Download - 2 1 2 1 4 2 4 2s 4x2 4 2 4x2x R x 1 R S dx =2π x dx π x +1) dx π x π +4 3 · PDF filey = 1 4 x2 − 1 2 lnx ⇒ dy dx = x 2 ... 4x2 = x2 4 + 1 2 + 1 4x2 = x 2 + 1 2x 2. So S =

Transcript

Page 1: 2 1 2 1 4 2 4 2s 4x2 4 2 4x2x R x 1 R S dx =2π x dx π x +1) dx π x π +4 3 · PDF filey = 1 4 x2 − 1 2 lnx ⇒ dy dx = x 2 ... 4x2 = x2 4 + 1 2 + 1 4x2 = x 2 + 1 2x 2. So S =

y = 1

4x2

−1

2ln x ⇒

dy

dx=

x

2−

1

2x⇒ 1 +

(

dy

dx

)2

= 1 +x2

4−

1

2+

1

4x2=

x2

4+

1

2+

1

4x2=

(

x

2+

1

2x

)2

. So

S =∫

4

12πx

√

(

x

2+

1

2x

)2

dx = 2π∫

4

1x

(

x

2+

1

2x

)

dx = π∫

4

1(x2 + 1) dx = π

[

1

3x3 + x

]4

1

= π[(

64

3+ 4

)

−

(

1

3+ 1

)]

= 24π

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