# Line integral conservative vector field

Let $F(x; y) = (-g(t)y; g(t)x)$, where $t = x^{2} + y^{2}$ and $g(t)$ is a function (of single variable) which is differentiable (class $C^1$) for $t > 0$.

I have to calculate the line integral on the unit circle, $F \cdot dr$, counter clockwise.

How does the information that $g(t)$ is differentiable help us ?

• you may use Stokes theorem, for which differentiability is needed – vidyarthi Nov 26 '16 at 9:14
• thanks but we are not supposed to do that because we did not go through it yet , this is on the problem set of green's theorem , line integrals and conservative vector fields. – Kasmir Khaan Nov 26 '16 at 9:16
• Then use Green's theorem with $\vec{r}=x\bar{i}+y\bar{j}$ – vidyarthi Nov 26 '16 at 9:18
• Can you please show me how ? i got stuck when i did that , because i end up taking double integral of 2 g(t) witch depend on the function g . – Kasmir Khaan Nov 26 '16 at 9:20

Hint. If $(x,y)$ is on the unit circle centred at the origin then $$\mathbf{F}(x; y) = (-g(t)y; g(t)x)=g(1)(-y,x).$$ Hence $$\int_{C}\mathbf{F}\cdot d\mathbf{r}=g(1)\int_0^{2\pi}\left(-\sin t,\cos t\right) \cdot\left(-\sin t,\cos t\right)dt.$$ P.S. If $g(1)\not=0$ then the vector field $\mathbf{F}$ is not conservative in a simply connected region which contains the origin. To be conservative we need that $$-g'(t)t_y\cdot y-g(t)=(-g(t)y)_y=(g(t)x)_x=g'(t)t_x\cdot x+g(t)$$ that is $-g(t)=g'(t)t$ which implies $g(t)=\frac{C}{t}$.
• To be conservative you need that $(-g(t)y)_y=(g(t)x)_x$. – Robert Z Nov 26 '16 at 10:46