414
SOLUTIONS DES EXERCICES
Directement
I =Io~ dx Io~ dy Io~ sin(x +y+ z) dz
1r
1r
z=1E.
= f 2 dx f 2 [-cos(x+y+z)]
2 dy
Jo
Jo
z=O
71"
71"
= fo 2 dx fo 2 ( cos(x +y) - cos(x +y+~) )dy
=Io~ [sin(x +y) - sin(x +y+ ~)]~:j dx
=Io~ (sin(x + ~) - sin(x + 7r) - sinx + sin(x + ~)) dx
=Io~ 2cosx dx = 2.
Exercice 5.26. I = Jfl xz dx dy dz ;
D = {(x,y,z) E JR 3 I y~ 0; z ~ 0; x+z ~ 1; y 2 ~ x}.
B étant l'ensemble des (x, y) tels que { 2 ~ ~ ~ 1 on a :
y °""'X°""' '
I = Jl (fol-x xz dz) dx dy =il x (fol-x z dz) dx dy =il x(l; x)
2 dx dy
=fol (l: x(l; x)
2 dx) dy = ~fol (l: (x - 2x 2 + x 3 ) dx) dy
111 [x2 2x3 x4] x=l
=- - - + -
dy
2 o 2
3
4 x=y2
1 [ 1 ( 1
4
2
6
1
8 )
= 2 lo 2(1 - Y ) - 3(l - Y ) + 4(l -y ) dy
= ! (! (1 - !) - ~ (1 - !) + ! (1 - !)) = 630 - 126 - 840 + 120 + 315 - 35
2 2
5
3
7
4
9
8x5x7x9
l= ~ = _!_,
2520
315
Exercice 5.27. I = Jf l xyz dx dy dz;
{
x2 y2 z2
}
D = (x,y,z) E JR 3 I x ~ 0; y~ 0; z ~ 0; 2 + 2 + 2 ~ 1 .
a
a
a
On fait le changement de variable x = aX, y= bY, z = bZ; le Jacobien de ce