26.3. METHODS FOR TRIPLE INTEGRALS 529
Example 26.3.4 Find∫
π
0∫ 3y
0∫ y+z
0 cos(x+ y) dxdzdy.
The inside integral is∫ y+z
0 cos(x+ y) dx = 2coszsinycosy+2sinzcos2 y−sinz−siny.Now this has to be integrated.∫ 3y
0
∫ y+z
0cos(x+ y) dxdz
=∫ 3y
0
(2coszsinycosy+2sinzcos2 y− sinz− siny
)dz
=−1−16cos5 y+20cos3 y−5cosy−3(siny)y+2cos2 y.
Finally, this last expression must be integrated from 0 to π . Thus∫π
0
∫ 3y
0
∫ y+z
0cos(x+ y) dxdzdy
=∫
π
0
(−1−16cos5 y+20cos3 y−5cosy−3(siny)y+2cos2 y
)dy =−3π
Example 26.3.5 Here is an iterated integral:∫ 2
0∫ 3− 3
2 x0
∫ x2
0 dzdydx. Write as an iteratedintegral in the order dzdxdy.
The inside integral is just a function of x and y. (In fact, only a function of x.) The orderof the last two integrals must be interchanged. Thus the iterated integral which needs to bedone in a different order is ∫ 2
0
∫ 3− 32 x
0f (x,y) dydx.
As usual, it is important to draw a picture and then go from there.
3− 32 x = y
3
2Thus this double integral equals
∫ 3
0
∫ 23 (3−y)
0f (x,y) dxdy.
Now substituting in for f (x,y),
∫ 3
0
∫ 23 (3−y)
0
∫ x2
0dzdxdy.
Example 26.3.6 Find the volume of the bounded region determined by 3y+ 3z = 2,x =16− y2,y = 0,x = 0.