Where did I do wrong in solution of $\int_C\left(x \mathrm{e}^{y^2}-2 y\right) \mathrm{d} x+\left(x^2-1\right) y \mathrm{e}^{y^2} \mathrm{~d} y$

Question

I had a wrong answer when solving $$ \int_C\left(x \mathrm{e}^{y^2}-2 y\right) \mathrm{d} x+\left(x^2-1\right) y \mathrm{e}^{y^2} \mathrm{~d} y $$ where $$ C:y=\sqrt{2x-x^2} $$ My first thought was, let $$ P=x \mathrm{e}^{y^2}-2 y, Q=\left(x^2-1\right) y \mathrm{e}^{y^2} $$ and we got $$ \begin{array}{l} \frac{\partial P}{\partial y}=2 y x e^{y^2}-2 \\ \frac{\partial Q}{\partial x}=2 x y e^{y 2} \end{array} $$ consider removing out $$ \int_C -2y \mathrm{d} x $$ so the rest applies $$ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} $$, thus the integral is independent from $ C $. Let the integral path be the straight line beneath the circle which means y=0, then the integral equals $$\int_{0}^{2} x \mathrm{d} x =2$$ And $$ \int_C -2y \mathrm{d} x =\int_0^2 -2\sqrt{2x-x^2} \sqrt{1+y'^{2}}=\int_{0}^{2} -2 \mathrm{d} x =-4$$, that makes the result -2, but the correct answer is $2-\pi$, where did I do wrong?

Answer 1

Let $L$ be the line segment joining $(0,0)$ and $(2,0)$ and $C$ the given curve $y=\sqrt{2x-x^2}$, and $D$ the region bounded by $C\cup L$.

With $P=xe^{y^2}-2y$, $Q=(x^2-1)ye^{y^2}$, and $\mathcal P=P+2y$, we have $Q_x-P_y=2$ and $Q_x-\mathcal P_y=0$.

By Green's theorem,

$$\oint_{C\cup L} \mathcal P\,dx+Q\,dy = \iint_D Q_x-\mathcal P_y\,dA = 0$$

$$\implies \oint_{C\cup L} P\,dx+Q\,dy = \oint_{C\cup L} -2y \, dx$$

Now

$$\begin{align*} \int_C y \, dx &= \int_0^2 \sqrt{2x-x^2} \, dx & (y=\sqrt{2x-x^2})\\[1ex] \int_L y \, dx &= \int_0^2 0 \, dx & (y=0)\\[1ex] \implies \oint_{C\cup L} -2y\,dx &= -\pi \end{align*}$$