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Triple Integrals in Cylindrical
Coordinates

Triple Integrals in Cylindrical Coordinates
In this section we want do take a look at triple integrals done completely in Cylindrical
Coordinates
...
The following are the conversion formulas for
cylindrical coordinates
...
We will be able to show in the Change of Variables section of
this chapter that,

dV = r dz dr dθ

The region, E, over which we are integrating becomes,

E = {( x, y, z ) | ( x, y ) ∈ D, u1 ( x, y ) ≤ z ≤ u2 ( x, y )}

= {( r ,θ , z ) | α ≤ θ ≤ β , h1 (θ ) ≤ r ≤ h2 (θ ) , u1 ( r cos θ , r sin θ ) ≤ z ≤ u2 ( r cos θ , r sin θ )}

Note that we’ve only given this for E’s in which D is in the xy-plane
...

In terms of cylindrical coordinates a triple integral is,
β h2 (θ ) u2 ( r cosθ ,r sinθ )
f ( x, y, z ) dV =
r f ( r cos θ , r sin θ , z ) dz dr dθ
α h1 (θ ) u1 ( r cosθ ,r sin θ )

∫∫∫

∫ ∫

E

∫

Don’t forget to add in the r and make sure that all the x’s and y’s also get converted over into
cylindrical coordinates
...


Example 1 Evaluate

∫∫∫ y dV

where E is the region that lies below the plane z = x + 2 above

E

the xy-plane and between the cylinders x 2 + y 2 = 1 and x 2 + y 2 = 4
...

We’ll start out by getting the range for z in terms of cylindrical coordinates
...

2π

2

∫∫∫ y dV = ∫ ∫ ∫
0

1

r cosθ + 2

0

( r sin θ ) r dz dr dθ

E

=∫

2π

=∫

2π

0

∫

2

∫

2

1

0

1

r sin θ ( r cos θ + 2 ) dr dθ
2

1 3
r sin ( 2θ ) + 2r 2 sin θ dr dθ
2

2π

⌠ ⎛1 4
2 3
⎞
= ⎮ ⎜ r sin ( 2θ ) + r sin θ ⎟ dθ
3
⎠1
⌡0 ⎝ 8
2

2π

15
14
⌠
=⎮
sin ( 2θ ) + sin θ dθ
⌡0 8
3
2π

14
⎛ 15
⎞
= ⎜ − cos ( 2θ ) − cos θ ⎟
3
⎝ 16
⎠0
=0

Just as we did with double integral involving polar coordinates we can start with an iterated
integral in terms of x, y, and z and convert it to cylindrical coordinates
...


Solution
Here are the ranges of the variables from this iterated integral
...
Since the range of y’s is −1 ≤ y ≤ 1 we know that we have the complete
right half of the disk of radius 1 centered at the origin
...


r2 ≤ z ≤ r
On a side note notice that the lower bound here is an elliptic paraboloid and the upper bound is a
cone
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The integral is,

1

∫ ∫

−1 0

1− y 2

x2 + y 2

∫x + y
2

2

xyz dz dx dy = ∫

=∫

π 2
−π 2

π 2
−π 2

∫ ∫ r r ( r cos θ )( r sin θ ) z dz dr dθ
1

r

2

0

1

r

∫ ∫r
0

3
zr
cos θ sin θ dz dr dθ
2



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