Question

Is the integral ſ A. dr path independent if A = (2x - y)i + (x + y)j?

Answer 1

The integral is path-independent if we can find a scalar function f such that grad(f ) = A. This requires

$\dfrac{\partial f}{\partial x}=2x-y$

$\dfrac{\partial f}{\partial y}=x+y$

Take the first PDE and integrate both sides with respect to x to get

$f(x,y)=x^2-xy+g(y)$

where g is assumed to be a function of y alone. Differentiating this with respect to x gives

$\dfrac{\partial f}{\partial y}=-x+\dfrac{\mathrm dg}{\mathrm dy}=x+y\implies\dfrac{\mathrm dg}{\mathrm dy}=2x+y$

which would mean g is *not* a function of only y, but also x, contradicting our assumption. So the integral is path-dependent.

Parameterize the unit circle (call it C) by the vector function,

$\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j$

with t between 0 and 2π.

Note that this parameterization takes C to have counter-clockwise orientation; if we compute the line integral of A over C, we can multiply the result by -1 to get the value of the integral in the opposite, clockwise direction.

Then

$\mathrm d\mathbf r=-\sin t\,\mathbf i+\cos t\,\mathbf j$

and the (counter-clockwise) integral over C is

$\displaystyle\int_C\mathbf A\cdot\mathrm d\mathbf r$

$\displaystyle=\int_0^{2\pi}((2\cos t-\sin t)\,\mathbf i+(\cos t+\sin t)\,\mathbf j)\cdot(-\sin t\,\mathbf i+\cos t\,\mathbf j)\,\mathrm dt$

$\displaystyle=\int_0^{2\pi}1-\sin t\cos t\,\mathrm dt=2\pi$

and so the integral in the direction we want is -2π.

By the way, that the integral doesn't have a value of 0 is more evidence of the fact that the integral is path-dependent.