04 em 07 emcved uda2 s4 material de estudio

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f: R ⊆ ℝ2 → ℝ f

R = {(x, y) ∈ ℝ2 │a ≤ x ≤ b, a ≤ y ≤ b} n lim ∑m i=1 ∑j=1 f(xij∗ , yij∗ )∆A

m,n→∞

m

n

∬ f(x, y)dA = lim ∑ ∑ f(xij∗ , yij∗ )∆A m,n→∞

R n ∑m i=1 ∑j=1 f(xij∗ , yij∗ )∆A

i=1 j=1


𝑧 𝑅ij

𝑖𝑗 f(xij∗ , yij∗ )

(xij∗ , yij∗ ) ∈ 𝑅ij 𝑅i 𝑅

𝑧 = 𝑓(x, y) 𝑅

f: R ⊆ ℝ2 → ℝ y ≤ d} 𝑘 ∈ ℝ

g: R ⊆ ℝ2 → ℝ

R = {(x, y) ∈ ℝ2 │a ≤ x ≤ b, c ≤

f(x, y) ≥ 0 ∬R f(x, y)dA = V

V

𝑧 = f(x, y) ∬R[f(x, y) + g(x, y)dA] = ∬R f(x, y)dA + ∬R g(x, y) dA ∬R kf(x, y)dA = k ∬R f(x, y)dA f(x, y) ≥ g(x, y)

∬R f(x, y)dA ≥ ∬R g(x, y)dA

R = R1 ∪ R 2 ∪ R 3 … . R m f Ri m ∬R f(x, y)dA = ∬R f(x, y)dA + ∬R f(x, y)dA + ⋯ + ∬R f(x, y)dA = ∑i=1 ∬R f(x, y)dA i = 1, … m 1

2

m

i

f: R = {(x, y) ∈ ℝ2 │a ≤ x ≤ b, c ≤ y ≤ d} → ℝ 𝑧 = f(x, y) 𝑥 = 𝑥0 z = f(x0 , y)


xy f(x, y) y

x d

A(x0 ) = âˆŤ f(x0, y)dy c

x

y

b

A(y0 ) = âˆŤ f(x, y0 )dy a

� = f(x, y)

b d

V = âˆŤ[âˆŤ f(x, y)dy dx] a

c

z = f(x, y) A(x) d âˆŤc f(x0,

y)dy

b

A(y) = âˆŤa f(x, y0 )dy

z = f(x, y) = seny + cosx � = [0, π] × [0, π] π

Ď€

âˆŤ0 [âˆŤ0 seny + cosx dx]dy


x π

∫ (seny + cosx) dx = xseny + senx│π0 = πseny 0

y

π

∫ πsenydy = −πcosy│π0 = 2π 0

π

π

∫0 [∫0 senx + cosy dx]dy = 2π

z = f(x, y) = seny + cosx z

xy

R

f: R ⊆ ℝ2 → ℝ

R = [a, b]X[c, d]

b d

d

b

∬ f(x, y)dA = ∫ ∫ f(x, y)dydx = ∫ ∫ f(x, y)dxdy . R

a c

c

a


𝑧 = 2𝑥 2 + 4𝑦 𝑅 = [0,4]𝑋[0,4] b

d

∫a ∫c f(x, y)dydx 4

4

∫ [∫ 2x 2 + 4y dy]dx 0

0

4

∫(2x 2 + 4y) dy = 2yx 2 + 2y 2 │40 = 8(x 2 + 4) 0 4 1 896 ∫ 8(x 2 + 4)dx = 8 ( x 3 + 4x) │40 = 3 3 0

d

b

∫c ∫a f(x, y)dxdy 4

4

∫ [∫ 2x 2 + 4y dx]dy 0

0

4

2 16 ∫ 2x 2 + 4y dx = x 3 + 4xy│40 = (3y + 8) 3 3 0

4

∫ 0

16 16 3 2 896 (3y + 8)dy = ( y + 8𝑦) = 3 3 2 3

z = 2x 2 + 4y

D

R = {(x, y)|a ≤ x ≤ b, 𝜙1 (x) ≤ y ≤ 𝜙2 (x)}

xy

R = {(x, y)|a ≤ x ≤ b, 𝜙1 (x) ≤ y ≤ 𝜙2 (x)}


D

f: D → ℝ D b

f

𝜙2 (x) f(x, y)dxdy 1 (x)

∬D f(x, y)dA = ∫a ∫𝜙

D b

𝜙1 (x) f(x, y)dxdy 2 (x)

∬D f(x, y)dA = ∫a ∫𝜙

𝜙1 (x) ≤ y ≤ 𝜙2 (x)

𝜙2 (x) ≤ y ≤ 𝜙1 (x)

3x 2 + y y=x x=1

y=0

b 𝜙2 (x)

∬ f(x, y)dA = ∫ ∫ [f(x, y)dy]dx D

a 𝜙1 (x)

y xy

𝜙1 𝜙2


𝜙1 (x) = 0 xy

𝜙2 (x) = x y=x x=1

y=

0

x=1 x

x=y

x=0

b

𝜙 (x)

∫a ∫𝜙 2(x) f(x, y)dxdy 1

1

x

1 1 1 1 11 ∫[∫ 3x 2 + y dy]dx = ∫ (3x 2 y + y 2 )|x0 ] dx = ∫ x 2 (6x + 1)dx = 2 2 0 12 0 0

0

D

xy R = {(x, y)|c ≤ y ≤ d, 𝜓1 (y) ≤ x ≤ 𝜓2 (y)} D

f: D → ℝ D

f d 𝜓2 (y)

∬ f(x, y)dA = ∫ ∫ [f(x, y)dx]dy D

c 𝜓1 (y)

D


f(x, y) = 2xy y = −2 y =

2

x = 2 − y ,x = y 𝜓1 , 𝜓2 𝜓1 (y) = y

y = −2 y =

xy 𝜓2 (y) = 2 − y 2


d

𝜓 (y)

∫c ∫𝜓 2(y) [f(x, y)dx]dy 1

1

y

1

∫ [∫ −2

D

2−y2

1

y

[2𝑥𝑦 𝑑𝑥]𝑑𝑦 = ∫ (𝑥 2 𝑦)|2−y2 ] = ∫ − 𝑦(𝑦 4 − 5𝑦 2 + 4) 𝑑𝑦 = − −2

−2

R R = {(x, y)|a ≤ x ≤ b, 𝜙1 (x) ≤ y ≤ 𝜙2 (x)}

9 4

R = {(x, y)|c ≤ y ≤ d, 𝜓1 (y) ≤ x ≤

𝜓2 (y)} D

f: D → ℝ D b 𝜙2 (x)

f

D

d 𝜓2 (y)

∬ f(x, y)dA = ∫ ∫ f(x, y)dxdy = ∫ ∫ [f(x, y)dx]dy D

cosy

2

R

a 𝜙1 (x)

c 𝜓1 (y)

(0,0) (0,1)

f(x, y) = (1,1)


𝜙1 (x) = x

𝜙2 (x) = 1 b 𝜙2 (x) ∫a ∫𝜙 (x) f(x, y)dxdy 1

y x=0

x=1

1

x

1

∫ [∫ cosy 2 dy]dx 0

x

𝜓1 (y) 𝜓2 (y)

x 𝜓1 (y), = 0

𝜓2 (y) = y y y=0

y=1

d 𝜓2 (y) ∫c ∫𝜓 (y) [f(x, y)dx]dy 1 1

y

∫ [∫ cosy 2 dx]dy 0

1

y

1

0

1

y

∫0 [∫0 cosy 2 dx]dy = ∫0 xcosy 2 │0 = ∫0 ycosy 2 𝑑𝑦 = 1 1 1 y 1 ∫ [∫ cosy 2 dy]dx = ∫ [∫ cosy 2 dx]dy = sen(1) 2 0 x 0 0

f

g

∬R[f(x, y) + g(x, y)]dA = ∬R f(x, y)dA + ∬R g(x, y)dA ∬R kf(x, y)dA = k ∬R f(x, y)dA k ∈ ℝ f(x, y) ≥ g(x, y) R = R1 + R 2 m ≤ f(x, y) ≤ M

(x, y) ∈ 𝑅

∬R f(x, y)dA ≥ ∬R g(x, y)dA

∬R f(x, y)dA = ∬R f(x, y)dA + ∬R f(x, y)dA 1

(x, y) ∈ R

2

m ∗ A(R) ≤ ∬R f(x, y)dA ≤ M ∗ A(R)


âˆŹR ecosxseny dA

R = {(x, y)│x 2 + y 2 = 9}

−1 ≤ seny ≤ 1

−1 ≤ cosx ≤ 1

−1 ≤ cosxseny ≤ 1 e−1 ≤ ecosxseny ≤ e1 m = e−1 M = e1

A(R) = π(3)2 = 9π

9Ď€ ≤ âˆŹ ecosxseny ≤ 9Ď€e e đ?‘…


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