Gryno formulė: Skirtumas tarp puslapio versijų

58 pridėti baitai ,  prieš 7 metus
:<math>y=R\sin(t),</math>
:<math>x=R\cos(t);</math>
:<math>dy=R\cos(t) \; dt,</math>
:<math>dx=-R\sin(t) \; dt.</math>
:Todėl
:<math>P(x, y)=x-y=R\cos(t)-R\sin(t), </math>
:Pasinaudodami internetiniu integratoriumi, [http://integrals.wolfram.com/index.jsp?expr=-%28cos%28x%29-sin%28x%29%29+sin%28x%29&random=false gauname], [http://integrals.wolfram.com/index.jsp?expr=%28cos%28x%29%2Bsin%28x%29%29+cos%28x%29&random=false kad]
:<math>\oint_L(x-y)dx+(x+y)dy=-R^2\int_0^{2\pi}(\cos(t)-\sin(t))\sin(t)\; dt+ R^2\int_0^{2\pi}(\cos(t)+\sin(t))\cos(t)\; dt=</math>
:<math>\frac{1R^2}{4}(2x-\sin(2x)+\cos(2x)+1)|_0^{2\pi}+\frac{1R^2}{4}(2x+\sin(2x)-\cos^2(2x))|_0^{2\pi}=</math>
:<math>\frac{1R^2}{4}(2\cdot 2\pi-\sin(2\cdot 2\pi)+\cos(2\cdot 2\pi)+1)-\frac{1R^2}{4}(2\cdot 0-\sin(2\cdot 0)+\cos(2\cdot 0)+1)+\frac{1R^2}{4}(2\cdot 2\pi+\sin(2\cdot 2\pi)-\cos^2(2\cdot 2\pi))-\frac{1R^2}{4}(2\cdot 0+\sin(2\cdot 0)-\cos^2(2\cdot 0))=</math>
:<math>\frac{1R^2}{4}(4\pi-\sin(4\pi)+\cos(4\pi)+1)-\frac{1R^2}{4}(0-\sin(0)+\cos(0)+1)+\frac{1R^2}{4}(4\pi+\sin(4\pi)-\cos^2(4\pi))-\frac{1R^2}{4}(0+\sin(0)-\cos^2(0))=</math>
:<math>=\frac{1R^2}{4}(4\pi-0+1+1)-\frac{1R^2}{4}(0-0+1+1)+\frac{1R^2}{4}(4\pi+0-1^2)-\frac{1R^2}{4}(0+0-1^2)=</math>
:<math>=\frac{1R^2}{4}(4\pi+2)-\frac{1R^2}{4}\cdot 2+\frac{1R^2}{4}(4\pi-1)+\frac{1R^2}{4}=</math>
:<math>=R^2(\pi+\frac{1}{2}-\frac{1}{2}+\pi-\frac{1}{4}+\frac{1}{4})=</math>
:<math>=2\pi R^2.</math>
 
 
* Taikydami Gryno formulę, apskaičiuokime kreivinį integralą
907

pakeitimai