Ejercicios sobre ecuaciones diferenciales de segundo orden, y dobles integrales resueltos, integrales polares.

1. Unidad I
Matemáticas aplicadas a la Ingeniería
Química I
Mtro. Octavio Flores Siordia
Ilse Alejandra Gutiérrez Salvador
28/04/2014
UNIVERSIDAD DE GUADALAJARA
CENTRO UNIVERSITARIO DE LA CIÉNEGA

3. Derivadas Parciales
 Dependen de más de una variable independiente
𝑓′(𝑥) = lim
∆𝑥 → 0
𝑓( 𝑥 + ∆𝑥 )– 𝑓(𝑥)
∆𝑥
Pendiente de una recta tangente a la función.
Variables que se encuentran en una dirección y/o en otra.
𝑓(𝑥, 𝑦) = 𝑥3
+ 𝑥2
𝑦3
− 2𝑦2
𝜕𝑓
𝜕𝑥
= 𝑓𝑥 (𝑥, 𝑦) =
𝜕
𝜕𝑥
(𝑥3
+ 𝑥2
𝑦3
− 2𝑦2) =
𝜕
𝜕𝑥
𝑥3
+
𝜕
𝜕𝑥
𝑥2
𝑦3
−
𝜕
𝜕𝑥
2𝑦2
= 3𝑥2
+ 2𝑥𝑦3
𝜕
𝜕𝑥
|1,2 = 3(1)2
+ 2(1)(2)3
= 3 + 16
= 19
𝜕𝑓
𝜕𝑦
= 𝑓𝑦 (𝑥, 𝑦) =
𝜕
𝜕𝑦
(𝑥3
+ 𝑥2
𝑦3
− 2𝑦2) =
𝜕
𝜕𝑦
𝑥3
+
𝜕
𝜕𝑦
𝑥2
𝑦3
−
𝜕
𝜕𝑦
2𝑦2
= 3𝑥2
𝑦2
− 4𝑦
𝜕
𝜕𝑦
|3,4 = 3(3)2
(4)2
− 4(4) = 432 − 16
= 416

4. f(x,y) = 4-x2-2y2
𝜕𝑓
𝜕𝑥
|1,1 =
𝜕
𝜕𝑥
(4 − 𝑥2
− 2𝑦2)|1,1 = −2𝑥 |1,1 = −2(1) = −2
𝜕𝑓
𝜕𝑦
|1,1
=
𝜕
𝜕𝑦
(4 − 𝑥2
− 2𝑦2)|1,1
= −4𝑦|1,1
= −4(1) = −4
f(x,y) = sin(𝑥2
𝑦)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
sin(𝑥2
𝑦) = cos(𝑥2
𝑦)
𝜕
𝜕𝑥
𝑥2
𝑦 =
2𝑥𝑦 cos(𝑥2
𝑦)
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
sin(𝑥2
𝑦) = cos(𝑥2
𝑦)
𝜕
𝜕𝑥
𝑥2
𝑦 =
𝑥2
cos(𝑥2
𝑦)
f(x,y) = cos (
𝑥
1+𝑦
)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
sin (
𝑥
1 + 𝑦
) = cos (
𝑥
1 + 𝑦
)
𝜕
𝜕𝑥
(
𝑥
1 + 𝑦
) =
(
1
1 + 𝑦
) cos (
𝑥
1 + 𝑦
)
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
sin (
𝑥
1 + 𝑦
) = cos (
𝑥
1 + 𝑦
)
𝜕
𝜕𝑦
(
𝑥
1 + 𝑦
) =
(
𝑥
(1 + 𝑦)2
) cos (
𝑥
(1 + 𝑦)
)

5. f(x,y) =
1
√𝑥2+ 𝑦2
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
1
√𝑥2 + 𝑦2
= −
1
2
(𝑥2
+ 𝑦2)−
3
2
𝜕
𝜕𝑥
(𝑥2
+ 𝑦2) =
−𝑥
(𝑥2 𝑦2)3/2
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
1
√𝑥2 + 𝑦2
= −
1
2
(𝑥2
+ 𝑦2)−
3
2
𝜕
𝜕𝑦
(𝑥2
+ 𝑦2) =
−𝑦
(𝑥2 𝑦2)3/2
f(x,y) = Arctan (x,y)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
tan−1(𝑥𝑦) =
1
𝑥2 𝑦2 + 1
𝜕
𝜕𝑥
(𝑥𝑦) =
𝑦
𝑥2 𝑦2 + 1
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
tan−1(𝑥𝑦) =
1
𝑥2 𝑦2 + 1
𝜕
𝜕𝑥
(𝑥𝑦) =
𝑥
𝑥2 𝑦2 + 1

6. f(x,y) = ln (x+3y)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
ln(𝑥 + 3𝑦) =
1
𝑥 + 3𝑦
𝜕
𝜕𝑥
(𝑥 + 3𝑦) =
1
𝑥 + 3𝑦
f(x,y,z) = ℮ 𝑥𝑦
ln 𝑧
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
℮ 𝑥𝑦
ln 𝑧 = ln 𝑧
𝜕
𝜕𝑥
℮ 𝑥𝑦
= ln 𝑧 ℮ 𝑥𝑦
𝜕
𝜕𝑥
(𝑥𝑦) =
𝑦 ln 𝑧 ℮ 𝑥𝑦
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
℮ 𝑥𝑦
ln 𝑧 = ln 𝑧
𝜕
𝜕𝑦
℮ 𝑥𝑦
= ln 𝑧 ℮ 𝑥𝑦
𝜕
𝜕𝑦
(𝑥𝑦) =
𝑥 ln 𝑧 ℮ 𝑥𝑦
𝜕𝑓
𝜕𝑧
=
𝜕
𝜕𝑧
℮ 𝑥𝑦
ln 𝑧 = ℮ 𝑥𝑦
𝜕
𝜕𝑧
ln 𝑧 =
℮ 𝑥𝑦
𝑧

7. f(x,y) = x cos(𝑥𝑦)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
x cos(𝑥𝑦) = 𝑥
𝜕
𝜕𝑥
cos(𝑥𝑦) + cos(𝑥𝑦)
𝜕
𝜕𝑥
𝑥 = −𝑥 sin(𝑥𝑦)
𝜕
𝜕𝑥
(𝑥𝑦)
+ cos( 𝑥𝑦) =
−𝑥𝑦 sin(𝑥𝑦) + cos(𝑥𝑦)
f(x,y) = ln(𝑥 + 3𝑦)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
ln(𝑥 + 3𝑦) =
1
𝑥 + 3𝑦
𝜕
𝜕𝑥
(𝑥 + 3𝑦) =
1
𝑥 + 3𝑦
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
ln(𝑥 + 3𝑦) =
1
𝑥 + 3𝑦
𝜕
𝜕𝑦
(𝑥 + 3𝑦) =
3
𝑥 + 3𝑦
f(x) = 𝑥 𝑥
𝜕
𝜕𝑥
𝑥 𝑥
= → 𝑦 = 𝑥 𝑥
ln 𝑦 = ln 𝑥 𝑥
= 𝑥 ln 𝑥
𝜕
𝜕𝑥
ln 𝑦 =
𝜕
𝜕𝑥
𝑥 ln 𝑥 = 𝑥
𝜕
𝜕𝑥
ln 𝑥 + ln 𝑥
𝜕
𝜕𝑥
𝑥

8. 1
𝑦
𝜕𝑦
𝜕𝑥
= 1 + ln 𝑥
𝜕𝑦
𝜕𝑥
= 𝑦(1 + ln 𝑥)
𝜕
𝜕𝑥
𝑥 𝑥
= 𝑥 𝑥
(1 + ln 𝑥)
f(x,y) = 𝑥 𝑦
𝜕
𝜕𝑥
𝑥 𝑦
=
𝑦𝑥 𝑦−1
𝜕
𝜕𝑦
𝑥 𝑦
𝑧 = 𝑥 𝑦
ln 𝑧 = ln 𝑥 𝑦
= 𝑦 ln 𝑥
𝜕
𝜕𝑦
ln 𝑧 = ln 𝑥
𝜕
𝜕𝑦
𝑦
1
𝑧
𝜕
𝜕𝑦
= ln 𝑥
𝜕
𝜕𝑦
𝑧 = 𝑧 ln 𝑥
𝜕
𝜕𝑦
𝑥 𝑦
= 𝑥 𝑦
ln 𝑥

9. W =
℮ 𝑣
𝑢+𝑣2
𝜕
𝜕𝑣
℮ 𝑣
𝑢 + 𝑣2
=
( 𝑢 + 𝑣2)
𝜕
𝜕𝑣
℮ 𝑣
− ℮ 𝑣 𝜕
𝜕𝑣
( 𝑢 + 𝑣2)
(𝑢 + 𝑣2)
2
=
( 𝑢 + 𝑣2) ℮ 𝑣
− 2𝑣 ℮ 𝑣
(𝑢 + 𝑣2)2
𝜕
𝜕𝑢
℮ 𝑣
𝑢 + 𝑣2
= ℮ 𝑣 𝜕
𝜕𝑢
( 𝑢 + 𝑣2)
−1
=
−℮ 𝑣
(𝑢 + 𝑣2)2
R(p,q) = tan−1
(𝑝𝑞2
)
𝜕
𝜕𝑝
tan−1
( 𝑝𝑞2) =
1
𝑝2 𝑞4 + 1
𝜕
𝜕𝑝
𝑝𝑞2 =
𝑞2
𝑝2 𝑞4 + 1
𝜕
𝜕𝑞
tan−1
( 𝑝𝑞2) =
1
𝑝2 𝑞4 + 1
𝜕
𝜕𝑞
𝑝𝑞2 =
2𝑝𝑞
𝑝2 𝑞4 + 1

10. U(r, ) = sin(𝑟 cos 𝜃)
𝜕𝑈
𝜕𝑟
=
𝜕
𝜕𝑟
sin(𝑟 cos 𝜃) = cos(𝑟 cos 𝜃)
𝜕
𝜕𝑟
𝑟 cos 𝜃 =
cos(𝑟 cos 𝜃) cos 𝜃
𝜕
𝜕𝑟
𝑟 =
cos 𝜃 cos(𝑟 cos 𝜃)
𝜕𝑈
𝜕
=
𝜕
𝜕
sin(𝑟 cos 𝜃) = cos(𝑟 cos 𝜃)
𝜕
𝜕𝑟
𝑟 cos 𝜃 =
cos(𝑟 cos 𝜃) 𝑟
𝜕
𝜕𝜃
cos 𝜃
−rsin 𝜃 cos(𝑟 cos 𝜃)
Segundas Derivadas
f(x,y)
(1)
𝜕2
𝑓
𝜕𝑥2
=
𝜕
𝜕𝑥
(
𝜕𝑓
𝜕𝑥
)
(2)
𝜕2
𝑓
𝜕𝑦2
=
𝜕
𝜕𝑦
(
𝜕𝑓
𝜕𝑦
)
(3)
𝜕2
𝑓
𝜕𝑥𝜕𝑦
=
𝜕
𝜕𝑥
(
𝜕𝑓
𝜕𝑦
)
(4)
𝜕2
𝑓
𝜕𝑦𝜕𝑥
=
𝜕
𝜕𝑦
(
𝜕𝑓
𝜕𝑥
)

11. (3) y (4) siempre dan el mismo resultado.
f(x,y) = 𝑥3
+ 𝑥2
𝑦3
− 2𝑦2
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
(𝑥3
+ 𝑥2
𝑦3
− 2𝑦2)
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
𝑥3
+ 𝑦3
𝜕
𝜕𝑥
𝑥2
− 2
𝜕
𝜕𝑥
𝑦2
𝜕𝑓
𝜕𝑥
= 3𝑥2
+ 2𝑥𝑦3
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
(𝑥3
+ 𝑥2
𝑦3
− 2𝑦2)
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
𝑥3
+ 𝑥2
𝜕
𝜕𝑦
𝑦3
− 2
𝜕
𝜕𝑦
𝑦2
𝜕𝑓
𝜕𝑦
= 3𝑥2
𝑦2
− 4𝑦
𝜕2
𝑓
𝜕𝑥2
=
𝜕
𝜕𝑥
(
𝜕𝑓
𝜕𝑥
) =
𝜕
𝜕𝑥
(3𝑥2
+ 2𝑥𝑦3)
𝜕2
𝑓
𝜕𝑥2
= 3
𝜕
𝜕𝑥
𝑥2
+ 2𝑦3
𝜕
𝜕𝑥
𝑥
𝜕2
𝑓
𝜕𝑥2
= 6𝑥 + 2𝑦3
𝜕2
𝑓
𝜕𝑦2
=
𝜕
𝜕𝑦
(
𝜕𝑓
𝜕𝑦
) =
𝜕
𝜕𝑦
(3𝑥2
𝑦2
− 4𝑦)
𝜕2
𝑓
𝜕𝑦2
= 3𝑥2
𝜕
𝜕𝑦
𝑦2
− 4
𝜕
𝜕𝑦
𝑦

12. 𝜕2
𝑓
𝜕𝑦2
= 6𝑥2
𝑦 − 4
𝜕2
𝑓
𝜕𝑥𝜕𝑦
=
𝜕
𝜕𝑥
(
𝜕𝑓
𝜕𝑦
) =
𝜕
𝜕𝑥
(3𝑥2
𝑦2
− 4𝑦)
𝜕2
𝑓
𝜕𝑥𝜕𝑦
= 6𝑥𝑦2
𝜕2
𝑓
𝜕𝑦𝜕𝑥
=
𝜕
𝜕𝑦
(
𝜕𝑓
𝜕𝑥
) = 3𝑥2
+ 2𝑥𝑦3
𝜕2
𝑓
𝜕𝑦𝜕𝑥
= 6𝑥𝑦2
Z= tan−1
(
𝑥+𝑦
1−𝑥𝑦
)
𝜕𝑧
𝜕𝑥
=
𝜕
𝜕𝑥
tan−1
(
𝑥 + 𝑦
1 − 𝑥𝑦
) =
1
(
𝑥 + 𝑦
1 − 𝑥𝑦
) 2 + 1
𝜕𝑧
𝜕𝑥
=
1
(
𝑥 + 𝑦
1 − 𝑥𝑦
) 2 +
1
1
(
1 − 𝑥𝑦
𝜕
𝜕𝑥
(𝑥 + 𝑦) − (𝑥 + 𝑦)
𝜕
𝜕𝑥
(1 − 𝑥𝑦)
(1 − 𝑥𝑦)2
)
𝜕𝑧
𝜕𝑥
=
(1 − 𝑥𝑦)2
(𝑥 + 𝑦)2 + (1 − 𝑥𝑦)2
(
1 − 𝑥𝑦 + 𝑥𝑦 + 𝑦2
(1 − 𝑥𝑦)2
)
𝜕𝑧
𝜕𝑥
=
1 + 𝑦2
(𝑥 + 𝑦)2 + (1 − 𝑥𝑦)2

13. f(x,y,z)= ℮ 𝑥𝑦𝑧2
f(x,y,z)=
𝜕3 𝑓
𝜕𝑥𝜕𝑦𝜕𝑧
𝜕
𝜕𝑥
℮ 𝑥𝑦𝑧2 𝜕
𝜕𝑥
(𝑥𝑦𝑧2) =
𝑦𝑧2
℮ 𝑥𝑦𝑧2
𝜕2
𝑓
𝜕𝑦𝜕𝑧
=
𝜕
𝜕𝑦
𝜕
𝜕𝑥
℮ 𝑥𝑦𝑧2
=
𝜕
𝜕𝑦
𝑦𝑧2
℮ 𝑥𝑦𝑧2
= 𝑧2
𝜕
𝜕𝑦
𝑦℮ 𝑥𝑦𝑧2
𝑧2
[𝑦
𝜕
𝜕𝑦
℮ 𝑥𝑦𝑧2
+ ℮ 𝑥𝑦𝑧2 𝜕
𝜕𝑦
𝑦 ] = 𝑧2
[𝑦𝑥𝑧2
℮ 𝑥𝑦𝑧2
+ ℮ 𝑥𝑦𝑧2
] =
= 𝑧2
[𝑦𝑥𝑧2
℮ 𝑥𝑦𝑧2
+ ℮ 𝑥𝑦𝑧2
]
𝜕3
𝑓
𝜕𝑥𝜕𝑦𝜕𝑧
=
𝜕
𝜕𝑧
𝜕2
𝜕𝑦𝜕𝑥
=
𝜕
𝜕𝑧
(𝑧2
[𝑦𝑥𝑧2
℮ 𝑥𝑦𝑧2
+ ℮ 𝑥𝑦𝑧2
])
𝜕3
𝑓
𝜕𝑥𝜕𝑦𝜕𝑧
= 𝑥𝑦
𝜕
𝜕𝑧
𝑧4
℮ 𝑥𝑦𝑧2
+
𝜕
𝜕𝑧
𝑧2
℮ 𝑥𝑦𝑧2
𝜕3
𝑓
𝜕𝑥𝜕𝑦𝜕𝑧
= 𝑥𝑦 [𝑧4
𝜕
𝜕𝑧
℮ 𝑥𝑦𝑧2
+ ℮ 𝑥𝑦𝑧2 𝜕
𝜕𝑧
𝑧4
] + [𝑧2
𝜕
𝜕𝑧
℮ 𝑥𝑦𝑧2
+ ℮ 𝑥𝑦𝑧2 𝜕
𝜕𝑧
𝑧2
]
𝜕3
𝑓
𝜕𝑥𝜕𝑦𝜕𝑧
= 2𝑥2
𝑦2
𝑧5
℮ 𝑥𝑦𝑧2
+ 4𝑥𝑦𝑧3
℮ 𝑥𝑦𝑧2
+ 2𝑥𝑦𝑧3
℮ 𝑥𝑦𝑧2
+ 2𝑧℮ 𝑥𝑦𝑧2

14. 𝜕3
𝑓
𝜕𝑥𝜕𝑦𝜕𝑧
= ( 2𝑥2
𝑦2
𝑧5
+ 6𝑥𝑦𝑧3
+ 2𝑧) ℮ 𝑥𝑦𝑧2
F(x) = a
Y= a
F(x,y) = a
Z=a

15. y = a
x=a
F(x) = x
y=x
F(x,y)

16. y=x
z=y
F(x) = x2
Y=x2

17. F(x,y)
Y = x2
F(x) = sen x

18. X + 2y + 3z = 6
0 0 z = 2 (0,0,2)
0 y 0 = 3 (0,3,0)
x 0 0 = 6 (6,0,0)
Derivadas Direccionales
Duf = ∇ 𝑓 𝑈
∇ 𝑓 =
𝑑𝑓
𝑑𝑥
| 𝑎𝑏 𝑖 +
𝑑𝑓
𝑑𝑦
| 𝑎𝑏 𝑗
Máxima derivada
∇ 𝑓
Dirección: ∇ 𝑓
Magnitud |∇ 𝑓|

19. Encontrar la máxima derivada de f(x,y) = x2y3 en el punto (2,3)
∇ 𝑓 =
𝑑𝑓
𝑑𝑥
|2,3 𝑖 +
𝑑𝑓
𝑑𝑦
|2,3 𝑗
∇ 𝑓 = 2𝑥𝑦3
|2,3 𝑖 + 3𝑥2
𝑦2
|2,3 𝑗
∇ 𝑓 = 2(2)(3)3
𝑖 + 3(2)2
(3)2
𝑗
∇ 𝑓 = 108𝑖 + 108𝑗
f(x,y,z) = ℮2𝑥+𝑦+3𝑧
∇ 𝑓 =
𝑑𝑓
𝑑𝑥
|0,0,0 𝑖 +
𝑑𝑓
𝑑𝑦
|0,0,0 𝑗 +
𝑑𝑓
𝑑𝑧
|0,0,0 𝑘
∇ 𝑓 = 2 ℮2𝑥+𝑦+3𝑧
|0,0,0 𝑖 + 1 ℮2𝑥+𝑦+3𝑧
|0,0,0 𝑗 + 3 ℮2𝑥+𝑦+3𝑧
|0,0,0 𝑘
∇ 𝑓 = 2𝑖 + 𝑗 + 3𝑘
Máxima derivada:
Dirección: 2𝑖 + 𝑗 + 3𝑘
Magnitud: √(2)2 + (1)2 + (3)2 = √14 = 3.74
Encontrar la derivada direccional f(x,y,z) = xyz2 en el punto (1,0,1)
Dirección i + j + k
Encontrar su máxima derivada direccional.

20. ∇ 𝑓 =
𝑑𝑓
𝑑𝑥
|1,0,1 𝑖 +
𝑑𝑓
𝑑𝑦
|1,0,1 𝑗 +
𝑑𝑓
𝑑𝑧
|1,0,1 𝑘
∇ 𝑓 = 𝑦𝑧2
|1,0,1 𝑖 + 𝑥𝑧2
|1,0,1 𝑗 + 2𝑥𝑦𝑧 |1,0,1 𝑘
∇ 𝑓 = 0𝑖 + 𝑗 + 𝑘
Máxima derivada:
Dirección: 0𝑖 + 𝑗 + 𝑘
Magnitud: √02 + (1)2 + (0)2 = √1 = 1
Duf = ∇ 𝑓 𝑈
𝑈 = 𝑖 + 𝑗 + 𝑘
|𝑈| = √3
𝑈 =
𝑖
√3
+
𝑗
√3
+
𝑘
√3
Duf = (0𝑖 + 𝑗 + 𝑘) (
𝑖
√3
+
𝑗
√3
+
𝑘
√3
)
Duf =
1
√3
Duf = 0.5773

21. f(x,y) = x2 y3-4y (2, -1) dirección 2i + 5j
Derivada direccional.
Máxima derivada direccional.
∇ 𝑓 =
𝑑𝑓
𝑑𝑥
|2,−1 𝑖 +
𝑑𝑓
𝑑𝑦
|2,−1 𝑗
∇ 𝑓 = 2𝑥𝑦3
|2,−1 𝑖 + 3𝑥2
𝑦2
− 4|2,−1 𝑗
∇ 𝑓 = 2(2)(−1)3
|2,−1 𝑖 + 3(2)2
(−1)2
− 4|2,−1 𝑗
∇ 𝑓 = −4𝑖 + 8𝑗
Máxima derivada:
Dirección: −4𝑖 + 8𝑗
Magnitud: √(−4)2 + (8)2 = √80 = 8.94
Duf = ∇ 𝑓 𝑈
𝑈 = 2𝑖 + 5𝑗
|𝑈| = √(2)2 + (5)2 = √29

22. 𝑈 =
2𝑖
√29
+
5𝑗
√29
Duf = (−4𝑖 + 8𝑗) (
2𝑖
√29
+
5𝑗
√29
)
Duf =
32
√29
Duf = 5.94
Siempre:
Máxima derivada direccional > Derivada direccional.
Máximos y Mínimos
- Función
Máximo Mínimo
F(c) > f(a) y f(b) F(c) = < f(a) y f(b)

23. - Primera Derivada
F’(a) F’(b)
+ a - → Máximo - a + → Mínimo
- Segunda derivada
F’(c) =
>, 0 Mínimo
<, 0 Máximo
𝜕𝑓
𝜕𝑥
= 0
𝜕𝑓
𝜕𝑦
= 0
- Resolver sistema
- Encontrar puntos críticos (a,b)
𝐷 =
𝜕2
𝑓
𝜕𝑥2
𝜕2
𝑓
𝜕𝑦2
| 𝑎,𝑏 − (
𝜕2
𝑓
𝜕𝑥𝜕𝑦
)
2
| 𝑎,𝑏
𝐷 > 0 𝑦
𝜕2
𝑓
𝜕𝑥2
> 0 𝑀í𝑛𝑖𝑚𝑜
𝐷 > 0 𝑦
𝜕2
𝑓
𝜕𝑥2
< 0 𝑀á𝑥𝑖𝑚𝑜

24. F(x,y) = x4+y4-axy +1
Encontrar, máximos, mínimos, puntos silla.
𝜕𝑓
𝜕𝑥
= 0
𝜕𝑓
𝜕𝑥
= 4𝑥3
− 4𝑦 = 0
𝑦 = 𝑥3
𝜕𝑓
𝜕𝑦
= 0
𝜕𝑓
𝜕𝑦
= 4𝑥3
− 4𝑦 = 0
4𝑥3
− 4𝑥 = 0
4(𝑥3
)3
− 4𝑥 = 0
4𝑥9
− 4𝑥 = 0
4𝑥(𝑥8
− 1) = 0
𝑥8
= 1
X= +1
X= -1
Si:
X= 0 y=0
x=1 y=1
X= -1 y=-1

25. Puntos críticos:
(0,0)
(1,1)
(-1,-1)
𝜕2
𝑓
𝜕𝑥2
= 12𝑥
𝜕2
𝑓
𝜕𝑦2
= 12𝑦2
𝜕2
𝑓
𝜕𝑥𝜕𝑦
= −4
(a,b)
(
𝜕2
𝑓
𝜕𝑥2
) | 𝑎,𝑏 (
𝜕2
𝑓
𝜕𝑦2
) | 𝑎,𝑏 (
𝜕2
𝑓
𝜕𝑥2
) (
𝜕2
𝑓
𝜕𝑦2
) − (
𝜕2
𝑓
𝜕𝑥𝜕𝑦
)
2 Máximos
Mínimos
Puntos Silla
0,0 0 0 -(-4)2 = -16 D= -16 Punto silla
1,1 12 12 (12)(12) -(-4)2 = 128 D= 128 Mínimo
-1,-1 12 12 (12)(12) -(-4)2 = 128 D= 128 Mínimo
a) 𝑥2
+ 3𝑥𝑦 + 𝑦2
b) 𝑥2
+ 2𝑥𝑦 + 𝑦2
c) 𝑥2
+ 𝑥𝑦 + 𝑦2
a) F(x,y) = 𝑥2
+ 3𝑥𝑦 + 𝑦2
𝜕𝑓
𝜕𝑥
= 0
2𝑥 + 3𝑦 = 0 → 𝑥 = −
3
2
𝑦
𝜕𝑓
𝜕𝑦
= 0
3𝑥 + 2𝑦 = 0

26. 3(−
3
2
𝑦) + 2𝑦 = 0
(−
9
2
+ 2) 𝑦 = 0
𝑦 = 0  𝑥 = 0
Punto crítico:
(0,0)
𝜕2
𝑓
𝜕𝑥2
= 2
𝜕2
𝑓
𝜕𝑦2
= 2
𝜕2
𝑓
𝜕𝑥𝜕𝑦
= 3
D= (2)(2) – (3)2 = 4-9 = -5
D<0 → Punto silla (0,0)
b) 𝑥2
+ 2𝑥𝑦 + 𝑦2
𝜕𝑓
𝜕𝑥
= 0
2𝑥 + 2𝑦 = 0 → 𝑥 = −𝑦
𝜕𝑓
𝜕𝑦
= 0
2𝑥 + 2𝑦 = 0
2(−𝑦) + 2𝑦 = 0

27. 0 = 0
𝜕2
𝑓
𝜕𝑥2
= 2
𝜕2
𝑓
𝜕𝑦2
= 2
𝜕2
𝑓
𝜕𝑥𝜕𝑦
= 2
D= (2)(2) – (2)2 = 0
No existe información para concluirla
c) 𝑥2
+ 𝑥𝑦 + 𝑦2
𝜕𝑓
𝜕𝑥
= 0
2𝑥 + 𝑦 = 0 → 𝑦 = −2𝑥
𝜕𝑓
𝜕𝑦
= 0
2𝑦 + 𝑥 = 0
2(−2𝑥) + 𝑥 = 0
−3𝑥 = 0
𝑥 = 0
𝑦 = 0  𝑥 = 0
Punto crítico:
(0,0)

28. 𝜕2
𝑓
𝜕𝑥2
= 2
𝜕2
𝑓
𝜕𝑦2
= 2
𝜕2
𝑓
𝜕𝑥𝜕𝑦
= 1
D= (2)(2) – (1)2 = 3
D>0
𝜕2
𝑓
𝜕𝑥2
> 0  𝑀í𝑛𝑖𝑚𝑜
F(x,y) = x + y + 1/xy
Encontrar, máximos, mínimos, puntos silla
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
𝑥 +
1
𝑦
𝜕
𝜕𝑥
𝑥−1
= 1 −
1
𝑥2 𝑦
= 0
1 =
1
𝑥2 𝑦
𝑦 =
1
𝑥2

29. Integrales Dobles
 Tipo 1
𝐴 = ∫ ∫ 𝑑𝑦 𝑑𝑥
𝑓(𝑥)
𝑔(𝑥)
𝑏
𝑎
 Tipo 2
𝐴 = ∫ ∫ 𝑑𝑥 𝑑𝑦
𝑓(𝑥)
𝑔(𝑥)
𝑑
𝑐

30. Encontrar el área entre las parábolas
y= 2x2 y= 1 + x2
2𝑥2
= 1 + 𝑥2
𝑥2
= 1 → 𝑥 = ±1
𝐴 = ∫
𝑥=1
𝑥=−1
𝐴 = ∫ [(1 + 𝑥2) − 2𝑥2] 𝑑𝑥
1
−1
𝐴 = ∫ [(1 + 𝑥2)] 𝑑𝑥
1
−1
= ∫ 𝑑𝑥
1
−1
− ∫ 𝑥2
𝑑𝑥
1
−1
𝐴 = 𝑥|−1
1
−
𝑥3
3
𝑥|−1
1
𝐴 = 1 − (1) −
1
3
+ (−
1
3
) =

31. 4
3
𝑈2
∫ ∫ 𝑑𝑥 𝑑𝑦
√ 𝑦
0
4
0
= ∫ [𝑦2
∫ 𝑥 𝑑𝑥
√ 𝑦
0
]
4
0
𝑑𝑦 = ∫ 𝑦2
𝑥2
2
|0
√ 𝑦
4
0
𝑑𝑦
= ∫ 𝑦2
[
(√ 𝑦)
2
2
−
02
2
]
4
0
𝑑𝑦 = ∫ 𝑦2
[
𝑦
2
]
4
0
=
1
2
∫ 𝑦3
𝑑𝑦
4
0
=
1
2
𝑦4
4
|0
4
=
1
8
[(4)2
− 02
)] =
32
∫ ∫ (𝑥 − 𝑦)𝑑𝑦 𝑑𝑥
2
2𝑥
1
0
= ∫ 𝑥 [∫ 𝑑𝑦 −
2
2𝑥
∫ 𝑦𝑑𝑦
2
2𝑥
]
1
0
𝑑𝑥 = ∫ [𝑥𝑦 |2𝑥
2
1
0
−
𝑦2
2
|2𝑥
2
]𝑑𝑥
∫ [𝑥(2 − 2𝑥) − (
22
2
−
4𝑥2
2
)] 𝑑𝑥
1
0
∫ (2𝑥 − 2𝑥2
− 2 + 2𝑥2)𝑑𝑥
1
0
= −1
∫ ∫ cos(𝑠3
) 𝑑𝑡 𝑑𝑠
𝑠2
0
1
0
= ∫ [cos(𝑠3
) ∫ 𝑑𝑡 ] 𝑑𝑠
𝑠2
0
1
0
= ∫ [cos(𝑠3) (𝑠2
− 0)𝑑𝑠
1
0
= ∫ 𝑠2
cos(𝑠3) 𝑑𝑠
1
0
=
1
3
𝑠𝑒𝑛(𝑠3) |0
1

32. =
1
3
𝑠𝑒𝑛(1)
∫ ∫
𝑦
𝑥5 + 1
𝑑𝑦𝑑𝑥
𝑥2
0
1
0
= ∫ [
1
𝑥5 + 1
1
0
𝑦2
2
|0
𝑥2
]𝑑𝑥
=
1
2
∫ [
1
𝑥5 + 1
1
0
(𝑥2)2
]𝑑𝑥 =
1
2
∫ [
𝑥4
𝑥5 + 1
1
0
=
1
10
∫
𝑑𝑢
𝑢
1
0
=
1
10
ln(𝑥5
+ 1) |0
1
=
1
10
ln(2)
∫ ∫ 𝑥 𝑑𝑦𝑑𝑥
𝑠𝑒𝑛(𝑥)
0
П
0
= ∫ [𝑥 ∫ 𝑑𝑦] 𝑑𝑥
𝑠𝑒𝑛(𝑥)
0
П
0
= ∫ 𝑥 (𝑦|0
𝑠𝑒𝑛𝑥)𝑑𝑥
П
0
= ∫ 𝑥 𝑠𝑒𝑛(𝑥) 𝑑𝑥
П
0
−𝑥 cos(𝑥) |0
П
+ 𝑠𝑒𝑛 𝑥|0
П
=
П
∫ ∫ 𝑥3
𝑑𝑦𝑑𝑥
ln 𝑥
0
℮
0
= ∫ ( 𝑥3
℮
0
∫ 𝑑𝑦)
ln 𝑥
0
𝑑𝑥 = ∫ 𝑥3
𝑦|0
ln 𝑥
℮
0
𝑑𝑥 = ∫ 𝑥3
ln 𝑥 𝑑𝑥
℮
0
=
𝑥4
4
ln(𝑥) |0
℮
−
1
16
𝑥4
|0
℮
=
3
16
℮4

33. Cambio de orden de integración
∫ ∫ 𝑑𝑥𝑑𝑦
𝑥=√ 𝑦
𝑥=0
𝑦=4
𝑦=0
∫ ∫ 𝑑𝑦𝑑𝑥
4
𝑥2
2
0

34. ∫ ∫ 𝑑𝑦𝑑𝑥
𝑦=2
𝑦=2𝑥
𝑥=1
𝑥=0
∫ ∫ 𝑑𝑥𝑑𝑦
𝑥=
𝑦
2
𝑥=0
𝑦=2
𝑦=0

35. ∫ ∫ 𝑑𝑦𝑑𝑥
𝑠𝑒𝑛 𝑥
0
П
0
∫ ∫ 𝑑𝑥𝑑𝑦
П
2
𝐴𝑟𝑐𝑠𝑒𝑛 𝑦
1
0

36. Dobles Integrales
∬ 𝑦2
𝑑𝐴
𝐷
∫ ∫ 𝑦2
𝑦
−𝑦−2
1
−1
𝑑𝑥𝑑𝑦
D = (x,y) -1<y<1 , -y-2<x<y
∫ ∫ 𝑦2
𝑦
−𝑦−2
1
−1
𝑑𝑥𝑑𝑦 = ∫ (𝑦2
∫ 𝑑𝑥)
𝑦
−𝑦−2
1
−1
𝑑𝑦 = ∫ (𝑦2
1
−1
𝑥|−𝑦−2
𝑦
)𝑑𝑦
∫ (𝑦2)(2𝑦 + 2)
1
−1
𝑑𝑦 = ∫ (2𝑦3
+ 2𝑦2)𝑑𝑦
1
−1
=
4
3
𝑈2
∬
𝑦
𝑥5 + 1
𝑑𝐴
𝐷
D= (x,y) 0<x< П 0<y<x2
∫ ∫
𝑦
𝑥5 + 1
𝑥2
0
𝑑𝑦𝑑𝑥
П
0
∫ ∫
𝑦
𝑥5 + 1
𝑥2
0
𝑑𝑦𝑑𝑥
П
0
= ∫ (
1
𝑥5 + 1
∫ 𝑦
𝑥2
0
𝑑𝑦) 𝑑𝑥
П
0

37. ∫ (
1
𝑥5 + 1
𝑦2
2
|0
𝑥2
) 𝑑𝑥
П
0
=
1
2
∫ (
𝑥4
𝑥5 + 1
) 𝑑𝑥
П
0
=
1
10
∫ (
𝑑𝑢
𝑢
) 𝑑𝑢
П
0
1
10
ln(𝑥5
+ 1)|0
𝑥2
=
0.5726
∫ ∫ 𝑥𝑐𝑜𝑠𝑦
𝑥2
0
𝑑𝑦𝑑𝑥
1
0
= ∫ (𝑥 ∫ 𝑐𝑜𝑠𝑦
𝑥2
0
𝑑𝑦) 𝑑𝑥
1
0
= ∫ (𝑥𝑠𝑒𝑛 𝑦|0
𝑥2
)𝑑𝑥
1
0
= ∫ (𝑥𝑠𝑒𝑛 (𝑥2)
1
0
= −
1
2
cos(𝑥2) |0
1
=
0.2298
∬ 𝑦2
𝑑𝐴
𝐷
D región triangular cuyos vértices son: (0,1), (1,2), (4,1)
𝑚 =
∆𝑥
∆𝑦
=
2 − 1
1 − 0
= 1
𝑦 = 𝑚𝑥 + 𝑏
𝑦 = 1 + 𝑥
𝑚 =
∆𝑥
∆𝑦
=
1 − 2
4 − 1
= −
1
3

38. 𝑦 = −
1
3
𝑥 +
7
3
∫ ∫ 𝑦2
++1
0
𝑑𝑦𝑑𝑥
1
0
+ ∫ ∫ 𝑦2
−
1
3
𝑥+
7
3
1
𝑑𝑦𝑑𝑥
4
1
∫
𝑦3
3
|1
𝑥+1
𝑑𝑥
1
0
+ ∫
𝑦3
3
|1
−
1
3
𝑥+
7
3
𝑑𝑥
4
1
=
1
3
∫ (𝑥3
+ 3𝑥2
+ 3𝑥 + 1 − 1)
1
0
𝑑𝑥 +
1
3
∫ (−
1
27
𝑥3
+
7
9
𝑥2
−
49
9
𝑥 +
343
9
− 1) 𝑑𝑥 =
4
1
11
3
∬(𝑥2
+ 2𝑦)𝑑𝐴
Y = x y = x3 x>= 0
∫ ∫ (𝑥2
+ 2𝑦)𝑑𝑦𝑑𝑥
𝑥
𝑥3
1
0
∬ 𝑥𝑦2
𝑑𝐴
𝐷
D (x=0) 𝑥 = √1 − 𝑦2

39. ∫ ∫ 𝑥𝑦2
𝑑𝑥𝑑𝑦
√1−𝑦2
0
=
1
−1
∫ [𝑦2
( 𝑥 ∫ 𝑑𝑥)]𝑑𝑦
√1−𝑦2
0
1
−1
= ∫ (
𝑦2
2
1
−1
𝑥2
|0
√1−𝑦2
) 𝑑𝑦
∫ (
𝑦2
2
1
−1
) (√1 − 𝑦2)
2
𝑑𝑦 =
1
2
∫ (𝑦2
− 𝑦4
)𝑑𝑦
1
−1
=
1
6
𝑦3
|−1
1
−
1
10
𝑦5
|−1
1
=
2
15
Área:
R = cos 2
𝐴 = ∫ ∫ 𝑟𝑑𝑟𝑑𝜃
𝑟
𝑟
𝜃
𝜃
𝐴 = 8 ∫ ∫ 𝑟𝑑𝑟𝑑𝜃
𝐶𝑜𝑠 2𝜃
0
П
4
0
= ∫
𝑟2
2
П
4
0
|0
𝐶𝑜𝑠 2𝜃
=
8
2
∫ cos2
2
П
4
0
𝑑
= 2𝜃|0
П
4
+
1
2
𝑠𝑒𝑛(4𝜃)|0
П
4
=
П
2

40. Integrales Polares
∬ 𝑠𝑒𝑛 ()𝑑𝐴
𝑅
R= fuera circulo: r= 2
Dentro cardiode: r= 2+2 cos 
∫ ∫ 𝑟𝑠𝑒𝑛()
𝑟=2+2cos()
𝑟=2
П
2
0
𝑑𝑟𝑑
∫ ∫ 𝑟𝑠𝑒𝑛()
𝑟=2+2cos()
𝑟=2
П
2
0
𝑑𝑟𝑑 = ∫ 𝑠𝑒𝑛() ∫ 𝑟
2+2cos()
2
𝑑𝑟𝑑
П
2
0
=
= ∫ 𝑠𝑒𝑛()[8𝑐𝑜𝑠 + 4 cos2

П
2
0
]𝑑 = 8 ∫ 𝑠𝑒𝑛()𝑐𝑜𝑠 𝑑 + 4 ∫ 𝑠𝑒𝑛() cos2

П
2
0
П
2
0
𝑑
= 4𝑠𝑒𝑛2
 |0
П
2
−
4
3
cos3
 |0
П
2
=
16
3
𝑥2
+ 𝑦2
+ 𝑧2
= 𝑎2
𝑧 = ±√𝑎2 − 𝑥2 − 𝑦2
𝑉 = 2 ∬ √𝑎2 − 𝑥2 − 𝑦2
𝑅
𝑑𝐴
𝑟2
= 𝑥2
+ 𝑦2


41. 𝑉 = 2 ∫ ∫ √ 𝑎2 − 𝑟2
9
0
П
2
0
𝑟𝑑𝑟𝑑 =
8
3
∫ 𝑎3
𝑑
П
2
0
=
4
3
𝑎3
П
∫ ∫ 𝑟2
𝑟 𝑑𝑟𝑑
1
0
П
2
0
= ∫ ∫ 𝑟3
𝑑𝑟𝑑
1
0
П
2
0
= ∫
𝑟4
4
|0
1
𝑑
П
2
0
=
1
4
∫ 𝑑
П
2
0
=
1
8
П
∫ ∫ 𝑟 𝑑𝑟𝑑
2
0
𝐴𝑟𝑐𝑡𝑔2
0𝐴𝑟𝑐𝑡𝑔
1
3
= ∫
𝑟2
2
𝐴𝑟𝑐𝑡𝑔2
0𝐴𝑟𝑐𝑡𝑔
1
3
|0
2
𝑑 =
1
2
∫ (2)2
𝐴𝑟𝑐𝑡𝑔2
0𝐴𝑟𝑐𝑡𝑔
1
3
𝑑 =
2 ∫ 𝑑 =
𝐴𝑟𝑐𝑡𝑔2
0𝐴𝑟𝑐𝑡𝑔
1
3
1.5707
Encontrar el área de las gráficas polares.
1) R= 3 + 3 sin t
2 ∫ ∫ 𝑟 𝑑𝑟𝑑
3+3𝑠𝑒𝑛()
0
3
2
П
П
2
= ∫ 𝑟2
|0
3+3𝑠𝑒𝑛()
3
2
П
П
2
𝑑 = ∫
3
2
П
П
2
(3 + 3𝑠𝑒𝑛())
2
𝑑 =
∫
3
2
П
П
2
(9 + 18 𝑠𝑒𝑛  + 9 𝑠𝑒𝑛2
 )𝑑 =

42. 9П +
9
2
П
2) R = cos 
2 ∫ ∫ 𝑟 𝑑𝑟𝑑
2+cos
0
П
0
= ∫ 𝑟2
П
0
|0
2+cos
= ∫ (2 + cos )
П
0
2
= ∫ (4 + 4 cos  + cos2
)
П
0
𝑑 =
14.1371
3) R = 2 sen 
2 ∫ ∫ 𝑟 𝑑𝑟𝑑
2 𝑠𝑒𝑛 
0
1
6
П
0
+ 2 ∫ ∫ 𝑟 𝑑𝑟𝑑
1
0
П
2
П
6
=
∫ 𝑟2
1
6
П
0
|0
2 𝑠𝑒𝑛 ()
𝑑 + ∫ 𝑟2
П
2
П
6
|0
1
𝑑
= ∫ (2 𝑠𝑒𝑛 ())2
1
6
П
0
𝑑 + ∫ 𝑑
П
2
П
6
= ∫ 4 𝑠𝑒𝑛2
()
1
6
П
0
𝑑 + |П
6
П
2
=
1.2283

43. 4) R= 5 cos 3
2 ∫ ∫ 4 𝑟 𝑑𝑟𝑑
5 𝑐𝑜𝑠 3
0
1
6
П
0
= 4 ∫ 𝑟2
1
6
П
0
|0
5 cos3
= 4 ∫ (5 cos 3)2
1
6
П
0
=
100 ∫ cos2
3
1
6
П
0
=
26.1799
5) Z = √𝑥2 + 𝑦2
Z= 0
𝑍 = 4 ∫ ∫ √𝑥2 + 𝑦2
5
0
𝑟𝑑𝑟𝑑
П
2
0
= 4 ∫ ∫ √ 𝑟2
5
0
𝑟𝑑𝑟𝑑
П
2
0
= 4 ∫ ∫ 𝑟2
5
0
𝑑𝑟𝑑
П
2
0
=
4
3
∫ 𝑟3
|0
5
𝑑
П
2
0
=
4
3
∫ (5)3
𝑑
П
2
0
=
500
3
∫ 𝑑
П
2
0
=
261.7993

44. 6) 𝑥2
+ 𝑦2
= 4 𝑧 = √9 − 𝑥2 − 𝑦2 𝑧 = 0
𝑟 = 4 ∫ ∫ √9 − 𝑥2 − 𝑦2
2
0
𝑟𝑑𝑟𝑑
П
2
0
= 4 ∫ ∫ √9 − 𝑟2
2
0
𝑟𝑑𝑟𝑑
П
2
0
= −2 ∫ ∫ 𝑢
1
2
2
0
𝑑𝑢𝑑
П
2
0
= −
4
3
∫ 𝑢
3
2|0
2
𝑑
П
2
0
= −
4
3
∫ (9 − 𝑟2)(
3
2
)
|0
2
𝑑
П
2
0
= −23.4153
7) 𝑥2
+ 𝑦2
= 1 𝑧 = √16 − 𝑥2 − 𝑦2
𝑥2
+ 𝑦2
= 9 𝑧 = 0
= 4 ∫ ∫ √16 − 𝑥2 − 𝑦2
3
0
𝑟𝑑𝑟𝑑
П
2
0
= 4 ∫ ∫ √16 − 𝑟2
3
0
𝑟𝑑𝑟𝑑
П
2
0
= −2 ∫ ∫ 𝑢
1
2
3
0
𝑑𝑢𝑑
П
2
0
= −
4
3
∫ 𝑢
3
2|0
3
𝑑
П
2
0
= −
4
3
∫ (16 − 𝑟2)(
3
2
)
|0
3
𝑑
П
2
0
= 82.87

45. 25) 𝑦 = √9 − 𝑥2 𝑦2
= 9 − 𝑥2
𝑥2
+ 𝑦2
= 9
∫ ∫ √𝑥2 + 𝑦2
√9−𝑥2
0
𝑑𝑦𝑑𝑥
3
−3
= 2 ∫ ∫ √ 𝑟2
3
0
𝑟𝑑𝑟𝑑
П
2
0
= 2 ∫ ∫ 𝑟2
3
0
𝑑𝑟𝑑
П
2
−3
=
2
3
∫ 𝑟3
|0
3
𝑑
П
2
0
= 18 ∫ 𝑑
П
2
0
=
9П