Alguém pode me ajudar nessa questão?
Anexos:
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Soluções para a tarefa
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Del teorema de Green obtenemos la siguiente fórmula para hallar el área encerrada por una curva cerrada orientada positivamente

Entonces tenemos que parametrizar cada curva [Haga la gráfica]
1) Parametrizando

2) Parametrizando

Ambos casos con![t\in[0,1] t\in[0,1]](https://tex.z-dn.net/?f=t%5Cin%5B0%2C1%5D)
3) Cálculo del área
![\displaystyle
2A=\oint_{\gamma_1^+}xdy-ydx+\oint_{\gamma_2^+}xdy-ydx\\ \\ \\
2A=\oint_{\gamma_1^+}(at^2)\,d[a(1-t)^2]-[a(1-t)^2]\,d(at^2)+\\ \\ +\oint_{\gamma_2^+}(a-at)\,d(at)-(at)\,d(a-at)\\ \\ \\
2A=\int_{0}^1-2a^2t^2(1-t)\,dt-2a^2t(1-t)^2\,dt+\\ \\ +\int_{0}^1a(a-at)\,dt+a^2t\,dt\\ \\ \\
2A=\int_{0}^1a^2(2t^2-2t+1)\,dt\\ \\ \\
2A=\dfrac{2a^2}{3}\\ \\ \\
\boxed{A=\dfrac{a^2}{3}}
\displaystyle
2A=\oint_{\gamma_1^+}xdy-ydx+\oint_{\gamma_2^+}xdy-ydx\\ \\ \\
2A=\oint_{\gamma_1^+}(at^2)\,d[a(1-t)^2]-[a(1-t)^2]\,d(at^2)+\\ \\ +\oint_{\gamma_2^+}(a-at)\,d(at)-(at)\,d(a-at)\\ \\ \\
2A=\int_{0}^1-2a^2t^2(1-t)\,dt-2a^2t(1-t)^2\,dt+\\ \\ +\int_{0}^1a(a-at)\,dt+a^2t\,dt\\ \\ \\
2A=\int_{0}^1a^2(2t^2-2t+1)\,dt\\ \\ \\
2A=\dfrac{2a^2}{3}\\ \\ \\
\boxed{A=\dfrac{a^2}{3}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0A2A%3D%5Coint_%7B%5Cgamma_1%5E%2B%7Dxdy-ydx%2B%5Coint_%7B%5Cgamma_2%5E%2B%7Dxdy-ydx%5C%5C+%5C%5C+%5C%5C%0A2A%3D%5Coint_%7B%5Cgamma_1%5E%2B%7D%28at%5E2%29%5C%2Cd%5Ba%281-t%29%5E2%5D-%5Ba%281-t%29%5E2%5D%5C%2Cd%28at%5E2%29%2B%5C%5C+%5C%5C+%2B%5Coint_%7B%5Cgamma_2%5E%2B%7D%28a-at%29%5C%2Cd%28at%29-%28at%29%5C%2Cd%28a-at%29%5C%5C+%5C%5C+%5C%5C%0A2A%3D%5Cint_%7B0%7D%5E1-2a%5E2t%5E2%281-t%29%5C%2Cdt-2a%5E2t%281-t%29%5E2%5C%2Cdt%2B%5C%5C+%5C%5C+%2B%5Cint_%7B0%7D%5E1a%28a-at%29%5C%2Cdt%2Ba%5E2t%5C%2Cdt%5C%5C+%5C%5C+%5C%5C%0A2A%3D%5Cint_%7B0%7D%5E1a%5E2%282t%5E2-2t%2B1%29%5C%2Cdt%5C%5C+%5C%5C+%5C%5C%0A2A%3D%5Cdfrac%7B2a%5E2%7D%7B3%7D%5C%5C+%5C%5C+%5C%5C%0A%5Cboxed%7BA%3D%5Cdfrac%7Ba%5E2%7D%7B3%7D%7D%0A)
Entonces tenemos que parametrizar cada curva [Haga la gráfica]
1) Parametrizando
2) Parametrizando
Ambos casos con
3) Cálculo del área
tpseletricista:
obrigrado!
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