Evaluate the integral $\int_{\Gamma}\sqrt{x^2+y^2}dx+y\log(x+\sqrt{x^2+y^2})dy$

Question

I have to evaluate this integral $$\int_\Gamma\omega=\int_{\Gamma}\sqrt{x^2+y^2}dx+y\log(x+\sqrt{x^2+y^2})dy,$$ where $\Gamma = \{(x, y) \,|\, x^2+y^2=1 \text{ and } 0 \leq x \leq 1 \}.$ Graphically, it's the right half of the circle of radius 1 centered at the origin. I want to prove it using Barrow's rule. I can use it because I'm integrating a closed exact 1-form inside a connected region. I've found its potential function $$\psi(x,y)=\frac{x}{2}\sqrt{x^2+y^2}+\frac{y^2}{2}\log \biggl(x+\sqrt{x^2+y^2} \biggr)-\frac{y^2}{4}.$$ So, applying Barrow's rule for a parametrization $\alpha(t)$ of $\Gamma$ on the interval $a \leq t \leq b,$ we have $$\int_\Gamma\omega=\psi(\alpha(b))-\psi(\alpha(a)).$$ My main problem is that I don't know what $a$ and $b$ have to be for this specific $\Gamma$. I guess it's not just $(0,-1)$ to $(0,1)$ because that gives me a result of $0$. Must I divide it in two parts -- the $y$-positive and the $y$-negative part -- and multiply the result by 2? (By symmetry, each of these integrals evaluates to half the result given by going from $(0,-1)$ to $(0,1).$) I will thank any help.

Answer 1

As a sanity check, we know that we are integrating the right half of the circle, so the line integral becomes

$$\int_0^1 \sqrt{x^2 + 1 - x^2}\: dx + \int_1^0 \sqrt{x^2 + 1 - x^2} \:dx + \int_{-1}^1 y\log\left(\sqrt{1-y^2} + \sqrt{1-y^2+y^2}\right)\:dy$$

$$\int_0^1 (1-1)\:dx + \int_{-1}^1 y\log\left(1+\sqrt{1-y^2}\right)\:dy = 0$$

since the integral on the right is an odd function. This makes it explicit which parts of the integral cancel out with which (in this case, the $x$ and $y$ integrals canceled themselves out separately).