Line integral and parametrization

Question

A vector field $F= (\sin y) i + x(1+\cos y)j$. Evaluate the line integral over the cirxular path $x^2+y^2=a^2,z=0$? Answer: $\pi a^2$. I know how to do it using parametric form of circle but i really want to know how to do it without parametrization, since it doesn't matter therefore answer should be same with both methods. I am getting integrals of $\sin (\cos u),\tan u\cos(\sin u)$ without using parametrization, and these integrals involve some $H$ and $J$ as i saw how to integrate them. But my actual problem is please someone verify without using parametrization you are getting $\pi a^2$.

Answer 1

but i really want to know how to do it without parametrization

I assume you mean by using a theorem to avoid having to compute the line integral directly (via a parametrization of the circle). Since the path is closed, you can apply Green's theorem.

With the notation from that page, you have $L=\sin y$ and $M = x\left( 1+\cos y\right)$ so: $$\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y} = \frac{\partial \left( x\left( 1+\cos y\right) \right)}{\partial x}-\frac{\partial \left( \sin y \right)}{\partial y} = 1+\cos y - \cos y = 1$$ This makes the calculations very easy; integrating over the disc $D$ bounded by the path (circle $C$): $$\int_C \vec F \cdot \mbox{d} \vec r = \iint_D \left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right) \,\mbox{d}x\,\mbox{d}y = \iint_D 1 \,\mbox{d}x\,\mbox{d}y = \mbox{Area}(D)=\pi a^2$$