Question

determine a integral dupla variando 0 a 1 e de 0 a raiz de 1-y^2 de (x+y)dxdy

Answer 1

Resposta:

0 a 1 ∫  0 a √(1-y^2) ∫  (x+y)  dx   dy

0 a 1 ∫  0 a √(1-y^2) [ (x²/2+xy) ]    dy

0 a 1 ∫  [ (√(1-y^2)²/2+y√(1-y^2) ]    dy

0 a 1 ∫  [ (1-y^2)/2+y√(1-y²) ]    dy

0 a 1 ∫   (1-y^2)/2   dy + 0 a 1 ∫  y√(1-y²)     dy

*****  0 a 1 ∫   (1-y^2)/2 dy

0 a 1 [ y/2 -y³/6 ] =1/2-1/6 =3/6 -1/6 =2/6=1/3   ---   (i)

***** 0 a 1 ∫  y√(1-y²)     dy

u=1-y²  ==>du=-2y dy

0 a 1 ∫  y√u     du/(-2y)

0 a 1 ∫  (-1/2)*√u     du

0 a 1 [  (-1/2)*u^(3/2)  /(3/2) ]

0 a 1 [  (-1/3)*u^(3/2)   ]

u =  1-y²

0 a 1 [  (-1/3)*(1-y²)^(3/2)   ]  =-[-1/3]=1/3   -- (ii)

(i)+(ii)

1/3+1/3 =2/3