Evaluate the line integral $\oint_ CH\cdot n\mathrm{d}s$ through the sum of the integrals over the sides of : $$∮ _ H · n\mathrm{d} = ∫ _{1} H · n\mathrm{d} + ∫ _{2} H · n\mathrm{d} + ∫ _{3}H · n + ∫ _{4} H · n\mathrm{d}$$ where 1, 2, 3 and 4 are the four sides of . enter image description here
The link has the diagram needed. Please help. I have used a variant of green's theorem and used $\mathrm{d}s = \mathrm{d}x~\hat{\mathrm{i}} + \mathrm{d}y~\hat{\mathrm{j}}$ which then allows me to use $n\mathrm{d}s = \mathrm{d}y~\hat{\mathrm{i}} - \mathrm{d}x~\hat{\mathrm{j}}$. After this working out the equation of each line allows me to substitute and do the integration with respect to $\mathrm{d}x$. I have done this and got 4 as my answer but when I do the double integral to check my answers don't match. $H = x~\hat{\mathrm{i}}+y~\hat{\mathrm{j}}$
You have made a mistake in your line integral somewhere as the sum should be zero and not $4$.
Please note that if the vector field is $\vec H = (x,y)$,
$Q_x = P_y = 0$ where $P = x, Q = y$.
So the line integral over a closed curve will be zero.
Line integral $\displaystyle = \oint_C (Q_x - P_y) \ dA = 0$
Line integral over $S2$ in anticlockwise direction,
$r(t) = (1,2) + (0-1, 1-2)t = (1-t,2-t), 0 \leq t \leq 1$
Line integral over $S2 = \displaystyle \int_0^1 (1-t,2-t) \cdot (-1,-1) \ dt = - 2$
Line integral over $S4$ in anticlockwise direction,
$r(t) = (1,0) + (2-1, 1-0)t = (1+t,t), 0 \leq t \leq 1$
Line integral over $S4 = \displaystyle \int_0^1 (1+t,t) \cdot (1,1) \ dt = 2$
So the sum over $S2$ and $S4$ is zero. Also line integral along $S1$ and $S3$ is zero. Please do the working and check.