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<math>\oint_L(x-y)dx+(x+y)dy=\iint_D[1-(-1)]dxdy=2\iint_D dxdy=2s=2\int_0^{2\pi}d\phi \int_0^R\rho d\rho=</math>
<math>=\int_0^{2\pi}\rho^2|_0^R d\phi=R^2\int_0^{2\pi}d\phi=R^2\phi|_0^{2\pi}=2\pi R^2.</math>
:'''Patikrinimas'''. Iš apskritimo lygties <math>x^2+y^2=R^2</math> gauname:
:<math>y=\sqrt{R^2-x^2},</math>
:<math>x=\sqrt{R^2-y^2}.</math>
:Todėl
:<math>P(x, y)=x-y=x-\sqrt{R^2-x^2}, </math>
:<math>Q(x, y)=x+y=\sqrt{R^2-y^2}+y.</math>
:Pasinaudodami internetiniu integratoriumi, [http://integrals.wolfram.com/index.jsp?expr=x-%28R%5E2+-+x%5E2%29%5E%281%2F2%29&random=false gauname], [http://integrals.wolfram.com/index.jsp?expr=x%2B%28R%5E2+-+x%5E2%29%5E%281%2F2%29&random=false kad]
:<math>\oint_L(x-y)dx+(x+y)dy=\int_0^R (x-\sqrt{R^2-x^2}) dx + \int_0^R (\sqrt{R^2-y^2}+y) dy=</math>
:<math>\left(-\frac{1}{2}x\sqrt{R^2-x^2}+\frac{1}{2} R^2 \arctan\frac{x\sqrt{R^2-x^2}}{x^2-R^2}+\frac{x^2}{2} \right)|_0^R+\frac{1}{2}\left( y(\sqrt{R^2-y^2}+y) +R^2\arctan\frac{y}{R^2-y^2} \right)|_0^R=</math>
:<math>\left(-\frac{1}{2} R \sqrt{R^2-R^2}+\frac{1}{2} R^2 \arctan\frac{R\sqrt{R^2-R^2}}{R^2-R^2}+\frac{R^2}{2} \right)+\frac{1}{2}\left( R(\sqrt{R^2-R^2}+R) +R^2\arctan\frac{R}{R^2-R^2} \right)=</math>
:<math>\left(\frac{1}{2} R^2 \arctan(\infty)+\frac{R^2}{2} \right)+\frac{1}{2}\left( R(0+R) +R^2\arctan(\infty) \right)=</math>
:<math>\left(\frac{1}{2} R^2 \pi+\frac{R^2}{2} \right)+\frac{1}{2}\left( R^2 +R^2 \pi \right)=</math>
:<math> R^2 \pi+R^2=R^2(\pi+1).</math>
:Riba <math>\lim_{x\to R}\arctan\frac{x\sqrt{R^2-x^2}}{x^2-R^2}=\lim_{x\to R}\arctan\frac{x}{\sqrt{x^2-R^2}}=\lim_{z\to 0}\arctan\frac{R}{z}=\arctan(\infty)=\pi.</math> Beje, <math>\arctan(0)=0.</math>
:Patikrinimas nesuveikė.
* Taikydami Gryno formulę, apskaičiuokime kreivinį integralą
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