Currently I'm taking vector calculus lectures and recently, I was given some problems about line integrals of some vector field $\vec F$ along a region. And among which most of the regions stated is CIRCLE.
Out of boredom, I attempt to generalize the result of line integrals of arbitrary vector field $\vec F:[u(x,y),v(x,y)]$ along a circle $C: x^2+y^2=r^2$, where $u$ and $v$ are scalar functions of $(x,y)$. (You may assume $u$ and $v$ are continuous and differentiable.)
Here is my partial statement:
Let $u(x,y)=\vec p(x)\cdot \vec q(y)=[p_1(x),p_2(x),\cdots,p_n(x)]\cdot[q_1(y),q_2(y),\cdots,q_n(y)]$, where $p_i(x)$ and $q_i(y)$ are scalar function of $x$ and $y$ respectively. Similarly, we denote $v(x,y)=\vec\alpha(x)\cdot\vec\beta(y)$.
Denote $\frac{\partial b}{\partial y}=\beta$.
\begin{align} I&=\int_C\vec F(\vec r)d\vec r=\iint_R\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)dxdy \\&=\iint_R\vec\alpha\prime(x)\cdot\vec\beta(y)-\vec p(x)\cdot \vec q\prime(y)dxdy \\&=\int^r_{-r}[\vec\alpha\prime(x)\cdot\vec b(y)-\vec p(x)\cdot \vec q(y)]^{\sqrt{r^2-x^2}}_{-\sqrt{r^2-x^2}}dx \\&=\int^r_{-r}\left(\vec\alpha\prime(x)\cdot\vec b(\sqrt{r^2-x^2})-\vec p(x)\cdot \vec q(\sqrt{r^2-x^2})\right)-\left(\vec\alpha\prime(x)\cdot\vec b(-\sqrt{r^2-x^2})-\vec p(x)\cdot \vec q(-\sqrt{r^2-x^2})\right)dx \end{align}
Let
\begin{align} J_1&=\int^r_{-r}\vec\alpha\prime(x)\cdot\vec b(\sqrt{r^2-x^2})dx\\&=\int^r_{-r}\vec b(\sqrt{r^2-x^2})d[\vec\alpha(x)]\\&=[\vec\alpha(x)\cdot\vec b(\sqrt{r^2-x^2})]^r_{-r}-\int^r_{-r}\vec\alpha(x)\cdot\vec\beta(\sqrt{r^2-x^2})d[\sqrt{r^2-x^2}]\\&=[\vec\alpha(r)-\vec\alpha(-r)]\cdot\vec b(0)-\int^r_{-r}v(x,\sqrt{r^2-x^2})d{\sqrt{r^2-x^2}} \end{align}
Similarly,
\begin{align} J_2&=\int^r_{-r}\vec\alpha\prime(x)\cdot\vec b(-\sqrt{r^2-x^2})dx\\&=[\vec\alpha(r)-\vec\alpha(-r)]\cdot\vec b(0)+\int^r_{-r}v(x,-\sqrt{r^2-x^2})d{\sqrt{r^2-x^2}} \end{align}
Also
$$K_1=\int^r_{-r}\vec p(x)\cdot\vec q(\sqrt{r^2-x^2})dx=\int^r_{-r}u(x,\sqrt{r^2-x^2})dx$$ $$K_2=\int^r_{-r}\vec p(x)\cdot\vec q(-\sqrt{r^2-x^2})dx=\int^r_{-r}u(x,-\sqrt{r^2-x^2})dx$$
Bringing these integrals together and we get:
\begin{align} I&=J_1-K_1-J_2+K_2\\&=\int^r_{-r}\left[u(x,-\sqrt{r^2-x^2})-u(x,\sqrt{r^2-x^2})\right]dx-\int^r_{-r}\left[v(x,\sqrt{r^2-x^2})+v(x,-\sqrt{r^2-x^2})\right]d{\sqrt{r^2-x^2}} \end{align}
This is what I've attempt. But can this expression be further simplified?
Current aim: Give a simpler form without integral signs. Smarter approaches are also welcomed.