Section 8.3 Applications to Physics and Engineering

(1)

Section 8.3 Applications to Physics and Engineering

764 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

8.3 Applications to Physics and Engineering

1. The weight density of water is  = 625 lbft³. (a)  =  ≈ (625 lbft³)(3 ft) = 1875 lbft²

(b)  =  ≈ (1875 lbft²)(5 ft)(2 ft) = 1875 lb. ( is the area of the bottom of the tank.) (c) By reasoning as in Example 1, the area of the th strip is 2 (∆) and the pressure is  = . Thus,

 =3

0  · 2  ≈ (625)(2)3

0   = 1251 2²3

0= 1259 2

= 5625 lb.

2. (a)  =  = (820 kgm³)(98 ms²)(15 m) = 12,054 Pa ≈ 12 kPa

(b)  =   = (12,054 Pa)(8 m)(4 m) ≈ 386 × 10⁵N ( is the area at the bottom of the tank.) (c) The area of the th strip is 4(∆) and the pressure is  =  . Thus,

 =15

0  · 4  = (820)(98) · 415

0   = 32,144₁

2²32

0 = 16,072₉

4

≈ 362 × 10⁴N.

In Exercises 3–9,is the number of subintervals of length∆and^∗is a sample point in theth subinterval[−1 ]. 3. Set up a vertical x -axis as shown, with  = 0 at the water’s surface and  increasing in the

downward direction. Then the area of the th rectangular strip is 2 ∆ and the pressure on the strip is ^∗ (where  ≈ 625 lbft³). Thus, the hydrostatic force on the strip is

^∗_ · 2 ∆ and the total hydrostatic force ≈



=1

^∗_· 2 ∆. The total force

x x_i* w i 2 ft

8 ft

=2 3

0

11

 = lim

→∞



=1

^∗ · 2 ∆ =11

3  · 2  = 211

3   = 21 2²11

3 = (121 − 9) = 112 ≈ 7000 lb 4. Set up a vertical axis as shown. Then the area of the th rectangular strip is

2(^∗− 2) ∆.



By similar triangles, 

^∗_ − 2= 10

5, so = 2(^∗ − 2)



The pressure on the strip is ^∗_, so the hydrostatic force on the strip is ^∗ · 2(^∗− 2) ∆ and the total hydrostatic force on the plate ≈^

=1

^∗_ · 2(^∗− 2) ∆. The total force

 = lim

→∞



=1

^∗ · 2(^∗^− 2) ∆ =7

2  · 2( − 2)  = 27

2(²− 2) 

= 2₁

3³− ²7

2= 2₃₄₃

3 − 49

−₈

3− 4

= 2₂₀₀

3

= ⁴⁰⁰₃  ≈ ⁴⁰⁰3 (625) = 83333lb.

5. Set up a coordinate system as shown. Then the area of the th rectangular strip is 2

8²− (^∗)²∆. The pressure on the strip is = (12 − ^∗), so the hydrostatic force on the strip is (12 − ^∗) 2

64 − (^∗)²∆and the total hydrostatic force on the plate ≈^

=1(12 − ^∗^) 2

64 − (^∗)²∆.

8. Set up a vertical -axis as shown. Then the area of the th rectangular strip is 3^∗∆.



By similar triangles,

^∗_ =6

2, so = 3^∗.



The pressure on the strip is

(^∗ + 4), so the hydrostatic force on the strip is (^∗+ 4)3^∗∆and the hydrostatic force on the plate ≈^

=1

(^∗ + 4)3^∗∆. The total force

 = lim

→∞



=1

(^∗ + 4) 3^∗∆ =2

0 ( + 4) 3  = 32

0(²+ 4) 

= 31

3³+ 2²2

0= 38 3+ 8

= 32 = 313,600 N [ = 1000,  ≈ 98]

9. Set up a vertical -axis as shown. Then the area of the th rectangular strip is

∆ =

4 + 2 ·²3^∗_∆. The pressure on the strip is (^∗_ − 1), so the hydrostatic force on the strip is (^∗ − 1)

4 +⁴₃^∗

∆and the hydrostatic

force on the plate ≈^

=1

(^∗− 1)

4 +⁴₃^∗_∆. The total force

 = lim

→∞



=1

(^∗− 1)

4 +⁴₃^∗_

∆ =3

1 ( − 1) 4 +⁴₃

 = 3 1

₄

3²+⁸₃ − 4



= ₄

9³+⁴₃²− 43 1 = 

(12 + 12 − 12) −₄

9+⁴₃ − 4

= ₁₂₈

9

≈ 889 lb [ ≈ 62.5]

10. Set up a vertical x-axis as shown. Then the area of the th rectangular strip is (4 − ^∗) ∆.

By similar triangles,

4 = 4 − ^∗^

4 , so = 4 − ^∗. The th

rectangular strip is (^∗− 1) m below the surface level of the water, so the pressure on the strip is (^∗_ − 1). The hydrostatic force on the strip is (^∗ − 1)(4 − ^∗) ∆and the hydrostatic force on the plate ≈^

=1

(^∗ − 1)(4 − ^∗) ∆. The total force

 = lim

→∞



=1

(^∗− 1)(4 − ^∗^) ∆ =2

1 ( − 1)(4 − )  = 2

1(−²+ 5 − 4) 

= 

−¹3³+⁵₂²− 42 1= 

−⁸3+ 10 − 8

−

−¹3+⁵₂− 4

=⁷₆ ≈ ⁷6· 1000 · 98 ≈ 114 × 10⁴N Note: If you let the water level correspond to  = 0, then  =1

0 (3 − ) .

11. Set up a vertical x-axis as shown. Then the area of the th rectangular strip is



(2 − ^∗) ∆.



By similar triangles, 

2 − ^∗

= 2

2, so =

(2 − ^∗)



The pressure on the strip is ^∗, so the hydrostatic force on the plate

≈



=1

^∗



(2 − ^∗^) ∆. The total force

SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING ¤ 767

 = lim

→∞



=1

^∗_ 

(2 − ^∗) ∆ = 





0 (2 − )  =



 0

2 − ²



=



²−¹3³ 0 = 



³−¹3³

= 



2³ 3



= ²₃²

12. (a) The solution is similar to the solution for Example 2. The pressure on a strip is approximately = 646(3 − ^^∗)and the total force is

 = lim

→∞



=1646(3 − ^∗^)2

9 − (^∗)²∆ = 12923

−3(3 − )

9 − ²

= 1292 · 33

−3

9 − ² − 12923

−3

9 − ²

= 3876 ·¹2(3)²− 0

the ﬁrst integral is the area of a semicircular disk with radius 3 and the second integral is 0 because the integrand is an odd function



= (17442) ≈ 5480 lb

(b) If the tank is half full, the surface of the milk is  = 0, so the pressure on a strip is approximately = 646(0 − ^∗^). The upper limit of integration changes from 3 to 0 and the total force is

 = 12920

−3(0 − )

9 − ² = 1292

1

3(9 − ²)^320

−3= 1292(9 − 0) = 11628 lb Note that this is about 21% of the force for a full tank.

13.By similar triangles, 8 4√

3=

^∗_ ⇒ ^= 2^∗

√3. The area of the th rectangular strip is 2^∗

√3 ∆and the pressure on it is  4√

3 − ^∗

.

 =

 4√ 3

0

 4√

3 −  2

√3 = 8

 4√ 3

0   −2

√3

 4√ 3

0

²

= 4

²4√ 3

0 − 2

3√ 3

³4√ 3

0 = 192 − 2

3√

364 · 3√

3 = 192 − 128 = 64

≈ 64(840)(98) ≈ 527 × 10⁵N 14. =2

0 (10 − )2√

4 − ²

= 202 0

√4 − ² − 2 0

√4 − ²2 

= 20¹₄(2²) − 4

0 ^12 [ = 4− ²,  = −2 ]

= 20 −²3

^324

0= 20 −¹⁶3 = 

20 −¹⁶3



= (1000)(98)

20 −¹⁶3

≈ 563 × 10⁵N

15. (a) The top of the cube has depth  = 1 m − 20 cm = 80 cm = 08 m.

 =  ≈ (1000)(98)(08)(02)²= 3136 ≈ 314 N (b) The area of a strip is 02 ∆ and the pressure on it is ^∗.

 =1

08(02)  = 02₁

2²1

08= (02)(018) = 0036 = 0036(1000)(98) = 3528 ≈ 353 N

 = lim

→∞



=1

^∗_ 

(2 − ^∗) ∆ = 





0 (2 − )  = 



 0

2 − ²



=



²−¹3³ 0 =



³−¹3³

=



2³ 3



=²₃²

12. (a) The solution is similar to the solution for Example 2. The pressure on a strip is approximately = 646(3 − ^^∗)and the total force is

 = lim

→∞



=1646(3 − ^^∗)2

9 − (^∗)²∆ = 12923

−3(3 − )

9 − ²

= 1292 · 33

−3

9 − ² − 12923

−3

9 − ²

= 3876 ·¹2(3)²− 0

the ﬁrst integral is the area of a semicircular disk with radius 3 and the second integral is 0 because the integrand is an odd function



= (17442) ≈ 5480 lb

(b) If the tank is half full, the surface of the milk is  = 0, so the pressure on a strip is approximately = 646(0 − ^^∗). The upper limit of integration changes from 3 to 0 and the total force is

 = 12920

−3(0 − )

9 − ² = 1292

1

3(9 − ²)^320

−3= 1292(9 − 0) = 11628 lb Note that this is about 21% of the force for a full tank.

13. By similar triangles, 8 4√

3= 

^∗_ ⇒ ^=2^∗

√3. The area of the th rectangular strip is2^∗

√3∆and the pressure on it is  4√

3 − ^∗

.

 =

 4√ 3

0

 4√

3 −  2

√3 = 8

 4√ 3

0   −2

√3

4√ 3

0

²

= 4

²4√ 3

0 − 2

3√ 3

³4√ 3

0 = 192 − 2

3√

364 · 3√

3 = 192 − 128 = 64

≈ 64(840)(98) ≈ 527 × 10⁵N 14.  =2

0 (10 − )2√

4 − ²

= 202 0

√4 − ² − 2 0

√4 − ²2 

= 20¹₄(2²) − 4

0 ^12 [ = 4− ²,  = −2 ]

= 20 −²3

^324

0 = 20 −¹⁶3 = 

20 −¹⁶3



= (1000)(98)

20 −¹⁶3

≈ 563 × 10⁵N

15. (a) The top of the cube has depth  = 1 m − 20 cm = 80 cm = 08 m.

 =  ≈ (1000)(98)(08)(02)²= 3136 ≈ 314 N (b) The area of a strip is 02 ∆ and the pressure on it is ^∗.

 =1

08(02)  = 02₁

2²1

08= (02)(018) = 0036 = 0036(1000)(98) = 3528 ≈ 353 N

768 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 16. The height of the dam is  =√

70²− 25² cos 30^◦= 15√ 19√

3 2

.

The width of the trapezoid is  = 50 + 2.

By similar triangles,25

 = 

 −  ⇒  = 25

( − ). Thus,

 = 50 + 2 ·25

( − ) = 50 +50

 ·  −50

 ·  = 50 + 50 −50

 = 100 −50

 . From the small triangle in the second ﬁgure, cos 30^◦=∆

 ⇒

 = ∆ sec 30^◦= 2 ∆√ 3.

 =

  0





100 −50



 2

√3 =200

√3

  0

  − 100

√ 3

  0

²

= 200

√3

² 2 −100

√ 3

³

3 =200² 3√

3 = 200(625) 3√

3 ·12,825

4 ≈ 771 × 10⁶lb

17. (a) The area of a strip is 10 ∆ and the pressure on it is .

 =1

0 10  = 101 2²1

0= 10 ·¹2 = 5

= 5(1000)(98) = 49 000N = 49 × 10⁴N

(b)  =3

0 10  = 10₁

2²3

0= 10 ·⁹2 = 45 = 45(1000)(98) = 441 000N ≈ 44 × 10⁵N.

(c) For the ﬁrst 1 m, the length of the side is constant at 20 m. For 1   ≤ 3, we can use similar triangles to ﬁnd the length :



20= 3 − 

2 ⇒  = 20 ·3 − 

2 = 10(3 − ).

 =1

0  20  +3

1  10(3 − )  = 20₁

2²1

0+ 103

1(3 − ²) 

= 20₁

2

+ 10₃

2²−¹3³3

1= 10 + 10₂₇

2 − 9

−₃

2−¹3



= 10 + 10₁₀

3

=¹³⁰₃  = ¹³⁰₃ (1000)(98)N ≈ 424 667 N ≈ 42 × 10⁵N

(d) For any right triangle with hypotenuse on the bottom, sin  = ∆

hypotenuse ⇒ hypotenuse = ∆ csc  = ∆

√20²+ 2²

2 =√101 ∆.

 =3

1 10√

101  = 10√ 101 1

2²3 1= 5√

101 (9 − 1)

= 40√

101(1000)(98) ≈ 3 939 551 N ≈ 39 × 10⁶N

18. Partition the interval [ ] by points as usual and choose ^∗ ∈ [^−1 ]for each . The th horizontal strip of the immersed plate is approximated by a rectangle of height ∆and width (^∗), so its area is ≈ (^∗) ∆. For small

∆, the pressure on the th strip is almost constant and ≈ ^∗by Equation 1. The hydrostatic force acting on the

th strip is = ≈ ^∗(^∗) ∆. Adding these forces and taking the limit as  → ∞, we obtain the hydrostatic force on the immersed plate:

 = lim

→∞



=1

= lim

→∞



=1

^∗_(^∗) ∆=

() 

19. From Exercise 18, we have  =

()  =94

70 64() . From the table, we see that ∆ = 04, so using Simpson’s Rule to estimate  , we get

 ≈ 64^043 [70(70) + 4(74)(74) + 2(78)(78) + 4(82)(82) + 2(86)(86) + 4(90)(90) + 94(94)]

=^256₃ [7(12) + 296(18) + 156(29) + 328(38) + 172(36) + 36(42) + 94(44)]

=^256₃ (48604) ≈ 4148 lb

20. (a) From Equation 8,  = _¹ 

()  ⇒  =

()  ⇒  = 

()  ⇒ () =

 ()  = by Exercise 18.

(b) For the ﬁgure in Exercise 10, let the coordinates of the centroid ( ) =

√ 2 0.

 = () =  

√2²= 

√2  2 ²=

√2 ³ 2 . 21. The moment  of the system about the origin is  =²

=1

= 11+ 22 = 6 · 10 + 9 · 30 = 330.

The mass  of the system is  =²

=1

= 1+ 2= 6 + 9 = 15.

The center of mass of the system is  =  =³³⁰₁₅ = 22.

22. The moment  is 11+ 22+ 33= 12(−3) + 15(2) + 20(8) = 154. The mass  is

1+ 2+ 3 = 12 + 15 + 20 = 47. The center of mass is  =  =¹⁵⁴₄₇.

23.  =

3

=1

= 6 + 5 + 10 = 21.

=

3

=1

= 6(5) + 5(−2) + 10(−1) = 10; ^=

3

=1

= 6(1) + 5(3) + 10(−2) = 1.

 =

 = 1

21and  =

 =10

21, so the center of mass of the system is 1 21¹⁰₂₁.

24. =

4

=1

= 6(−2) + 5(4) + 1(−7) + 4(−1) = −3, ^=

4

=1

= 6(1) + 5(3) + 1(−3) + 4(6) = 42,

and  =⁴

=1

= 16, so  =

 = 42 16=21

8 and  = 

 = −3

16; the center of mass is ( ) =21 8 −16³

.

th strip is = ≈ ^∗(^∗) ∆. Adding these forces and taking the limit as  → ∞, we obtain the hydrostatic force on the immersed plate:

 = lim

→∞



=1

= lim

→∞



=1

^∗_(^∗) ∆=

() 

19. From Exercise 18, we have  =

()  =94

70 64() . From the table, we see that ∆ = 04, so using Simpson’s Rule to estimate  , we get

 ≈ 64^043 [70(70) + 4(74)(74) + 2(78)(78) + 4(82)(82) + 2(86)(86) + 4(90)(90) + 94(94)]

=^256₃ [7(12) + 296(18) + 156(29) + 328(38) + 172(36) + 36(42) + 94(44)]

=^256₃ (48604) ≈ 4148 lb

20. (a) From Equation 8,  = _¹ 

()  ⇒  =

()  ⇒  = 

()  ⇒ () =

 ()  = by Exercise 18.

(b) For the ﬁgure in Exercise 10, let the coordinates of the centroid ( ) =

√ 2 0.

 = () =  

√2²= 

√2  2 ²=

√2 ³ 2 . 21. The moment  of the system about the origin is  =²

=1

= 11+ 22 = 6 · 10 + 9 · 30 = 330.

The mass  of the system is  =²

=1

= 1+ 2= 6 + 9 = 15.

The center of mass of the system is  =  =³³⁰₁₅ = 22.

22. The moment  is 11+ 22+ 33= 12(−3) + 15(2) + 20(8) = 154. The mass  is

1+ 2+ 3 = 12 + 15 + 20 = 47. The center of mass is  =  =¹⁵⁴₄₇.

23.  =

3

=1

= 6 + 5 + 10 = 21.

=

3

=1

= 6(5) + 5(−2) + 10(−1) = 10; ^=

3

=1

= 6(1) + 5(3) + 10(−2) = 1.

 =

 = 1

21and  =

 =10

21, so the center of mass of the system is 1 21¹⁰₂₁.

24. =

4

=1

= 6(−2) + 5(4) + 1(−7) + 4(−1) = −3, ^=

4

=1

= 6(1) + 5(3) + 1(−3) + 4(6) = 42,

and  =⁴

=1

= 16, so  =

 = 42 16=21

8 and  = 

 = −3

16; the center of mass is ( ) =21 8 −16³

.

1

(2)

SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING ¤ 771 29.  =1

0(^12− ²)  =

2

3^32−¹3³1 0=₂

3−¹3

− 0 = ¹3.

 =_¹ 1

0 (^12− ²)  = 31

0(^32− ³) 

= 3

2

5^52−¹4⁴1 0= 32

5−¹4

= 33 20

=₂₀⁹.

 = _¹ 1 0 1 2



(^12)²− (²)²

 = 3₁

2

 1

0( − ⁴) 

= ³₂1

2²−¹5⁵1 0=³₂1

2−¹5

=³₂3 10

=₂₀⁹.

Thus, the centroid is ( ) =₉

20₂₀⁹.

30. The curves intersect when 2 − ²=  ⇔ 0 = ²+  − 2 ⇔ 0 = ( + 2)( − 1) ⇔  = −2 or  = 1.

 =1

−2(2 − ²− )  =

2 −¹3³−¹2²1

−2=⁷₆ −

−¹⁰3

= ⁹₂.

 =_¹ 1

−2(2 − ²− )  =²9

1

−2(2 − ³− ²) 

=²₉

²−¹4⁴−¹3³1

−2=²₉₅

12−⁸3

= −¹2.

 = ¹

1

−2 1

2[(2 − ²)²− ²]  =²9·¹2

1

−2(4 − 5²+ ⁴) 

= ¹₉

4 −⁵3³+¹₅⁵1

−2=¹₉38 15−

−¹⁶15

= ²₅.

Thus, the centroid is ( ) = (−¹2²₅).

31.  =4

0 (cos  − sin )  =

sin  + cos 4

0 =√

2 − 1.

 = ⁻¹4

0 (cos  − sin ) 

= ⁻¹

(sin  + cos ) + cos  − sin 4

0 [integration by parts]

= ⁻¹ 4

√2 − 1

=

1 4√

2 − 1

√2 − 1 .

 = ⁻¹4 0

1

2(cos² − sin²)  = _2¹ 4

0 cos 2  =_4¹ 

sin 24

0 = 1

4 = 1

4√

2 − 1 . Thus, the centroid is ( ) =

 √ 2 − 4 4√

2 − 1  1 4√

2 − 1



≈ (027 060).

32.  =1

0 ³ +2

1(2 − )  =1 4⁴1

0+

2 −¹2²2 1

= ¹₄+ (4 − 2) − 2 −¹2

=³₄.

 =_¹1

0 (³)  +2

1 (2 − ) 

= ⁴₃1

0 ⁴ +2

1(2 − ²) 

=⁴₃₁

5⁵1 0+

²−¹3³2 1



= ⁴₃1 5+

4 −⁸3

− 1 −¹3



=⁴₃13 15

=⁵²₄₅.

[continued]

 = _¹1 0

1

2(³)² +2 1

1

2(2 − )²

=²₃1

0 ⁶ +2

1( − 2)²

= ²₃1 7⁷1

0+1

3( − 2)³2 1



=²₃₁

7 − 0 + 0 +¹3

= ²₃₁₀

21

= ²⁰₆₃.

Thus, the centroid is ( ) =₅₂

45²⁰₆₃.

33. The curves intersect when 2 −  = ² ⇔ 0 = ²+  − 2 ⇔ 0 = ( + 2)( − 1) ⇔  = −2 or  = 1.

 =1

−2(2 −  − ²)  =

2 −¹2²−¹3³1

−2=⁷₆ −

−¹⁰3

= ⁹₂.

 = _¹ 1

−2 1

2[(2 − )²− (²)²]  = ²₉·¹2

1

−2(4 − 4 + ²− ⁴) 

= ¹₉

4 − 2²+¹₃³−¹5⁵1

−2=¹₉₃₂

15−

−¹⁸⁴15

=⁸₅.

 = _¹ 1

−2(2 −  − ²)  = ²₉1

−2(2 − ²− ³) 

=²₉

²−¹3³−¹4⁴1

−2=²₉5 12−⁸3

= −¹2.

Thus, the centroid is ( ) = (⁸₅ −¹2).

34. An equation of the line is  = −³2 + 3.  = ¹₂(2)(3) = 3, so  =  = 4(3) = 12.

= 2 0 1 2

−³2 + 32

 =¹₂2 0

9

4²− 9 + 9

 =¹₂(4)3

4³−⁹2²+ 92

0= 2(6 − 18 + 18) = 12.

= 2 0 

−³2 + 3

 = 2 0

−³2²+ 3

 = 4

−¹2³+³₂²2

0= 4(−4 + 6) = 8.

 =

 = 8

12 =2

3and  = 

 =12

12= 1. Thus, the center of mass is ( ) =₂

3 1. Since  is constant, the center of mass is also the centroid.

35. The quarter-circle has equation  =√

4²− ²for 0 ≤  ≤ 4 and the line has equation  = −2.

 = ¹₄(4)²+ 2(4) = 4 + 8 = 4( + 2), so  =  = 6 · 4( + 2) = 24( + 2).

= 4 0 1 2

√16 − ²2

− (−2)²

 = ¹₂4

0 (16 − ²− 4)  = ¹2(6)

12 −¹3³4 0= 3

48 −⁶⁴3

= 80.

= 4 0 √

16 − ²− (−2)

 = 4 0 √

16 − ² + 4

0 2  = 6

−¹3(16 − ²)^324 0+ 6

²4 0

= 6

0 +⁶⁴₃+ 6(16) = 224.

 =

 = 224

24( + 2)= 28

3( + 2)and  =

 = 80

24( + 2) = 10 3( + 2). Thus, the center of mass is

 28

3( + 2) 10 3( + 2)



≈ (182 065).

36. We’ll use  = 8, so ∆ =^^{− }_ = ⁸^{− 0}₈ = 1.

 =8

0  ()  ≈ ¹⁰=¹₃[ (0) + 4 (1) + 2 (2) + 4 (3) + 2 (4) + 4 (5) + 2 (6) + 4 (7) +  (8)]

≈¹3[0 + 4(20) + 2(26) + 4(23) + 2(22) + 4(33) + 2(40) + 4(32) + 0]

=¹₃(608) = 2026 or ³⁰⁴₁₅

Now 8

0   ()  ≈¹3[0 · (0) + 4 · 1 · (1) + 2 · 2 · (2) + 4 · 3 · (3)

+ 2 · 4 · (4) + 4 · 5 · (5) + 2 · 6 · (6) + 4 · 7 · (7) + 8 · (8)]

≈¹3[0 + 8 + 104 + 276 + 176 + 66 + 48 + 896 + 0]

=¹₃(2672) = 8906 or ¹³³⁶₁₅ , so  =_¹ 8

0  ()  ≈ 439.

Also, 8

0 [ ()]² ≈ ¹3[0²+ 4(20)²+ 2(26)²+ 4(23)²+ 2(22)²+ 4(33)²+ 2(40)²+ 4(32)²+ 0²]

= ¹₃(17688) = 5896, so  =_¹ 8 0

1

2[ ()]² ≈ 145.

Thus, the centroid is ( ) ≈ (44 15).

37.  =1

−1[(³− ) − (²− 1)]  =1

−1(1 − ²) 

odd-degree terms drop out



= 21

0(1 − ²)  = 2

 −¹3³1 0= 22

3

=⁴₃.

 =_¹ 1

−1(³−  − ²+ 1)  = ³₄1

−1(⁴− ²− ³+ ) 

=³₄1

−1(⁴− ²)  = ³₄· 21

0(⁴− ²) 

=³₂₁

5⁵−¹3³1 0= ³₂

−15²

= −¹5.

 = _¹ 1

−1 1

2[(³− )²− (²− 1)²]  = ³₄·¹2

1

−1(⁶− 2⁴+ ²− ⁴+ 2²− 1) 

= ³₈· 21

0(⁶− 3⁴+ 3²− 1)  = ³4

1

7⁷−³5⁵+ ³− 1 0= ³₄

−¹⁶35

= −¹²35.

Thus, the centroid is ( ) =

−¹5 −¹²35

.

38. The curves intersect at  =  ≈ −1315974 and  =  ≈ 053727445.

 =

[(2 − ²) − ^]  =

2 −¹3³− ^

≈ 1452014.

 =_¹ 

(2 − ²− ^)  = _¹

²−¹4⁴− ^+ ^



≈ −0374293

 = _¹ 

 1

2[(2 − ²)²− (^)²]  = _2¹ 

(4 − 4²+ ⁴− ^2) 

= _2¹ 

4 −⁴3³+¹₅⁵−¹2^2

≈ 1218131 Thus, the centroid is ( ) ≈ (−037 122).

39. Choose - and -axes so that the base (one side of the triangle) lies along the -axis with the other vertex along the positive -axis as shown. From geometry, we know the medians intersect at a point²₃ of the way from each vertex (along the median) to the opposite side. The median from  goes to the midpoint₁

2( + ) 0of side , so the point of intersection of the medians is₂

3 ·¹2( + )¹₃

=₁

3( + )¹₃ .

This can also be verified by finding the equations of two medians, and solving them simultaneously to find their point of intersection. Now let us compute the location of the centroid of the triangle. The area is  =¹₂( − ).

[continued]

45. A cone of height  and radius  can be generated by rotating a right triangle about one of its legs as shown. By Exercise 39,  =¹₃, so by the Theorem of Pappus, the volume of the cone is

 =  =1

2· base · height

· (2) =¹2 · 21 3

=¹₃².

46. From the symmetry in the ﬁgure,  = 4. So the distance traveled by the centroid when rotating the triangle about the -axis is  = 2 · 4 = 8. The area of the triangle is  =¹₂ =¹₂(2)(3) = 3. By the Theorem of Pappus, the volume of the resulting solid is  = 3(8) = 24.

47. The curve  is the quarter-circle  =√

16 − ², 0 ≤  ≤ 4. Its length  is¹4(2 · 4) = 2.

Now ⁰= ¹₂(16 − ²)^−12(−2) = −

√16 − ² ⇒ 1 + (⁰)²= 1 + ²

16 − ² = 16 16 − ² ⇒

 =

1 + (⁰)² = 4

√16 − ² so

 = 1





  = 1 2

 4 0

4(16 − ²)^−12 = 4 2



−(16 − ²)^124 0= 2

(0 + 4) = 8

 and

 = 1





  = 1 2

4 0

16 − ²· 4

√16 − ²  = 4 2

 4 0

 = 2





4 0= 2

(4 − 0) = 8

. Thus, the centroid is

8

8





. Note that the centroid does not lie on the curve, but does lie on the line  = , as expected, due to the symmetry of the curve.

48. (a) From Exercise 47, we have  = (1)

  ⇔  = . The surface area is

 =

2  = 2  = 2() = (2), which is the product of the arc length of  and the distance traveled by the centroid of .

(b) From Exercise 47,  = 2 and  =_⁸. By the Second Theorem of Pappus, the surface area is

 = (2) = 2(2 ·⁸) = 32.

A geometric formula for the surface area of a half-sphere is  = 2². With  = 4, we get  = 32, which agrees with our ﬁrst answer.

49. The circle has arc length (circumference)  = 2. As in Example 7, the distance traveled by the centroid during a rotation is

 = 2. Therefore, by the Second Theorem of Pappus, the surface area is

 =  = (2)(2) = 4²