
[tex]F(t)\cdot dr=-81\cos2t+\frac{81}{2}\sin2 t[/tex]
[tex]\int\limits^a_b {F\cdot \, dr=\int\limits^{2\pi}_0 {-81\cos2t+\frac{81}{2}\sin2 t} \, dt[/tex]
[tex]=[\frac{-81\sin2t}{2}-\frac{81cos2t}{4}]^{2\pi}_0\\=\frac{-81}{4}+\frac{81}{4}=0[/tex]
Hence the circulation of F is 0.
