Ejercicios de Teorema de Green


Ejercicio 13 (resuelto con teorema de Green):

Ejercicio 15:

Para validar el teorema de Green, la curva C debe ser simple y cerrada en una región D, y M y N deben tener derivadas parciales continuas sobre esa región:

Las curva es simple cerrada y las derivadas parciales son continuas puesto que x e y se pueden definir en todo el dominio.

8 lines Line 1: 0 is less than or equal to y is less than or equal to 1 Line 2: the negative square root of y is less than or equal to x is less than or equal to the square root of y Line 3: the integral from c to of x y squared minus x squared y equals the integral from D to of the integral from to of open paren Q over x minus P over y close paren Line 4: equals the integral from D to of the integral from to of open bracket negative 2 x y minus 2 x y close bracket Line 5: equals negative 4 the integral from 0 to 1 of the integral from the negative square root of y to the square root of y of x y Line 6: equals negative 4 the integral from 0 to 1 of y times open bracket open paren the fraction with numerator the square root of y and denominator 2 close paren squared minus open paren the fraction with numerator negative the square root of y squared and denominator 2 close paren close bracket Line 7: equals negative 4 the integral from 0 to 1 of y of 0 Line 8: equals 0

Ejercicio 16:

Mediante el teorema de Green, evalúe

1 lines Line 1: the integral from c to of the square root of 1 plus x cubed plus 2 x y d y

donde C es el triángulo con vértices (0,0); (1,0); (1,3):


10 lines Line 1: La ecuaci modifying above o with acute n para la linea que une dos puntos es colon Line 2: the fraction with numerator y sub 2 minus y sub 1 and denominator x sub 2 minus x sub 1 equals the fraction with numerator y sub 2 minus y sub 1 and denominator x sub 2 minus x sub 1 Line 3: the fraction with numerator y minus 0 and denominator x minus 0 equals the fraction with numerator 3 minus 0 and denominator 1 minus 0 equals is greater than y equals 3 x Line 4: 0 is less than or equal to x is less than or equal to 1 Line 5: 0 is less than or equal to y is less than or equal to 3 x Line 6: the integral from c to of the square root of 1 plus x cubed plus 2 x y d y equals the integral from to of the integral from D to of open bracket 2 x y over x minus the fraction with numerator the of open paren the square root of 1 plus x cubed close paren and denominator y close bracket Line 7: equals the integral from to of the integral from D to of open bracket 2 y minus 0 close bracket Line 8: equals the integral from 0 to 1 of the integral from 0 to 3 x of 2 y d y d x Line 9: equals the integral from 0 to 1 of 9 x squared Line 10: equals 3

Ejercicio 17: 

Utilice el teorema de Green para evaluar

the integral from to of x squared y minus x y squared d y

donde C es la circunferencia x²+y²=4 y orientación sentido contrario a las manecillas del reloj.


10 lines Line 1: Esta curva cumple las caracter modifying above i with acute sticas para aplicar el teorema de Green colon Line 2: the integral from to of x squared y minus x y squared d y equals the integral from to of the integral from D to of open bracket the fraction with numerator the of open paren negative x y squared close paren and denominator x minus the fraction with numerator the of open paren x squared y close paren and denominator y close bracket Line 3: the integral from to of x squared y minus x y squared d y equals the integral from to of the integral from D to of negative y squared minus x squared Line 4: Parametrizamos con polares colon Line 5: x squared plus y squared equals r squared Line 6: 0 is less than or equal to r is less than or equal to 2 Line 7: 0 is less than or equal to normal theta is less than or equal to 2 pi Line 8: equals negative the integral from 0 to 2 pi of the integral from 0 to 2 of r cubed Line 9: equals negative the integral from 0 to 2 pi of 4 pi Line 10: equals negative 8 pi

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