5.43(b)

Calcule el trabajo realizado por el campo en una particula en el semicirculo superior

x
2

y
2

=
4

{\displaystyle x^2 + y^2 = 4}

 recorriendolo una vez en sentido antihorario.

Tenemos que la curva a recorrer es

x
2

y
2

=
4

{\displaystyle x^2 + y^2 = 4}

, por lo que el radio es

2

{\displaystyle 2}

. Como nos movemos en
sentido antihorario, vamos de

0

{\displaystyle 0}

 a

π

{\displaystyle \pi }

.

Entonces, parametrizamos la curva como sigue:

# x
2
cos
⁡
t

{\displaystyle x=2\cos t}

# y
2
sin
⁡
t

{\displaystyle y=2\sin t}

Necesitamos a

r
(
t
)

{\displaystyle r(t)}

 en forma de vector, por lo que expresamos la curva como

r
(
t
# )
2
cos
⁡
t

i

2
sin
⁡
t

j

{\displaystyle r(t)=2\cos t{\textbf {i}}+2\sin t{\textbf {j}}}

.
Usaremos la definición del libro para calcular el trabajo realizado en un campo:

∫

C

F
⋅
d
# r

∫

a

b

F
(
r
(
t
)
)
⋅

r
′

(
t
)
d
t

{\displaystyle \int \_{C}F\cdot dr=\int \_{a}^{b}F(r(t))\cdot r'(t)dt}

Entonces procedemos a encontrar

F
(
r
(
t
)
)

{\displaystyle F(r(t))}

 y

r
′

(
t
)

{\displaystyle r'(t)}

. sabiendo que la ecuación del campo vectorial es

F
(
x
,
y
# )
(
x
−
y
)

i

x
y

j

{\displaystyle F(x,y)=(x-y){\textbf {i}}+xy{\textbf {j}}}

:

F
(
r
(
t
)
)

=
(
2
cos
⁡
t
−
2
sin
⁡
t
)

i

2
cos
⁡
t
2
sin
⁡
t

j

=
(
2
cos
⁡
t
−
2
sin
⁡
t
)

i

(
4
(
cos
⁡
t
sin
⁡
t
)
)

j

=
(
2
cos
⁡
t
−
2
sin
⁡
t
)

i

(
2
sin
⁡
(
2
t
)
)

j

=
(
2
cos
⁡
t
−
2
sin
⁡
t
,
2
sin
⁡
(
2
t
)
)

{\displaystyle {\begin{aligned}F(r(t))&=(2\cos t-2\sin t){\textbf {i}}+2\cos t2\sin t{\textbf {j}}\\\\&=(2\cos t-2\sin t){\textbf {i}}+(4(\cos t\sin t)){\textbf {j}}\\\\&=(2\cos t-2\sin t){\textbf {i}}+(2\sin(2t)){\textbf {j}}\\\\&=(2\cos t-2\sin t,2\sin(2t))\\\\\end{aligned}}}

Ahora buscamos

r
′

(
t
)

{\displaystyle r'(t)}

 y derivamos:

r
(
t
)

=
2
cos
⁡
t

i

2
sin
⁡
t

j

⇒

r
′

(
t
)

=
−
2
sin
⁡
t

i

2
cos
⁡
t

j

=
(
−
2
sin
⁡
t
,
2
cos
⁡
t
)

{\displaystyle {\begin{aligned}r(t)&=2\cos t{\textbf {i}}+2\sin t{\textbf {j}}\\\\\Rightarrow r'(t)&=-2\sin t{\textbf {i}}+2\cos t{\textbf {j}}\\\\&=(-2\sin t,2\cos t)\\\\\end{aligned}}}

Entonces calculamos la integral:

# W

∫

C

F
⋅
d
r

=

∫

a

b

F
(
r
(
t
)
)
⋅

r
′

(
t
)
d
t

=

∫

0

π

(
2
cos
⁡
t
−
2
sin
⁡
t
,
2
sin
⁡
(
2
t
)
)
⋅
(
−
2
sin
⁡
t
,
2
cos
⁡
t
)
d
t

=

∫

0

π

(
−
4
sin
⁡
t
cos
⁡
t
4

sin

2

⁡
t
4
sin
⁡
2
t
cos
⁡
t
)
d
t

=

∫

0

π

(
−
2
sin
⁡
2
t
4

sin

2

⁡
t
2
sin
⁡
3
t
2
sin
⁡
t
)
d
t

=
−

∫

0

π

2
sin
⁡
2
t
d
t

∫

0

π

4

sin

2

⁡
t
d
t

∫

0

π

2
sin
⁡
3
t
d
t

∫

0

π

2
sin
⁡
t
d
t

=
cos
⁡
2
t
4

∫

0

π

1
−
cos
⁡
2
t

2

d
t
2

∫

0

π

sin
⁡
3
t
d
t
−
2
cos
⁡
t

=
cos
⁡
2
t
2

∫

0

π

1
−
cos
⁡
2
t
d
t
2

∫

0

π

sin
⁡
u

3

d
u
−
2
cos
⁡
t

=
cos
⁡
2
t
2
t
−
sin
⁡
2
t

2
3

∫

0

π

sin
⁡
u
d
u
−
2
cos
⁡
t

=
cos
⁡
2
t
2
t
−
sin
⁡
2
t
−

2
3

cos
⁡
3
t
−
2
cos
⁡
t

=
cos
⁡
2
t
2
t
−
sin
⁡
2
t
−

2
3

cos
⁡
3
t
−
2
cos
⁡
t

0

π

=
[
(
cos
⁡
2
π
2
π
−
sin
⁡
2
π
−

2
3

cos
⁡
3
π
−
2
cos
⁡
π
)
−
(
cos
⁡
2
(
0
)
2
(
0
)
−
sin
⁡
2
(
0
)
−

2
3

cos
⁡
3
(
0
)
−
2
cos
⁡
0
)
]

=
[
(
1
2
π
−
0

2
3

2
)
−
(
1
−

2
3

−
2
]

=
[
2
π

11
3

5
3

# ]
2
π

16
3

{\displaystyle {\begin{aligned}W=\int \_{C}F\cdot dr&=\int \_{a}^{b}F(r(t))\cdot r'(t)dt\\\\&=\int \_{0}^{\pi }(2\cos t-2\sin t,2\sin(2t))\cdot (-2\sin t,2\cos t)dt\\\\&=\int \_{0}^{\pi }(-4\sin t\cos t+4\sin ^{2}t+4\sin 2t\cos t)dt\\\\&=\int \_{0}^{\pi }(-2\sin 2t+4\sin ^{2}t+2\sin 3t+2\sin t)dt\\\\&=-\int \_{0}^{\pi }2\sin 2tdt+\int \_{0}^{\pi }4\sin ^{2}tdt+\int \_{0}^{\pi }2\sin 3tdt+\int \_{0}^{\pi }2\sin tdt\\\\&=\cos 2t+4\int \_{0}^{\pi }{\frac {1-\cos 2t}{2}}dt+2\int \_{0}^{\pi }\sin 3tdt-2\cos t\\\\&=\cos 2t+2\int \_{0}^{\pi }1-\cos 2tdt+2\int \_{0}^{\pi }{\frac {\sin u}{3}}du-2\cos t\\\\&=\cos 2t+2t-\sin 2t+{\frac {2}{3}}\int \_{0}^{\pi }\sin udu-2\cos t\\\\&=\cos 2t+2t-\sin 2t-{\frac {2}{3}}\cos 3t-2\cos t\\\\&=\cos 2t+2t-\sin 2t-{\frac {2}{3}}\cos 3t-2\cos t{\big |}\_{0}^{\pi }\\\\&=[(\cos 2\pi +2\pi -\sin 2\pi -{\frac {2}{3}}\cos 3\pi -2\cos \pi )-(\cos 2(0)+2(0)-\sin 2(0)-{\frac {2}{3}}\cos 3(0)-2\cos 0)]\\\\&=[(1+2\pi -0+{\frac {2}{3}}+2)-(1-{\frac {2}{3}}-2]\\\\&=[2\pi +{\frac {11}{3}}+{\frac {5}{3}}]=2\pi +{\frac {16}{3}}\end{aligned}}}