474b9020-9275-4886-a018-4b848abac415c213278c-5b13-441e-88f7-c30aacac385dbf9c8cfb-3123-438a-a941-8945554badac4.4.11<p>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 $$C:=x^2+y^2=1$$</p><p>We will follow the procedure:</p> <ul> <li> <p>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 $$\dfrac{d}{dx} F_2-\dfrac{d}{dy}F_1=y^2$$</p> </li> <li>Calculate the new integral, in the region $$D$$ (the area enclosed by the curve $$C$$). $$$\int\int_D y^2 \ dxdy= \left\{\begin{array}{c} \mbox{Change to polar coordinates } |J|=r \\ x=r\cdot \cos\theta \\ y=r\cdot\sin\theta \end{array}\right\}=$$$ $$$=\int_0^1\int_0^{2\pi} r^2\cdot\sin^2\theta\cdot r\cdot dr\cdot d\theta=\int_0^{2\pi}\int_0^1\Big(\dfrac{1-\cos(2\theta)}{2}\Big)\cdot r^3drd\theta=\dfrac{\pi}{4}$$$</li> </ul><p>$$\displaystyle I=\int_C \Big(\sqrt{1+x^2+y^2}, y\cdot(xy+\ln (x+\sqrt{1+x^2+y^2}))\Big) \cdot dL=\dfrac{\pi}{4}$$</p>7/18/16, 11:44 AM7/18/16, 11:44 AM