x=3·r·cos φ
y=4·r·sin φ ⇒ (x2/9)+(y2/16)=1 ⇒ r2=1 ⇒ r=1
Якобиан
abr=3·4·r
y=4√3·x
4·r·sin φ =4·√3·3·r·cos φ ⇒ tg φ =3√3 ⇒ φ =arctg 3√3
y=–4x/√3
4·r·sin φ =–4·3·r·cos φ/√3 ⇒ tg φ =–√3 ⇒ φ =2π/3
= ∫10 dr ∫ 2π/3arctg3√3(3·r·cos φ )2·(4·r·sin φ )2· (12 r)d φ =
=32·42·12 ∫10r5 dr ∫ 2π/3arctg3√3((1+cos 2φ )/2)·((1–cos2φ)/2 ) d φ =
=432∫10r5 dr ∫ 2π/3arctg3√3(1–cos2 2φ )d φ =
=432∫10r5 dr ∫ 2π/3arctg3√3(sin2 2φ )d φ =
=432∫10r5 dr ∫ 2π/3arctg3√3((1–cos4φ )/2)d φ =
=216·(r6/6)|10 ( φ –(1/4)sin4 φ )| 2π/3arctg3√3 =...