Let R ⊂ $ \mathbb{R}^{2}$ be the rectangle with oriented edges (0, 0) → (2, 0) → (2, 1) → (0, 1) → (0, 0). Let C denote the piecewise-smooth curve in the surface $z = 1+x^{2}+y^{2}$ whose projection onto the xy-plane is R. Then find $ \int_{C} x\ dy-y\ dz$. I guessed the integral around C would be same as the integral around R. By evaluating both of these integrals with the same integrand $ x\ dy -y\ dz $, I get 4. Is this a special case or Am I missing something in general?
Edit (my workings): $ Let\ w = x \ dy-y \ dz$. Then $ \int_{R} w= \int_{R_{1}} w \ + \int_{R_{2}} w \ + \int_{R_3} w \ +\int_{R_4} w$, where $ R_1,R_2,R_3,R_4$ are horizontal and vertical segments of contour. $R_{1}$ corresponds to (0, 0) → (2, 0), $R_2$ corresponds to (2, 0) → (2, 1) ands so on. Thus,
$\int_{R_1} w= 0$ $,since\ dy=0\ and \ dz =0 \ as \ z=0$,$ \ \int_{R_2}w =2\int_{0}^{1} dt=2,\ \int_{R_3}w=0 \ as\ dy=0, \ \int_{R_4}w=0\ as\ x=0.$
Similarly, $ \int_{C} w= \int_{C_{1}} w \ + \int_{C_{2}} w \ + \int_{C_3} w \ +\int_{C_4} w $ where $ R_1,R_2,R_3,R_4$ are projections of the contours $C_1,C_2,C_3,C_4$ on the xy plane respectively. Thus,
$ \int_{C_1} w= 0\ as\ y=0,\ \int_{C_2} w= 2\int_{0}^{1} dt-2\int_{0}^{1}t^{2} dt=\frac{4}{3},\ \int_{C_3} w=-2\int_{2}^{0}t\ dt=4,\ \int_{C_4}w =-2\int_{1}^{0}t^2\ dt=\frac{2}{3}.$
I had arithmetic mistakes in my previous answer. Now I can clearly see the integrals are not equal.