528 CHAPTER 26. THE RIEMANNN INTEGRAL ON Rp

x

z

y

R

Then∫

E f dV =∫

R∫ b(x,y)

a(x,y) f (x,y,z)dzdA It might be helpful to think of dV = dzdA. Now∫ b(x,y)a(x,y) f (x,y,z)dz is a function of x and y and so you have reduced the triple integral to a

double integral over R of this function of x and y. It is similar if the region in R3 were ofthe form {(x,y,z) : a(y,z)≤ x ≤ b(y,z)} or {(x,y,z) : a(x,z)≤ y ≤ b(x,z)}.

Example 26.3.2 Find the volume of the region E in the first octant between z = 1− (x+ y)and z = 0.

In this case, R is the region shown.

x

y

R

1 x

y

z

Thus the region E is between the plane z = 1− (x+ y) on the top, z = 0 on the bottom,and over R shown above. Thus∫

E1dV =

∫R

∫ 1−(x+y)

0dzdA =

∫ 1

0

∫ 1−x

0

∫ 1−(x+y)

0dzdydx =

16

Of course iterated integrals have a life of their own although this will not be exploredhere. You can just write them down and go to work on them. Here are some examples.

Example 26.3.3 Find∫ 3

2∫ x

3∫ x

3y (x− y) dzdydx.

The inside integral yields∫ x

3y (x− y) dz = x2 −4xy+3y2. Next this must be integratedwith respect to y to give

∫ x3(x2 −4xy+3y2

)dy=−3x2+18x−27. Finally the third integral

gives ∫ 3

2

∫ x

3

∫ x

3y(x− y) dzdydx =

∫ 3

2

(−3x2 +18x−27

)dx =−1.