Consider the line integral \(\int_C 2x\,dx + x^2y\,dy\), where \(C\) is the boundary of the unit square in the 1st quadrant. Evaluate the line integral (a) directly and (b) by using Green’s theorem.
Solution
1. The value depends on the orientation of \(C\). For the positive (counterclockwise) orientation,\[ \int_C 2x\,dx+x^2y\,dy=\frac12. \]For the clockwise orientation the value is \(-\frac12\).The source does not specify an orientation; the steps below compute the positively oriented integral, which is the orientation Green’s theorem uses.
2. Compute the positively oriented integral directly by traversing the four sides\[ (0,0)\to(1,0)\to(1,1)\to(0,1)\to(0,0). \]
3. The bottom side contributes \(1\).
Along \(y=0\), parameterize by \((x,0)\) for \(0\le x\le1\). Then \(dy=0\), so \[ \int_0^1 2x\,dx=1. \]
4. The right side contributes \(\frac12\).
Along \(x=1\), parameterize by \((1,y)\) for \(0\le y\le1\). Then \(dx=0\), so \[ \int_0^1 y\,dy=\frac12. \]
5. The top side contributes \(-1\), and the left side contributes \(0\).
Along the top side, \(y=1\) and \(x\) runs from \(1\) to \(0\), hence \[ \int_1^0 2x\,dx=-1. \] Along the left side, \(x=0\), so both terms vanish.
6. Therefore the direct computation gives\[ 1+\frac12-1+0=\frac12. \]
7. Green’s theorem gives the same value.
Set \(P(x,y)=2x\) and \(Q(x,y)=x^2y\). For the unit square \(D=[0,1]^2\) with positively oriented boundary, \[ \int_C P\,dx+Q\,dy =\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA =\int_0^1\int_0^1 2xy\,dx\,dy =\frac12. \]
8. Reversing the orientation changes the sign of every line integral, giving \(-\frac12\) for clockwise orientation.