Задача 76781 Вычислить криволинейный интеграл по...
Условие
Решение
[m]x=2cos^3t[/m]
[m]y=2sin^3t[/m]
[m]dx=2\cdot 3cos^2t\cdot (cost)1dt=6cos^2t\cdot (-sint)dt=[/m]
[m]dy=2\cdot 3sin^2t\cdot (sint)1dt=6sin^2t\cdot (cost)dt=[/m]
[m]x^2dy-y^2dx=(2cos^3t)^2\cdot 6sin^2t\cdot (cost)dt-(2sin^3t)^2\cdot 6cos^2t\cdot (-sint)dt=24\cdot cos^2t sin^2t\cdot (cos^5t+sin^5t)dt=[/m]
[m]\sqrt[3]{x^5}+\sqrt[3]{y^5}=\sqrt[3]{(2cos^3t)^5}+\sqrt[3]{2sin^3t)^5}= \sqrt[3]{32}(cos^5t+sin^5t)[/m]
[m]\frac{x^2dy-y^2dx}{\sqrt[3]{x^5}+\sqrt[3]{y^5}}=\frac{24\cdot cos^2t sin^2t\cdot (cos^5t+sin^5t)dt}{ \sqrt[3]{32}(cos^5t+sin^5t)}=[/m]=[m]\frac{24}{ 2\sqrt[3]{4}}cos^2tsin^2tdt=\frac{12}{\sqrt[3]{4}}cos^2tsin^2tdt[/m]
Пределы интегрирования:
[m]A(2;0)[/m]
[m]2cos^3t=2[/m]
[m]2sin^3t=0[/m] ⇒[m] t=0[/m]
[m]B(0;2)[/m]
[m]2cos^3t=0[/m]
[m]2sin^3t=2[/m] ⇒ [m] t=\frac{π}{2}[/m]
тогда
[m] ∫ _{L_{AB}}\frac{x^2dy-y^2dx}{\sqrt[3]{x^5}+\sqrt[3]{y^5}}= ∫^{\frac{π}{2}}_{0}\frac{12}{\sqrt[3]{4}}cos^2tsin^2tdt= [/m]
применяем формулы понижения степени:
[m]=∫^{\frac{π}{2}}_{0}\frac{12}{\sqrt[3]{4}}\frac{1+cos2t}{2}\cdot \frac{1-cos2t}{2}dt= [/m]
[m]=∫^{\frac{π}{2}}_{0}\frac{3}{\sqrt[3]{4}}(1-cos^22t)dt=∫^{\frac{π}{2}}_{0}\frac{3}{\sqrt[3]{4}}(1-\frac{1+cos4t}{2})dt=\frac{3}{\sqrt[3]{4}}∫^{\frac{π}{2}}_{0}(\frac{1}{2}-\frac{1}{2}cos4t)dt=\frac{3}{2\sqrt[3]{4}}(t-\frac{1}{4}sin4t)|^{\frac{π}{2}}_{0}=\frac{3}{2\sqrt[3]{4}}\cdot \frac{π}{2}=\frac{3π}{4\sqrt[3]{4}}[/m]
