# Work done by a force field $F$ via the line integrals

Let $F:\mathbb R^2\to \mathbb R^2$ be the force field with

$$F(x,y) = -\frac{(x,y)}{\sqrt{x^2 + y^2}}$$

the unit vector in the direction from $(x,y)$ to the origin. Calculate the work done against the force field in moving a particle from $(2a,0)$ to the origin along the top half of the circle $(x−a)^2+y^2=a^2$.

Okay, I tried to use the line integral and I set $x=a+a\cos(t)$, $y= a\sin(t)$ and $t\in [0,\pi]$. Then the work should be

$$\int_0^\pi F(r (t))(r)′dt$$

But I can't got the right answer!!

• Can you show us your calculations? – user99914 Nov 15 '13 at 3:38
• $y$ should be $y=a\sin t$ – Shuchang Nov 15 '13 at 3:44

Your vector field is conservative: $\nabla \times F = 0$. Thus the integral is path independent. This should simply your calculation considerably—choose the easy straight line path from $(2a,0)$ to $(0,0)$ and integrate.
$\vec{F} = -\nabla r$ where $r \equiv \sqrt{x^{2} + y^{2}}$ $$\color{#0000ff}{\large W} = \int_{\left(2a,0\right)}^{\left(0,0\right)}\vec{F}\cdot{\rm d}\vec{r} = \int_{\left(2a,0\right)}^{\left(0,0\right)}\left(-\nabla r\right)\cdot{\rm d}\vec{r} = \sqrt{\left(2a\right)^{2} + 0^{2}} - \sqrt{0^{2} + 0^{2}} = \color{#0000ff}{\large 2\left\vert a\right\vert}$$