Ejercicios de integrales de línea


Ejercicio 1:

14 lines Line 1: Verificamos si la curva de cap F es independiente de la trayectoria colon Line 2: r o t F equals 0 Line 3: r o t F equals x F Line 4: r o t F equals open paren x cubed minus x cubed close paren times i minus open paren 3 x squared y minus 3 x squared y close paren times j positive open paren 3 x squared z minus 3 minus 3 x squared z plus 3 close paren times k Line 5: r o t F equals 0 i minus 0 j plus 0 k Line 6: El campo cap F es conservativo period Por lo tanto es independiente de su trayectoria period Line 7: r of t equals open paren 1 minus t close paren times r sub 0 plus t r sub 1 Line 8: r of t equals open paren 1 minus t close paren is less than 0 comma 0 comma 2 is greater than positive t is less than 0 comma 3 comma 0 is greater than Line 9: r of t equals is less than 0 comma 3 t comma 2 minus 2 t is greater than Line 10: x equals 0 semicolon y equals 3 t semicolon z equals 2 minus 2 t Line 11: r prime of t equals is less than 0 comma 3 comma negative 2 is greater than Line 12: 0 is less than or equal to t is less than or equal to 1 Line 13: the integral from c to of F of r of t period r prime of t equals the integral from 0 to 1 of is less than 0 comma 3 comma negative 2 is greater than period is less than 9 t comma 0 comma 4 minus 4 t is greater than d t Line 14: the integral from 0 to 1 of open paren 8 t minus 8 close paren equals 8 times open bracket the fraction with numerator t squared and denominator 2 close bracket sub 0 to the first power minus 8 times open bracket t close bracket sub 0 to the first power equals negative 4

Ejercicio 2:



Ejercicio 11:










Ejercicio 12:


es la curva cuyo punto inicial open paren 0 comma 0 comma 2 close paren y el punto final open paren 0 comma 3 comma 0 close paren


Ejercicio 14:




Ejercicio 21:

a)


b)
22 lines Line 1: Verificamos si cap F es conservativo para poder aplicar el teorema fundamental de las integrales de l modifying above i with acute nea colon Line 2: F equals is less than 2 x e raised to the 2 y power comma 2 x squared e raised to the 2 y power plus 2 y cotangent z comma y squared cosine squared z is greater than Line 3: r o t F equals x F Line 4: r o t F equals open paren negative 2 y cosecant z plus 2 y cosecant squared z close paren times i minus 0 times j plus open paren 4 x e raised to the 2 y power minus 4 x e raised to the 2 y power plus 0 close paren times k Line 5: r o t F equals 0 Line 6: Como cap F es conservativo comma existe un funci modifying above o with acute n cap F equals nabla f period Hallaremos la funci modifying above o with acute n potencial para aplicar el teorema fundamental de las integrales de linea colon Line 7: f sub x of open paren x comma y comma z close paren equals 2 x e raised to the 2 y power semicolon f sub y of open paren x comma y comma z close paren equals 2 x squared e raised to the 2 y power plus 2 y cotangent z semicolon f sub z of open paren x comma y comma z close paren equals negative y squared cosecant squared z Line 8: f sub open paren x comma y comma z close paren equals the integral from to of 2 x e raised to the 2 y power equals e raised to the 2 y power x squared plus g sub open paren x comma y close paren Line 9: f sub y of open paren x comma y comma z close paren equals 2 x squared e raised to the 2 y power plus g sub y of open paren x comma y close paren minus is greater than 2 x squared e raised to the 2 y power plus 2 y cotangent z equals 2 x squared e raised to the 2 y power plus g sub y of open paren x comma y close paren Line 10: g sub y of open paren x comma y close paren equals 2 y cotangent z Line 11: the integral from to of g sub y of open paren x comma y close paren of d y equals the integral from to of 2 y cotangent z Line 12: g sub open paren x comma y close paren equals y squared cotangent z plus h sub open paren z close paren Line 13: f sub open paren x comma y comma z close paren equals e raised to the 2 y power x squared plus y squared cotangent z plus h sub open paren z close paren Line 14: f sub z of open paren x comma y comma z close paren equals 0 minus y squared cosecant squared z plus h sub z of z minus is greater than negative y squared cosecant squared z equals negative y squared cosecant squared z minus h sub z of z Line 15: h sub z of z equals 0 Line 16: h sub open paren z close paren equals K Line 17: f sub open paren x comma y comma z close paren equals e raised to the 2 y power x squared minus y squared cotangent z plus K Line 18: t equals 0 minus is greater than x equals 1 semicolon y equals 0 semicolon z equals 0 Line 19: t equals 2 pi minus is greater than x equals 1 semicolon y equals 0 semicolon z equals 0 Line 20: f sub open paren x sub 1 comma y sub 1 comma z sub 1 close paren equals open paren 1 comma 0 comma 0 close paren semicolon f sub open paren x sub 2 comma y sub 2 comma z sub 2 close paren equals open paren 1 comma 0 comma 0 close paren Line 21: the integral from to of F period d r equals the integral from to of f of d r equals f of open paren x sub 2 comma y sub 2 comma z sub 2 close paren minus f of open paren x sub 1 comma y sub 1 comma z sub 1 close paren Line 22: f sub open paren 1 comma 0 comma 0 close paren minus f sub open paren 1 comma 0 comma 0 close paren equals 0


Comentarios

Publicar un comentario

Entradas populares