358 CHAPTER 19. THE RIEMANNN INTEGRAL ON Rn

Example 19.3.5 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 19.3.6 Find the volume of the bounded region determined by 3y+ 3z = 2,x =16− y2,y = 0,x = 0.

In the yz plane, the first of the following pictures corresponds to x = 0.

3y+3z = 2

23

23

y (0,0,0)

z

x = 16− y2

Therefore, the outside integrals taken with respect to z and y are of the form∫ 2

30∫ 2

3−y0 dzdy,

and now for any choice of (y,z) in the above triangular region, x goes from 0 to 16− y2.Therefore, the iterated integral is∫ 2

3

0

∫ 23−y

0

∫ 16−y2

0dxdzdy =

860243

Example 19.3.7 Find the volume of the region determined by the intersection of the twocylinders, x2 + y2 ≤ 1 and x2 + z2 ≤ 1.