Calculate $$$\displaystyle I=\int_C \Big(\sqrt{1+x^2+y^2}, y\cdot(xy+\ln (x+\sqrt{1+x^2+y^2}))\Big) \cdot dL$$$ on the curve $$C:=x^2+y^2=1$$
See development and solution
Development:
We will follow the procedure:
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Calculate $$\dfrac{d}{dx} F_2-\dfrac{d}{dy}F_1$$ $$$\dfrac{d}{dx}F_2=\dfrac{d}{dx}(y\cdot(xy+\ln (x+\sqrt{1+x^2+y^2})))=y\cdot\Big(y+\dfrac{1}{\sqrt{1+x^2+y^2}} \Big)$$$ $$$\dfrac{d}{dy}F_1=\dfrac{d}{dx}(\sqrt{1+x^2+y^2})=\dfrac{y}{\sqrt{1+x^2+y^2}}$$$ so $$\dfrac{d}{dx} F_2-\dfrac{d}{dy}F_1=y^2$$
- Calculate the new integral, in the region $$D$$ (the area enclosed by the curve $$C$$). $$$\int\int_D y^2 \ dxdy= \left\{\begin{array}{c} \mbox{Change to polar coordinates } |J|=r \\ x=r\cdot \cos\theta \\ y=r\cdot\sin\theta \end{array}\right\}=$$$ $$$=\int_0^1\int_0^{2\pi} r^2\cdot\sin^2\theta\cdot r\cdot dr\cdot d\theta=\int_0^{2\pi}\int_0^1\Big(\dfrac{1-\cos(2\theta)}{2}\Big)\cdot r^3drd\theta=\dfrac{\pi}{4}$$$
Solution:
$$\displaystyle I=\int_C \Big(\sqrt{1+x^2+y^2}, y\cdot(xy+\ln (x+\sqrt{1+x^2+y^2}))\Big) \cdot dL=\dfrac{\pi}{4}$$