ΒΏAYUDAAA CON SUPERFICIES INTEGRALES TRIPLESSSS?
Calcule la superficie del paraboloide π(π₯, π¦) = 2 β π₯^2 β π¦^2 , con β1 β€ π₯ β€ 1,
β1 β€ π¦ β€ 1, sabiendo que π = β¬ β1 + 4x^2+4y^2
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f(x,y) = 2 - x^2 - y^2
f'x = -2 x
f'y = -2 y
S = β¬ β(1 + (f'x)^2 + (f'y)^2) dA
El Γ‘rea se toma sobre el cΓrculo
2 - x^2 - y^2 = 0
Queda
S = β¬[x_de_-1_a_1][y_de_-β(2 - x^2)_a_β(2 - x^2)]
β(1 + 4 x^2 + 4 y^2) dx dy
S = (2) β¬[x_de_-1_a_1][y_de_-β(2 - x^2)_a_β(2 - x^2)]
β( (1/4) + x^2 + y^2) dx dy
Por simetrΓa del integrando
S = (2)(2)(2) β¬[x_de_0_a_1][y_de_0_a_β(2 - x^2)]
β( (1/4) + x^2 + y^2) dx dy
Integramos con respecto a y
S = (8) Κ [x_de_0_a_1]
(1/2) ( y β((1/4) + x^2 + y^2) + ((1/4)+x^2) ln(y + β((1/4) + x^2 + y^2)) )
[y_de_0_a_β(2 - x^2)]
dx
S = (4) Κ [x_de_0_a_1]
β(2 - x^2) β((1/4) + x^2 + 2 - x^2) +
+ ((1/4)+x^2) ln(β(2 - x^2) + β((1/4) + x^2 + 2 - x^2))
- ((1/4)+x^2) ln(0 + β((1/4) + x^2 + 0^2))
dx
S = (4) Κ [x_de_0_a_1]
β(2 - x^2)β(9/4) +
((1/4)+x^2) ln(β(2 - x^2) + β(9/4)) - ((1/4)+x^2) ln(β((1/4) + x^2))
dx
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Parece mejor llevar a polares
S = β¬ β(1 + (f'x)^2 + (f'y)^2) dA
El Γ‘rea se toma sobre el cΓrculo
2 - x^2 - y^2 = 0
Queda
S = β¬[x_de_-1_a_1][y_de_-β(2 - x^2)_a_β(2 - x^2)]
β(1 + 4 x^2 + 4 y^2) dx dy
En polares
Queda
S = β¬[a_de_0_a_2pi][r_de_0_a_β2]
β(1 + 4 r^2) r dr da
S = (1/2) β¬[a_de_0_a_2pi][r_de_0_a_β2]
β(1 + 4 r^2) d(r^2) da
S = (1/8) β¬[a_de_0_a_2pi][r_de_0_a_β2]
(1 + 4 r^2)^(1/2) d(4 r^2) da
S = (1/8) * a [a_de_0_a_2pi] *
* (2/3) (1 + 4 r^2)^(3/2) [r_de_0_a_β2]
S = (1/8) (2 pi) (2/3) (1 + 4*2)^(3/2)
S = pi (1/6) (9)^(3/2)
S = pi (1/6) (27)
S = pi (9/2)
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