Evaluating Int. on V Along C Using Greens Theorem

[imagemania]

Homework Statement

V = (3y^2 - sin(x)) i +(6xy+√(y^4+1))j along teh closed path C defined by x^2 + y^2 =1, counterclockwise direction

Homework Equations

Greens Thoerem

The Attempt at a Solution

I am stuck on the limits part of the integration. I get so far into greens theorem to obtain:
∫∫ [6y√(y^4+1) -6y].dx.dy

But unsure how to go beyond this, i can't see any clear x or y limits. I could convert into polar, but that also looks very messy.

Any help/point in the right direction will be appreciated!

 

[vela]

Recheck your integrand. It looks complicated because you made a mistake.

The limits should be straightforward to get. Hint: What shape does the equation x²+y²=1 define?

 

[imagemania]

Ohh i see where i made teh mistake. I now get within the integral 6y - 6y = 0.

Is this correct?

thanks :)

 

[vela]

Yup. You could always do the line integral to verify. :)

 

[imagemania]

I've just thought, if you had any path then, regardless of its involvement, this must always be zero if the force is still the same. Take the same force and the equation xi + cos(x) j clockwise and x ≤ |π/2 |and y = 0.

If this is true and any path results in zero work done, then do we simply call this a conservative force?

 

[vela]

Green's theorem is a special case of Stoke's theorem, which says

$$\int_S (\nabla\times \mathbf{F})\cdot d\mathbf{A} = \oint_{\partial S} \mathbf{F}\cdot d\mathbf{r}$$

A conservative force satisfies ∇xF = 0, so the integral around any closed path will be 0. In this case, by applying Green's theorem, you found ∇xF does indeed vanish, so yes, it's a conservative force.


Note you can also look at this problem as being of the form ∫ (M(x,y) dx + N(x,y) dy). Since you had ∂M/∂y = ∂N/∂x, the integrand is an exact differential, meaning you can, in principle, find a function Φ(x,y) such that dΦ = M(x,y) dx + N(x,y) dy. The function Φ is the potential energy function (to within a minus sign), which, again, exists because F is conservative.