04 em 07 emcved uda2 s7 material de estudio

Page 1



C1

Sp f: R ⊆ �2 → �3

Ď•(u, v) = (x(u, v), y(u, v), z(u, v))

x(u, v), y(u, v), z(u, v)

(u0 , v0 ) ∈ Sp

u

Tu =

∂x ∂y ∂z (u , v )đ??˘ + (u0 , v0 )đ??Ł + (u0 , v0 )đ??¤ ∂u 0 0 ∂u ∂u v

Tv =

1

C1

∂x ∂y ∂z (u0 , v0 )đ??˘ + (u0 , v0 )đ??Ł + (u0 , v0 )đ??¤ ∂v ∂v ∂v

Los campos vectoriales son las funciones vectoriales que estudiaste en la unidad 1.


đ??˘ đ??Ł ∂x ∂y | |Tu Ă— Tv | = ∂u ∂u | ∂x ∂y ∂v ∂v

đ??¤ ∂z | ∂u| = ∂z ∂v

∂z ∂y ∂y ∂z ∂z ∂x ∂x ∂z ∂y ∂x ∂x ∂y = ( ( ) − ( )) đ??˘ − ( ( ) − ( )) đ??Ł + ( ( ) − ( )) đ??¤ ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u

Sp

Tu Ă— Tv = 0

Tu Ă— Tv

Sp = r(u, v) f: R ⊆ �3 → �

đ?‘†đ?‘?

f(x, y, z) ∈ � f

đ?‘†đ?‘?

âˆŹ f(x, y, z)dS = âˆŹ f(Ď•(u, v))|Tu Ă— Tv |dudv = Sp

Sp

∂(x, y) 2 ∂(y, z) 2 ∂(z, x) 2 = âˆŹ f(x(u, v), y(u, v), z(u, v))√[ ] +[ ] +[ ] dudv ∂(u, v) ∂(u, v) ∂(u, v) R

âˆŹđ?‘† 3đ?‘Ľ 2 đ?‘‘đ?‘† đ?‘?

x = senφcosθ y = senφsenθ

z = cosφ

0 ≤ θ ≤ 2Ď€, 0 ≤ φ ≤ Ď€

âˆŹS fdS p

−senθsenφ cosθcosφ =| | = −senφcosφ cosθsenφ senθcosφ ∂(z, x) = −sen2 φsenθ ∂(θ, φ) ∂(x,y) ∂(θ,φ)

∂(y,z) ∂(θ,φ)

= −sen2 φcosθ

|Tu Ă— Tv | = √[−senφcosφ]2 + [−sen2 φcosθ]2 + [−sen2 φsenθ]2 = |senφ|


2Ď€

Ď€

2Ď€

âˆŤ0 [âˆŤ0 3cos 2 θsen2 φsenφdφ]dθ = âˆŤ0 4cos2 θdθ = 4 Ď€

âˆŹS 3x 2 dS = 4Ď€ p

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k R

Sp

Sp A(Sp ) = âˆŹ |ru Ă— rv |dS R

C1

f

f(x, y) = z

(x, y) ∈ R |ru Ă— rv | = −

x = x y = y z = f(x, y)

A(Sp ) = âˆŹ √1 + [ R

∂f ∂f i − ∂y j ∂x

∂z 2 ∂z 2 ] + [ ] dS ∂x ∂y Ď•(r, θ) = (rcosθ, rsenθ, θ)

R = {0 ≤ r ≤ 1

0 ≤ θ ≤ Ď€}

∂(x,y) ∂(x,y) ∂(x,y) , , ∂(r,θ) ∂(r,θ) ∂(r,θ) ∂(x,y) ∂(r,θ)

=r

∂(y,z) ∂(r,θ)

= senθ

∂(z,x) ∂(r,θ)

= cosθ

|Ď•r Ă— Ď•θ | = √r 2 + 1

Ď€ 1 đ?œ‹ đ?œ‹ 1 1 âˆŤ [âˆŤ √r 2 + 1 dr] dθ = âˆŤ [ √r 2 + 1 + log(r + √r 2 + 1)]│0 dθ = Ď€ (√2 + log(1 + √2)) 2 0 0 0 2 1 Ď€(√2 + 2

log(1 + √2))


F

Sp =

r(u, v)

F

Sp

∬F ∙ dS = ∬ F ∙ (ru × rv ) dudv r

ru × rv = n

Sp

n F(x, y, z) = (x, y, z)

Sp

r(φ, θ) = (senφcosθ senφsenθ, cosφ) 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π

Sp F

rφ × rθ F(r(φ, θ))

F(r(φ, θ)) ∙ (ru × rv )

rφ × rθ = (sen2 φcosθ, sen2 φsenθ, senφcosφ) F(r(φ, θ)) = (cosθsenφ, senθsenφ, cosφ) F(r(φ, θ)) ∙ (ru × rv ) = cosφsen2 φcosθ + sen3 φsen2 θ + sen2 φcosφcosθ

π 2π 4 ∫ ∫ (cosφsen2 φcosθ + sen3 φsen2 θ + sen2 φcosφcosθ)dθdφ = = π 3 0 0


So (๐ ฅ, ๐ ฆ, ๐ ง) โ So n1

n2 n1 = โ n2

So

So โ ฌS F โ dS = โ ฌR F โ (ru ร rv )dS o

So r1 , r2

F C

1

r1 โ ฌ F โ dS = โ ฌ F โ dS r1

r2

Y si r2 la invierte serรก โ ฌ F โ dS = โ โ ฌ F โ dS r1

r2


F = (F1 , F2 , F3 ) C1

F3 (x, y, z)

F1 (x, y, z) F2 (x, y, z) z = f(x, y)

R

∂f ∂f − F2 + F3 ) dA ∂x ∂y

∬ F ∙ dS = ∬ (−F1 So

R

F(x, y, z) = (0, x, z) 𝑧= 0

∂f

∂f

∬S F ∙ dS = ∬R (−F1 ∂x − F2 ∂y + F3 ) dA o

F1 = 0 F2 = z = −𝑥 2 − 𝑦 2 + 1

F3 = y R

2

2

f(x, y) = z = −𝑥 − 𝑦 + 1

f(x, y) = z = 0

f(x, y) = z = −x 2 − y 2 + 1 ∂f ∂x

= −2x

∂f ∂y

= −2y

∬S F ∙ dS = ∬S F ∙ dS + ∬S F ∙ dS o

−x 2 − y 2 + 1

2

S1 =

S2 = 0

∬ (−F1 R

1

∂f ∂f − F2 + F3 ) dA = ∬ [−(0)(−2x) − (x)(−2y) + (−x 2 − y 2 + 1)] dA ∂x ∂y R ∬ [−2xy − x 2 − y 2 + 1] dA R


Ď•(r, θ) Ď•(r, θ) = (rcosθ, rsenθ) 2Ď€

∂(x,y) ∂(r,θ)

=đ?‘&#x;

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2Ď€

1

2Ď€

âˆŹ F ∙ dS = âˆŤ [âˆŤ (−2r 2 senθcosθ − r 2 + 1)dr]dθ = âˆŤ S1

0

0

0

1 4 (2 − 2senθcosθ)dθ = Ď€ 3 3

z=0

âˆŹ F ∙ dS = − âˆŹ zdA = − âˆŹ (0)dA = 0 S2

R

R 4

4

âˆŹS F ∙ dS = âˆŹS F ∙ dS + âˆŹS F ∙ dS = 3 Ď€ + 0 = 3 Ď€ o

1

2


r R ∈ ℝ2 C

1

P, Q

r ∮ Pdx + Qdy = ∬ ( r

R

∫R 2xydx + xydy

∂Q ∂P − ) dxdy ∂x ∂y

x2 + y2 = 9

R

x 2 + y 2 = 25

R

R

xy

P(x, y) = 2xy, Q(x, y) = xy ∂Q ∂P − ∂y ∂x

5

𝜋

∫ [∫ (𝑟𝑠𝑒𝑛𝜃 − 2𝑟𝑐𝑜𝑠𝜃)𝑟𝑑𝑟]𝑑𝜃 = ∫ − 0

3

=y

∂P ∂y

= 2x

= y − 2x x = rcos𝜃 y = rsen𝜃

𝜋

∂Q ∂x

0

∂(x,y) ∂(r,θ)

=r

98 196 (2𝑐𝑜𝑠𝜃 − 𝑠𝑒𝑛𝜃) 𝑑𝜃 = 3 3


∫R 2xydx + xydy =

R⊆ℝ ∂R

196 3

F(x, y) = P𝐢 + Q𝐣 r

2

R

rotF ∫ F ∙ dS = ∬ (rotF) ∙ 𝐤dA ∂R

R

divF

n=

(x′ (t)− y′(t)) √[x′(t)]2 +[y′(t)]2

𝑟(𝑡) = (𝑥(𝑡), 𝑦(𝑡)) ∫ F ∙ ndS = ∬ (divF)dA ∂R

∬R(rotF) ∙ 𝐤dA x=y

R

F(x, y) = (xy)i + (x − 2y)j x2 = y

x≥0 (rotF) ∙

k

Q(x, y) = xy

P(x, y) = x − 2y (rotF) ∙ 𝐤 =

∂Q ∂P − = y+2 ∂x ∂y x = x2

x=1 x=0 0≤x≤1

x2 ≤ y ≤ x

1 x 1 1 2 ∫ ∫ (y + 2)dydx = ∫ − x(x 3 + 3x − 4) dx = 2 5 0 x2 0

F(x, y) = ey x 2 i + ex j

∬R(divF)dA [0,1] × [0,1]

P(x, y) = ey x 2

Q(x, y) = ex divF =

∂P ∂Q + ∂y ∂x

= 2xey + 0


1

1

1

∬ 2xey dxdy = ∫ [∫ 2xey dx]dy = ∫ ey dy = e − 1 R

0

0

0

r R C

1

P, Q

R ∫ Pdx + Qdy = ∬ ( r

R

∂Q ∂P − ) ∂x ∂y

∫r xdx + 2x 2 ydy (0,0), (1,0), (1,1)

0≤y≤x P(x, y) = x ∂Q ∂P − ∂x ∂y

∂Q ∂x

Q(x, y) = x 2 y

= 4xy 1

x

1

∫ [∫ 4xydy]dx = ∫ 2x 3 dx = 0

0

0

1 2

0≤x≤1 = 4xy,

∂P ∂y

=0


z = f(x, y)

C

2

R (x, y) ∈ R

R F(x, y, z)

∂R ∬ rotF ∙ dS = ∬ (∇ × F) ∙ dS = ∫ F ∙ ds R

R∗

∂R∗ R

∫R −y 3 dx + ydy + zdz x2 + y2 = 1

z=1−x−y

R

∂R

C1


i ∂ ∇ Ă— F = || ∂x −xy 3

j k ∂ ∂ | = 3xy 2 đ??¤ = (0,0,3y 2 ) ∂y ∂z| y z

âˆŹ 3xy 2 ∙ đ?‘‘đ?‘† = âˆŹ 3xy 2 dxdy đ?‘…∗

đ?‘…

đ?‘… x = rcosθ y = rsenθ

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2

1 2Ď€ 1 3 âˆŤ [âˆŤ (3r 2 sen2 θ)rdθ]dr = âˆŤ 3 Ď€r 3 dr = Ď€ 4 0 0 0

R ∗

2

R∗

φ: R ⊆ â„? → Sp ∂R C

1

R

F

R âˆŹ (∇ Ă— F) ∙ dS = âˆŤ F ∙ dS R

∂R

F(x, y, z) = (2y 2 + x)đ??˘ + zđ??Ł + (x 2 + 4y)đ??¤ z = 9 − √x 2 + y 2

z=0

z = 9 − √x 2 + y 2 Ď•(x, y, z) = (rcosθ, rsenθ, 0)

âˆ‡Ă—F = |

0 ≤ r ≤ 3, 0 ≤ θ ≤ 2Ď€

i

j

k

∂ ∂x 2

∂ ∂y

∂ ∂z

z

x 2 + 4y

2y + x

| = (4 − 1) − (2x − 0) + (0 − 2y) = 3i − 2xj − 2yk

n i j k Ď•r Ă— Ď•θ = | cosθ senθ 0| = (0)i − (0)j + (r)k, −rsenθ rcosθ 0 rr Ă— rθ 1 n= = (0,0, r) = (0,0,1) |rr Ă— rθ | r

|Ď•r Ă— Ď•θ | = r


∇ × F ∙ n = (3, −2x, −2y) ∙ (0,0,1) = −2y = −2senθ

2π 3 2π 3 ∬ (∇ × F) ∙ dS = ∫ [∫ 2senθdr]dθ = ∫ 2 − 2 cos(3) dθ = 8πsen2 ( ) 2 R 0 0 0

ℝ3

F C1

∇×F=0 F

f

∇f = F ℝ2

F(x, y) = P(x, yi) + Q(x, y)j

∂P ∂Q = ∂y ∂x

F(x, y) = P(y 2 cosx + 2)i + Q(2ysenx + 2)j

∂P = 2ycosx ∂y ∂Q = 2ycosx ∂x



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