Integralen balioa

f(x,y,z)=z r:(0,1)-IR r(t)=(t*cost,t*sent,t)
1ºr´(t) 2ºraiz((sent+tcost)^2+(cost-tsent)^2+(1)^2)==raiz (2+t^2)

0∫ 1f(r(t))*||r`(t)||dt= 0∫ 1(t*raiz(2+t^2)dt)==

1/2*((2+t)^3/2)/(3/2) 0??1 ==1/3(3^(3/2) -2^(3/2))

valor integral //d (xy)dxdxy x2+y2 =4, x2+y2=9, x2—y2=4 y x2—y2=1.

escribo lo primero y lo segundo como d=[(x,y)ER^2/ valores]

u= x2+y2 v=x2—y y=raiz((u-v)/(2) ) x=raiz((u-v)/(2))

flujo(u,v)=(raiz((u-v)/(2)),raiz((u-v)/(2)) )

flujo =inyectiva flujo= clase 1 [det]flujo(u,v) distinto de 0

|1/4 *raiz((u+v)/(2) ) 1/4 raiz((u+v)/(2) )|

| |

|1/4*raiz((u-v)/(2) ) 1/4*raiz((u-v)/(2) |

=1/8 raiz((u+v)/(2)) *raiz((u-v)/(2)) distinto de 0

x2+y2=9 U=9 x2—y2=4 v=4 x2—y2=1 v=1

∫ ∫ d (xy)dxdxy=∫ ∫ s(raiz((u+v)/(2) ) *raiz((u-v)/(2) ) *|-1/8|dudv

=1/8∫ ∫ s dudv =1/8 area (s)=1/8 *15= 15/8

valor integral //d cos((x-y)/(x+y))dxdxy x=0,y=0, x+y=1

escribo lo primero y lo segundo como d=[(x,y)ER^2/ valores]

u=x-y v=x+y y=(u-v)/2 x=(u+v)/2 flujo(u,v)=[(u+v)/2 ,(u+v)/2 ]

flujo =inyectiva flujo= clase 1 [det]flujo(u,v) distinto de 0

u=x u=-v x+y=1-->v=1 s=[(u,v)EIR^2/ 0

∫ ∫ d cos((x-y)/(x+y)) dxdy=∫ ∫ s cos((u)/(v))*1/2dudv=

1/2*0∫ 1 ( -v∫ v ( cos(u)/(v) du)dv)= 1/2*0∫ 1 vsen(u)/(v) -v??v dv

=1/2* 0∫ 1 v*(sen(1)-sen(-1))dv =1/2*0∫ 1 v*sen(1) dv

=1/2*sen(1)*(v^2)/2 0??1 = sen(1)/2

Sea Φ(u,v) =(u+v,v-u2) y sea U={(u,v)∈R2/u≥0,v≥0, u+v≤2}

u+v=x v-u2=y u=raiz(x-y+1/4)+ 1/2 v=x-raiz(x-y+1/4)-1/2

F=(1)/(2*raiz(x-y+1/4))

| F*1 F*(-1)| =(1)/(4*(x-y+1/4) )+(1)/(2*raiz(x-y+1/4))+

|1- F -F | (1)/(4*(x-y+1/4))

= (1)/(2*(x-y+1/4) ) +(1)/(2*raiz(x-y+1/4))

∫ ∫ dydx

raiz(x-y+1/4)+ 1/2≥ 0 x-raiz(x-y+1/4)-1/2 ≥ 0

x-1/2≤ 2-1/2 x≤ 2 x≥ y

x-y +1/4≥ 1/4+x^2-x

Y≤ x

0∫ 2 0∫ x ( (1)/(2*(x-y+1/4) ) +(1)/(2*raiz(x-y+1/4)) ) dydx

0∫2 -1/2 ln(x-y+1/4) 0??x+ 1/2*0/2 (-(x-y+1/4)^-1/2+1)/(-1+2)/2

-ln (1/4)+1/2*ln(9/4)+0∫2 ( 1/2-raiz(x+(1)/(4)) dx

ln (4)+1/2*ln (9/4)+1-(1)/(2raiz(x+1/4)) 0??2

ln(4)+ 1/2*ln(9/4)+1-(1)/(2*raiz(4/4) -0+1

resultado= ln(4)+1/2*ln(9/4)+5/3

D = {(x,y)∈R2 / x≥0, y≥0, x2 + y2 ≤1, x 2 + y2 ≥2x}, ∬d 𝑦𝑑𝑥𝑑𝑥𝑦

x2+y2=2x -> x2-2x+y2=0 ->(x-1)^2+y^2=1

x2+y2=1 | Y=+raiz(2x-x^2) => x=1

x2+y2=2x| y=+raiz(1-x^2)=>

D = {(x,y)∈R2 / 0≤ x≤1/2, raiz(2x-x^2) ≤ y≤ raiz(1-x^2) }

∫∫ d (y)dxdy = 0∫(1)/(2) (raiz(2x-x^2))∫(raiz(1-x^2)) (y) dy )dx

0∫ (1)/(2) ( 1/2*y^2) raiz(2x-x^2)?? raiz(1-x^2) =

0∫ (1)/(2) ( 1/2*(1-x^2) -1/2*(2x-x^2)dx

=1/2*( x-x^2) 0??1/2 =1/2*(1/2-1/4)===1/8

( 1/2*(1-x^2) -1/2*(2x-x^2)dx

∬ raiz(𝑥 2 + 𝑦 2 )𝑑𝑥𝑑y D semicírculo definido x2 + y2 ≤2x y y≥0

x2+ y2=2x==>(x-1)^2+y =1

flujo(r,@)=(rcos@,rsen@)=(x,y)

x2+y2=2x==>r2=2rcos@==>r=2cos@

B[(r,@)EIR^2/ 0≤ @≤ (pi)/2 ,0≤ r≤ 2cos@]

flujo es de clase c1 /// flujo es inyectivo

///det flujo(r,@)=r distinto de =0

∫ ∫ d raiz(x^2+y^2) dxdy=∫ ∫ b r^2 drd@=

0∫(pi)/(2)( 0∫2cos@ r^2dr)d@=0∫(pi)/(2)(1/3r^3 0??2cos@)d@

0∫(pi)/(2) 8/3 cos^3*@ d@= 8/3 0∫(pi)/(2) cos@-cos@sen^2@)d@

=8/3(sen@-(sen^3*@)/3 (pi/2)??0 = 8/3*(1-1/3) =16/9