Part 3: Path Independence

If C₁ and C₂ are both curves in the region R beginning at point A and ending at point B, then ò_C1F·dr = U( B) -U(A) and ò_C2F·dr = U( B)-U( A) . Consequently, 

ó õ C1  F·dr = ó õ C2  F·dr

(2)

As a result, we say that ò_CF·dr is independent of path over the region R. Indeed, if F(x,y) is conservative and C is a curve that begins at A and ends at B, then theorem 2 is also written in the form

ó õ B A  F·dr = U( B) -U( A)

(3)

EXAMPLE 5    Use theorem 2 to evaluate the line integral

ó õ ( 1,2) ( 0,0)   2xdx+2ydy

Solution: To do so, we write the integral in the form ò_A^BF·dr, which yields

ó õ ( 1,2) ( 0,0)   2xdx+2ydy = ó õ ( 1,2) (0,0)   á 2x,2y ñ · á dx,dy ñ

so that the vector field is F( x,y) = á2x,2y ñ . Since M = 2x and N = 2y, we have M_y = 0 and N_x = 0, thus implying that F( x,y) = á2x,2y ñ is conservative. To find its potential, we integrate M = 2x with respect to x to obtain

U( x,y) = ó õ 2xdx = x2+C( y)

Since U_y = C^¢( y) and since N = 2y, we have C^¢( y) = 2y so that C( y) = y²+k. Thus, the potential is

U( x,y) = x2+y2+k

so that by theorem 2 and (3), we have

ó õ ( 1,2) ( 0,0)   2xdx+2ydy = (x2+y2+k) ( 1,2) ( 0,0) = ( 12+22+k) -( 02+02+k) = 5

LiveGraphics3d Applet

ó õ

C

F·dr »4.9996

Potential Difference: U(Black) - U(Green) = jac

Move the green dot to the beginning of the curve. Move the black dot to the end of the curve.
Does the potential difference match the work integral?

       

Conversely, theorem 2 can be used to means of computing potentials. In particular, if F( x,y,z) is conservative over an open connected solid S and if we require the potential for F to satisfy U( A) = 0, where A is a fixed point in S called ground, then theorem 2 implies that

U( x,y,z) = ó õ ( x,y,z) A  F·dr

(4)

In particular, U( x,y,z) is defined as a path independent line integral over F even when U( x,y,z) cannot be expressed in closed form.       

EXAMPLE 6    Show that F( x, y, z) = áy,x,sin( z²) ñ is conservative. Then express its potential as an integral over the parameter t by letting A = (0,0,0) and using the parameterized curve r( t) = á xt, yt, zt ñ , t in [ 0,1] .       

Solution: Identifying M = y, N = x, and P = sin( z²) leads to

curl  F =  ¶sin( z2) ¶y -  ¶x ¶z ,  ¶y ¶z -  ¶sin( z2) ¶x ,  ¶x ¶x -  ¶y ¶y = á0,0,1-1 ñ

Thus, F is conservative, and as a result, (4) implies that

U( x,y,z) = ó õ ( x,y,z) ( 0,0,0)   F·dr

(5)

Notice now that r( t) = áxt,yt,zt ñ implies that

M( xt,yt,zt) = yt,  N( xt,yt,zt) = xt,  and  P(xt,yt,zt) = sin( z2t2)

In addition, dr/dt = á x,y,z ñ , so that if we now pull back the line integral (5) to the parametrization t, we obtain

U( x,y,z) = ó õ 1 0  F·  dr dt dt = ó õ 1 0  á yt,xt,sin( z2t2) ñ · á x,y,z ñ   dt

Evaluating the inner product then leads to

U( x,y,z) = ó õ 1 0  [ x( xt) +y(yt) +zsin( z2t2) ] dt = ó õ 1 0  [ x2t+y2t+zsin( t2z2) ] dt

Moreover, although we cannot find U( x,y,z) in closed form, we can at least separate the integrand and integrate the first two terms:

U( x,y,z) = ( x2+y2) ó õ 1 0  tdt+z ó õ 1 0  sin( t2z2) dt = ( x2+y2)  t2 2 ê ê 1 0  +z ó õ 1 0  sin( t2z2) dt =  x2+y2 2 +z ó õ 1 0  sin( t2z2) dt