Алгебра: с решением этого интеграла

с решением этого интеграла

Ответы

katizh 15.10.2020 14:23

$\int\limits_{L}\, (x^2-y)\, dx-(x-y^2)\, dy \ \ ,\ \ \ L:\left\{\begin{array}{l}x=5cost\\y=-5sint\end{array}\right\ \ ,\ \ A(5;0)\ ,\ \ B(0;5)\\\\\\(x^2-y)\, dx-(x-y^2)\, dy=\\\\=(25cos^2t+5sint)\cdot (5cost)'\, dt-(5cost-25sin^2t)\cdot (-5sint)'\, dt=\\\\=(25cos^2t+5sint)\cdot (-5sint)\, dt-(5cost-25sin^2t)\cdot (-5cost)\, dt=\\\\=(-125cos^2t\cdot sint-25sin^2t+25cos^2t-125sin^2t\cdot cost)\, dt=\\\\=(-125cos^2t\cdot sint+25(cos^2t-sin^2t)-125sin^2t\cdot cost)\, dt=$

$=(-125cos^2t\cdot sint+25cos2t-125sin^2t\cdot cost)\, dt\\\\A(5;0):\ \ 5cost=5\ \ ,\ \ cost=1\ \ ,\ \ t=0\\\\B(0;5):\ \ 5cost=0\ \ ,\ \ cost=0\ \ ,\ \ t=\dfrac{\pi}{2}\\\\\\\int\limits _{L}\, (x^2-y)\, dx-(x-y^2)\, dy=\\\\=\int\limits^{\pi /2}_0\, (-125cos^2t\cdot sint+25cos2t-125sin^2t\cdot cost)\, dt=\\\\=-125\int\limits^{\pi /2}_0\, cos^2t\cdot sint\, dt+25\int\limits^{\pi /2}_0\, cos2t\, dt-125\int\limits^{\pi /2}_0\, sin^2t\cdot cost\, dt=$

$=-125\cdot \dfrac{-cos^3t}{3}\Big|_0^{\pi /2}+\dfrac{25}{2}\cdot sin2t\Big|_0^{\pi /2}-125\cdot \dfrac{sin^3t}{3}\Big|_0^{\pi /2}=\\\\\\=\dfrac{125}{3}\cdot (cos^3\dfrac{\pi}{2}-cos^30)+\dfrac{25}{2}\cdot (sin\pi -sin0)-\dfrac{125}{3}\cdot (sin^3\dfrac{\pi}{2}-sin^30)=\\\\\\=\dfrac{125}{3}\cdot (0-1)+\dfrac{25}{2}\cdot (0-0)-\dfrac{125}{3}\cdot (1-0)=-\dfrac{125}{3}-\dfrac{125}{3}=-\dfrac{250}{3}$