2.18

Encuentre la masa total y el centro de masa de una lámina cuya forma está determinada por las curvas

# y

x

,
# y
0
y
# x
4

{\displaystyle y={\sqrt {x}},y=0yx=4}

, y una función densidad

ρ
(
x
,
y
# )
k
x
y

{\displaystyle \rho (x,y)=kxy}

 donde

k

{\displaystyle k}

 es una constante.

Solución (suponiendo la constante

k

{\displaystyle k}

 positiva):

Calculamos primero la masa

m

{\displaystyle m}

 estableciendo los límites de integración en los que corren

x

{\displaystyle x}

 y

y

{\displaystyle y}

 como sigue:

m

=

∬

D

ρ
(
x
,
y
)
d
A

⇒

∫

0

4

∫

0

x

k
x
y
d
y
d
x

=

∫

0

4

k
x

[

1
2

y

2

0

x

]

d
x

=

∫

0

4

k
x

[

1
2

y

2

0

x

]

d
x

=

∫

0

4

k
x

[

x

2

2

−

0
2

]

d
x

=

∫

0

4

1
2

k

x

2

d
x

=

1
2

k

∫

0

4

x

2

d
x

=

1
2

k

[

1
3

x

3

0

4

]

=

1
2

k

[

4

3

3

−

0
3

]

=

1
2

k

[

64
3

]

=

64
6

# k

32
3

k

{\displaystyle {\begin{aligned}m&=\iint \limits \_{D}\rho (x,y)dA\\\\&\Rightarrow \int \_{0}^{4}\int \_{0}^{\sqrt {x}}kxydydx\\\\&=\int \_{0}^{4}kx\left[{\frac {1}{2}}y^{2}{\Big |}\_{0}^{\sqrt {x}}\right]dx\\\\&=\int \_{0}^{4}kx\left[{\frac {1}{2}}y^{2}{\Big |}\_{0}^{\sqrt {x}}\right]dx\\\\&=\int \_{0}^{4}kx\left[{\frac {{\sqrt {x}}^{2}}{2}}-{\frac {0}{2}}\right]dx\\\\&=\int \_{0}^{4}{\frac {1}{2}}kx^{2}dx\\\\&={\frac {1}{2}}k\int \_{0}^{4}x^{2}dx\\\\&={\frac {1}{2}}k\left[{\frac {1}{3}}x^{3}{\Big |}\_{0}^{4}\right]\\\\&={\frac {1}{2}}k\left[{\frac {4^{3}}{3}}-{\frac {0}{3}}\right]\\\\&={\frac {1}{2}}k\left[{\frac {64}{3}}\right]\\\\&={\frac {64}{6}}k={\frac {32}{3}}k\end{aligned}}}

Entonces la masa es

# m

32
3

k

{\displaystyle m={\frac {32}{3}}k}

.

Ahora calculamos el centro de masa con coordeandas

(

x
¯

,

y
¯

)

{\displaystyle ({\overline {x}},{\overline {y}})}

 usando las
fórmulas del libro, y tenemos:

x
¯

=

1
m

∬

D

x
ρ
(
x
,
y
)
d
A

=

1

32
k

3

∫

0

4

∫

0

x

k

x

2

y
d
y
d
x

=

3

32
k

∫

0

4

k

x

2

[

1
2

y

2

0

x

]

d
x

=

3

32
k

∫

0

4

k

x

2

[

(

x

)

2

2

−

0
2

]

d
x

=

3

32
k

∫

0

4

k

x

3

2

d
x

=

3
k

32
k

[

x

4

8

0

4

]

=

3
32

[

4

4

8

−

0
8

]

=

3
32

[

256
8

]

=

3
32

[
32
]

=
3

{\displaystyle {\begin{aligned}{\overline {x}}&={\frac {1}{m}}\iint \limits \_{D}x\rho (x,y)dA\\\\&={\frac {1}{\frac {32k}{3}}}\int \_{0}^{4}\int \_{0}^{\sqrt {x}}kx^{2}ydydx\\\\&={\frac {3}{32k}}\int \_{0}^{4}kx^{2}\left[{\frac {1}{2}}y^{2}{\Big |}\_{0}^{\sqrt {x}}\right]dx\\\\&={\frac {3}{32k}}\int \_{0}^{4}kx^{2}\left[{\frac {({\sqrt {x}})^{2}}{2}}-{\frac {0}{2}}\right]dx\\\\&={\frac {3}{32k}}\int \_{0}^{4}{\frac {kx^{3}}{2}}dx\\\\&={\frac {3k}{32k}}\left[{\frac {x^{4}}{8}}{\Big |}\_{0}^{4}\right]\\\\&={\frac {3}{32}}\left[{\frac {4^{4}}{8}}-{\frac {0}{8}}\right]\\\\&={\frac {3}{32}}\left[{\frac {256}{8}}\right]={\frac {3}{32}}\left[32\right]=3\\\\\end{aligned}}}

Ahora calculamos

y
¯

{\displaystyle \overline{y}}

:

y
¯

=

1
m

∬

D

y
ρ
(
x
,
y
)
d
A

=

1

32
k

3

∫

0

4

∫

0

x

k
x

y

2

d
y
d
x

=

3

32
k

∫

0

4

k
x

[

1
3

y

3

0

x

]

d
x

=

3

32
k

∫

0

4

k
x

[

(

x

)

3

3

−

0
3

]

d
x

=

3

32
k

∫

0

4

k

x

2

x

3

d
x

=

3
k

32
k

∫

0

4

x

5
2

3

d
x

=

3
32

[

2

x

7
2

21

0

4

]

=

3
32

[

2

x

7

21

0

4

]

=

3
32

[

2

x

3

x

21

0

4

]

=

3
32

[

2
⋅

4

3

4

21

−

2
⋅

0

3

0

21

]

=

3
32

[

256
21

]

=

768
672

=

8
7

Entonces, el centro de masa es

(

3
,

8
7

)

{\displaystyle \left(3,{\frac {8}{7}}\right)}

. La representación gráfica se ve de la siguiente forma: