Let $f(x,y) = 1$
Evaluate $ \int_C^ \! f(x,y) \, \mathrm{d}s \,\,$ where $C$ is straight-line segment from $(1,1,1)$ to $(0,0,0)$.
Using $(t,t),$ $1 \leqslant t \leqslant 0 $
$\qquad\qquad$ $\mathrm{d}s = \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } \,dt = \sqrt{1^2+1^2} = \sqrt2$
$\qquad\qquad$ $ \int_C^ \! f(x,y) \, \mathrm{d}s \,\, = \int_1^0 \sqrt2 \, \mathrm{d}t \,\,=
\,\, -\sqrt2$
$\,$
$\,$
Using $(1-t,1-t),$ $0 \leqslant t \leqslant 1 $
$\qquad\qquad$ $\mathrm{d}s = \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } \,dt = \sqrt{(-1)^2+(-1)^2} = \sqrt2$
$\qquad\qquad$ $ \int_C^ \! f(x,y) \, \mathrm{d}s \,\, = \int_0^1 \sqrt2 \, \mathrm{d}t \,\,=
\,\, \sqrt2$
I know $\sqrt2$ is the correct answer, from geometric interpretation, it is just the area of the wall.
Q1: What is the fault in the first calculation & what is the right way of doing it.
Q2: Does $ \int_{C1}^ \! f(x,y) \, \mathrm{d}s \, = -\int_{C2}^ \! f(x,y) \, \mathrm{d}s \, $ where C1 and C2 traces out same curve in opposite order. (Does the direction matters for line integrals in scalar fields)