Задача 35802 ...

kolya

13 апреля 2019 г. в 16:20

Вычислить площади фигур, ограниченные линиями:

{x = 6cost 5π/6 ≤ t ≤ π/6;
{ y = 6sint
ρ = 3sin6φ.

Вычислить длины дуг кривых:

{x = 3(2cost – cos2t), 0 ≤ t ≤ 2π;
y = 3(2sint – sin2t)
ρ = 2e^4φ/3, –π/2 ≤ φ ≤ π/2.

sova

13 апреля 2019 г. в 22:19

★

S=∫ ^β _α y(t)·x`(t)dt=

= ∫^π/6_5π/6 6sint·(6cost)`dt=

=–36 ∫^π/6_5π/6 sin²tdt=

=–36∫^π/6_5π/6 (1–cos2t)dt/2=

=–18·(t–(1/2)sin2t)|^π/6_5π/6=

=18·(4π/6)+9·(sin(2π/6)–sin(10π/6))=

=12π+9·((√3/2)–(–√3/2))= 12π+9√3

S=(1/2)∫ ^β _α ρ²( φ )d φ =

=6· ∫ ^π/6₀(3sin6 φ )²d φ =

=54 ∫ ^π/6₀(sin²6 φ )d φ=

=54 ∫ ^π/6₀(1–cos12 φ )d φ/2=

=27·( φ – (1/12)sin(12 φ ))^π/6₀=

=27·(π/6)–(1/12)sin2π+(1/12)sin0=

=9π/2

3.

x(t)=3·(2cost–cos2t)

y(t)=3·(2sint–sin2t)

L= ∫ t2t1√(x`(t))<sup>2</sup>+(y`(t))2dt

x`(t)=3·(–2sint+2sin2t)

y`(t)=3·(2cost–2cos2t)

(x`(t))²=9·(4sin²t–8sint·sin2t+4sin²2t)

(y`(t))²=9·(4cos^t–8costcos2t+4cos²2t)

(x`(t))<sup>2</sup>+(y`(t))²=9·(4·(sin²t+cos²t)–8sint·sin2t–8costcos2t+4·(sin²2t+cos²2t))=

=9·(8–8·(cos(2t–t))=9·8·(1–cost)=72·2sin²(t/2)=144sin²t/2


L= ∫ ^2π₀√144sin²(t/2)dt=

=12 ∫ ^2π₀sin(t/2)dt=

=12(–2cos(t/2))^2π₀=

=–24·(cosπ–cos0)= 48

4.
L= ∫ ^β _α √ρ²+(ρ`)²dφ

ρ=2e^{4φ /3}

ρ`=2e<sup>4 φ /3</sup>·(4 φ /3)`=2·(4/3)·e^{4 φ /3}=(8/3)e^{4 φ /3}

ρ²+(ρ`)²=(2e^{4φ /3})²+((8/3)e^{4φ /3})²=

=(e^{4 φ /3})²·(4+(64/9))= ((10/9)e^{4 φ /3})²

L= ∫ ^π/2 _{– π/2}(10/9)·e^{4 φ /3}d φ =

=(3/4)·(10/9)e^{4 φ /3}|^π/2_–π/2=

= (5/6)·(e^2π/3–e^–2π/3)