Consider the vector field K: \mathbb R^2 \to \mathbb R^2 with
K(x,y) =\begin{pmatrix} A y + e^{\sin(x)} \\
Bx + \sqrt{y^{C} + D} \end{pmatrix}
and the rectangle below with boundary curve \gamma = \partial R.
Compute
\displaystyle \oint_\gamma K d\gamma.
\displaystyle \oint_\gamma K d\gamma=
\displaystyle \oint_\gamma K d\gamma=
If a vector field looks so complicated as K(x,y) =\begin{pmatrix} Ay + e^{\sin(x)} \\
Bx + \sqrt{y^{C} + D} \end{pmatrix} =
\begin{pmatrix} P(x,y) \\ Q(x,y) \end{pmatrix}
it might be a good idea to apply Green's formula.
This states for a curve with positive orientation that \displaystyle \oint_{\gamma = \partial R} K d\gamma = \iint_R (Q_x - P_y)(x,y) dA.
Hier, the orientation of the boundary curve \gamma is positive.
Hier, the orientation of the boundary curve \gamma is negative.
Flipping the orientation gives a positive one for the curve \overline{\gamma}.
Therefore \displaystyle \oint_\gamma K d\gamma = - \oint_{\overline{\gamma}} K d\gamma = - \iint_R (Q_x - P_y)(x,y) dA.
The actual solution is the negative of \displaystyle \iint_R (Q_x - P_y)(x,y) dA.
We have Q(x,y)_x=(Bx + \sqrt{y^{C} + D} )_x = B and
P(x,y)_y=(A x + e^{\sin(x)})_x = A.
With substitution is becomes \displaystyle \iint_R (Q_x - P_y)(x,y) dA = \iint_R F \ dA.
The integral simplifies to
\displaystyle \iint_R FdA = F\iint_R 1 dA = F \cdot area of R.
= I we get
\displaystyle \oint_\gamma K d\gamma = S.
Using the area of the rectangle = I we get with the correct orientation
\displaystyle \oint_\gamma K d\gamma = - \oint_{\overline{\gamma}} K d\gamma = -S.