Section 15.4 Applications of Double Integrals 8. Find the mass and center of mass of the lamina that occupies the region

Section 15.4 Applications of Double Integrals

8. Find the mass and center of mass of the lamina that occupies the region D and has the given density function ρ.

D is the triangular region enclosed by the lines y = 0, y = 2x, and x + 2y = 1; ρ(x, y) = x Solution:

SECTION 15.4 APPLICATIONS OF DOUBLE INTEGRALS ¤ 553 5. =2

0

3−

2( + )   =2 0

 +¹₂²^=3−

=2 =2 0

(3 − ) +¹2(3 − )²−¹2²−¹8²



=2 0

−⁹8²+⁹₂

 =

−⁹8

₁

3³ +⁹₂2

0= 6,

=2 0

3−

2 (²+ )   =2 0

² +¹₂²=3−

=2  =2 0

₉

2 −⁹8³

 =⁹₂,

=2 0

3−

2 ( + ²)   =2 0

1

2²+¹₃³=3−

=2  =2 0

9 −⁹2 = 9.

Hence  = 6, ( ) =



 





=

3 43

2

 .

6.Here  =

( ) | 0 ≤  ≤ ²5 2 ≤  ≤ 1 − 2.

 =25 0

1−2

2    =25 0

1

2²=1−2

=2  = ¹₂25 0



(1 − 2)²−1 22



=¹₂25 0

15

4²− 4 + 1

 = ¹₂5

4³− 2²+ 25 0 =¹₂2

25−25⁸ +²₅

=₂₅²,

 =25 0

1−2

2  ·    =25 0

₁

3³=1−2

=2  = ¹₃25 0



(1 − 2)³−₁

23



= ¹₃25 0

−⁶⁵8³+ 12²− 6 + 1

 = ¹₃

−⁶⁵32⁴+ 4³− 3²+ 25 0 =¹₃

−250¹³ +₁₂₅³² −¹²25+²₅

= ₇₅₀³¹,

 =25 0

1−2

2  ·    =25 0 1

2²=1−2

=2  = ¹₂25 0 15

4²− 4 + 1



=¹₂25 0

₁₅

4³− 4²+ 

 = ¹₂₁₅

16⁴−⁴3³+¹₂²25 0 = ¹₂ ₃

125−375³² +₂₅²

=₇₅₀⁷ . Hence  =₂₅², ( ) =_31750

225 ^7750_225

=31 60₆₀⁷.

7.  =1

−1

1−²

0    = 1

−1

1

2²=1−²

=0  =¹₂1

−1(1 − ²)² =¹₂1

−1(1 − 2²+ ⁴) 

=¹₂

 −²3³+¹₅⁵1

−1=¹₂

1 −²3 +¹₅ + 1 −²3+¹₅

=₁₅⁸,

 =1

−1

1−²

0    = 1

−1

1

2²=1−²

=0  =¹₂1

−1 (1 − ²)² =¹₂1

−1( − 2³+ ⁵) 

= ¹₂1

2²−¹2⁴+¹₆⁶1

−1=¹₂1

2−¹2+¹₆−¹2+¹₂−¹6

= 0,

 =1

−1

1−²

0 ²  = 1

−1

₁

3³=1−²

=0  = ¹₃1

−1(1 − ²)³ = ¹₃1

−1(1 − 3²+ 3⁴− ⁶) 

=¹₃

 − ³+³₅⁵−¹7⁷1

−1=¹₃

1 − 1 +³5 −¹7 + 1 − 1 +³5−¹7

= ₁₀₅³².

Hence  =₁₅⁸, ( ) =

0^32105_815 

= 0⁴₇

.

8.The boundary curves intersect when  + 2 = ² ⇔ ²−  − 2 = 0 ⇔  = −1,  = 2. Thus here

 =

( ) | −1 ≤  ≤ 2 ²≤  ≤  + 2 .

 =2

−1

+2

² ²  = 2

−1²

=+2

=²  = 2

−1(³+ 2²− ⁴) 

= ₁

4⁴+²₃³−¹5⁵2

−1= ₄₄

15+¹³₆₀

= ⁶³₂₀,

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

13. A lamina occupies the part of the disk x²+ y² ≤ 1 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.

Solution:

554 ¤ CHAPTER 15 MULTIPLE INTEGRALS

 =2

−1

+2

² ³  = 2

−1³

=+2

=²  = 2

−1(⁴+ 2³− ⁵) 

= 1

5⁵+¹₂⁴−¹6⁶2

−1= 56 15−15²

= ¹⁸₅,

 =2

−1

+2

² ²   = 2

−1²1

2²=+2

=²  =¹₂2

−1²(²+ 4 + 4 − ⁴) 

=¹₂2

−1(⁴+ 4³+ 4²− ⁶)  = ¹₂₁

5⁵+ ⁴+⁴₃³−¹7⁷2

−1=¹₂₁₅₅₂

105 +₁₀₅⁴¹

= ⁵³¹₇₀.

Hence  = ⁶³₂₀, ( ) =

185

6320^53170_6320

=₈

7¹¹⁸₄₉ .

9.  =1 0

^−

0    =1 0 ₁

2²=^−

=0  = ¹₂1 0 

^−2

 =¹₂1

0 ^−2

integrate by parts with

 =   = ^−2



=¹₂

−¹4(2 + 1)^−21 0= −¹8

3⁻²− 1

=¹₈ −³8⁻²,

 =1 0

^−

0 ²   =1 0 ²1

2²=^−

=0  =¹₂1

0 ²^−2 [integrate by parts twice]

=¹₂

−¹4

2²+ 2 + 1

^−21 0= −¹8

5⁻²− 1

= ¹₈−⁵8⁻²,

 =1 0

^−

0 ²  =1 0 ₁

3³=^−

=0  =¹₃1

0 ^−3

=¹₃

−¹9(3 + 1)^−31 0= −27¹

4⁻³− 1

=₂₇¹ −27⁴⁻³.

Hence  = ¹₈

1 − 3⁻², ( ) =

₁

8

1 − 5⁻²

1

8(1 − 3⁻²)

1 27

1 − 4⁻³

1

8(1 − 3⁻²)



=

²− 5

²− 3 8

³− 4 27 (³− 3)

 .

10.Note that cos  ≥ 0 for −2 ≤  ≤ 2.

 =2

−2

cos 

0    =2

−2

₁

2²=cos 

=0  =¹₂2

−2cos²  =¹₂₁

2 +¹₄sin 22

−2=^₄,

=2

−2

cos 

0    =2

−2₁

2²=cos 

=0  =¹₂2

−2 cos² 

 integrate by parts with

 =   = cos² 



=¹₂

1

2 +¹₄sin 2^2

−2−2

−2

1

2 +¹₄sin 2



=¹₂

1

8²−¹8²−1

4²−¹8cos 22

−2

= ¹₂ 0 −1

16²+¹₈ −16¹²−¹8

= 0,

=2

−2

cos 

0 ²  =2

−2

1

3³=cos 

=0  =¹₃2

−2cos³  =¹₃2

−2(1 − sin²) cos  

[substitute  = sin  ⇒  = cos  ]

= ¹₃

sin  −¹3sin³2

−2=¹₃

1 −¹3+ 1 −¹3

=⁴₉.

Hence  = ^₄, ( ) = 0_4^49

= 0¹⁶_9.

11.( ) = ,  =

  =2 0

1

0 ( sin )    = 2

0 sin  1 0 ²

= 

− cos 2 0

₁

3³1

0= (1)₁

3

= ¹₃,

 =

 ·   =2 0

1

0 ( cos )( sin )    = 2

0 sin  cos  1 0 ³

= ₁

2sin²2 0

₁

4⁴1 0= ₁

2

 ₁

4

= ¹₈,

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

SECTION 15.4 APPLICATIONS OF DOUBLE INTEGRALS ¤ 555

 =

 ·   =2 0

1

0 ( sin )²   = 2

0 sin² 1 0 ³

= 1

2 −¹4sin 22 0

1 4⁴1

0=  4

 1 4

=₁₆^.

Hence ( ) =

8

3^16_3 

=₃

8^3₁₆. 12. ( ) = (²+ ²) = ²,  =2

0

1

0 ³  = ^₈,

=2 0

1

0 ⁴cos    =¹₅2

0 cos   =¹₅ sin 2

0 =¹₅,

=2 0

1

0 ⁴sin    = ¹₅2

0 sin   = ¹₅

− cos 2 0 =¹₅.

Hence ( ) = ₈

5_5⁸.

13. ( ) = 

²+ ²= ,

 =

( ) = 0

2

1  ·   

=  0 2

1 ² = ()₁

3³2 1=⁷₃,

=

( ) = 0

2

1( cos )()    = 

0 cos  2 1 ³

=  sin 

0

1 4⁴2

1= (0)15 4

= 0 [this is to be expected as the region and density function are symmetric about the y-axis]

=

( ) = 0

2

1( sin )()    = 

0 sin  2 1 ³

= 

− cos  0

₁

4⁴2

1= (1 + 1)₁₅

4

=¹⁵₂

Hence ( ) =

0_73^152

= 0_14⁴⁵.

14. Now ( ) =  

²+ ² = , so

 =

( ) = 0

2

1()    =  0 2

1  = ()(1) = ,

=

( ) = 0

2

1( cos )()    = 

0 cos  2 1  

=  sin 

0

1 2²2

1= (0)3 2

= 0,

=

( ) = 0

2

1( sin )()    = 

0 sin  2 1  

= 

− cos  0

₁

2²2

1= (1 + 1)₃

2

= 3.

Hence ( ) = 0_^3

= 0_³.

15. Placing the vertex opposite the hypotenuse at (0 0), ( ) = (²+ ²). Then

 = 0

− 

0 

²+ ²

  =  0

²− ³+¹₃( − )³

 = ₁

3³−¹4⁴−12¹ ( − )⁴

0 =¹₆⁴. [continued]

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

18. A lamina occupies the region inside the circle x²+ y² = 2y but outside the circle x²+ y²= 1. Find the center of mass if the density at any point is inversely proportional to its distance from the origin.

Solution:

556 ¤ CHAPTER 15 MULTIPLE INTEGRALS By symmetry, = =

0

− 

0 (²+ ²)   =  0

1

2( − )²²+¹₄( − )⁴



= 1

6²³−¹4⁴+₁₀¹⁵−20¹( − )⁵

0 =₁₅¹⁵ Hence ( ) =2

5²₅. 16.( ) = 

²+ ²= .

 =

 56

6

 2 sin  1



   = 

 56

6

[(2 sin ) − 1] 

= 

−2 cos  − 56

6 = 2√

3 −^3



By symmetry of  and () = , = 0, and

=56

6

2 sin 

1  sin    =¹₂56

6 (4 sin³ − sin ) 

=¹₂

−3 cos  +⁴3cos³56

6 =√ 3  Hence ( ) =



0 ^3√3

2(^3√3− )

 .

17. =

²( ) =3 1

4

1 ²· ²  = 3 1 4

1 ⁴ =  []³₁1 5⁵4

1= (2)1023 5

= 4092,

 =

²( )  =3 1

4

1 ²· ²  = 3

1 ²4

1 ² = 1 3³3

1

1 3³4

1= 26 3

(21) = 182,

and 0= + = 4092 + 182 = 5912.

18. =

²( ) =25 0

1−2

2 ²·    =25 0 ²₁

2²=1−2

=2  = ¹₂25

0 ²(¹⁵₄²− 4 + 1) 

=¹₂25

0 (¹⁵₄⁴− 4³+ ²)  =¹₂₃

4⁵− ⁴+¹₃³25 0 =₉₃₇₅¹⁶ ,

 =

²( )  =25 0

1−2

2 ²·    =25 0

₁

4⁴=1−2

=2  = ¹₄25 0

(1 − 2)⁴−16¹⁴



=¹₄25

0 (²⁵⁵₁₆⁴− 32³+ 24²− 8 + 1)  =¹4

51

16⁵− 8⁴+ 8³− 4²+ 25 0 =₃₁₂₅⁷⁸ , and 0= + =₉₃₇₅¹⁶ +₃₁₂₅⁷⁸ =₇₅².

19.As in Exercise 15, we place the vertex opposite the hypotenuse at (0 0) and the equal sides along the positive axes.

= 0

−

0 ²(²+ ²)   =  0

−

0 (²²+ ⁴)   =  0

₁

3²³+¹₅⁵=−

=0 

=  0

₁

3²( − )³+¹₅( − )⁵

 = ₁

3

₁

3³³−³4²⁴+³₅⁵−¹6⁶

−30¹( − )⁶

0 =₁₈₀⁷ ⁶,

= 0

−

0 ²(²+ ²)   =  0

−

0 (⁴+ ²²)   =  0

⁴ +¹₃²³=−

=0 

=  0

⁴( − ) +¹3²( − )³

 = ₁

5⁵−¹6⁶+¹₃₁

3³³−³4²⁴+³₅⁵−¹6⁶

0 =₁₈₀⁷ ⁶, and 0= + =₉₀⁷⁶.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

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