Given the path C which is a circle with equation $(x-9)^2 +y^2 =36$ in the anticlockwise direction. Give an example of a vector field $F(x,y)$ that satisfies $\oint_C \vec F \cdot dr = 11\pi$
I used the parametrization $(x,y)=(9+6cos(t),6sin(t))$ and I think I need to use green's theorem to find a function (for example $F(x, y) = x^2\mathbf{i} + xy\mathbf{j} $ ) that satisfies $\oint_C F(r(t))\cdot r'(t)\,dt =11\pi$
- Is my parametrization correct since the circle is centered off-origin?
- Is this only solvable by trial and error of multiple functions or is there something I'm missing?
Your parametrization is correct.
If the vector field is $(P, Q)$ then as per Green's theorem,
$\oint_C P \, dx + Q \, dy = \iint_D (Q_x - P_y) \,dA$
Given area of the circle is $36 \pi$, some easy examples of the vector field resulting in line integral of $11\pi$ will be where $Q_x - P_y = \frac{11}{36}$.
We can pick $\vec{F} = (c, \frac{11x}{36}) \,$ where $c$ is any constant or you can pick $(y, \frac{47x}{36})$.
For the first vector field, the line integral will turn out to -
$\int_0^{2\pi} (-6 c \sin t + 11 \cos^2t + \frac{33}{2} \cos t) \, dt = 11 \pi$
For the second vector field, the line integral will turn out to -
$\int_0^{2\pi} (47 \cos^2t - 36 \sin^2t + \frac{141}{2} \cos t) dt = 11 \pi$