Setting up a triple integral in cylindrical coordinates in two ways.

 It is possible to set up a triple integral in cylindrical coordinates in two ways. We are going to do that by solving an example.

Example 

Let E be the region bounded below by the the cone z = ⎷x² + y² and the paraboloid z = 2 - x³ - y². Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration.

a.dzdrdθ

b. drdzdθ

Solution

a. Integration following the order dzdrdθ

Let's find the intersection of the 2 surfaces. by equalizing their equation

 2 - x2 - y2 = √(x2 + y2)

We want to solve the equation:

2 − x² − y² = √(x² + y²)

2 −( x² +y²⁾ = √(x² + y²)

Since x² + y² = r²

r = √ (x² + y²)

Let's substitute x² + y² and √ (x² + y²) in the equation:

2 − r2 = r 

2 − r² − r = 0

r² + r − 2 = 0

:(r + 2) (r − 1) = 0

 r = −2 (not valid since r ≥ 0)

r = 1 (valid)

Substitute r in the expression: x² + y² = r²

 We finally have: x² + y² = 1

The equation represents a circle centered at the origin and of radius 1 The projection of the region E onto the xy plane is a circle centered at the origin and of radius 1.

Let's find the limits. For fixed θ and r, we have:

Limits for θ: 0 ≤ θ ≤ 2ℼ

Limits for r: 0 ≤ r ≤1

Limits for z:

The cone is the lower limit for z and z is the upper limit. From previous calculations we can write:

r ≤ z ≤2-r

The region E is then defined by:

Hence the integral becomes:

b) Integration in the order: drdzdθ. We adopt the following procedure 

• Fix a height z and angle θ.
• Integrate r first, over the region between the cone and paraboloid.
• Then integrate z and θ over their respective ranges.

 Let's determine the bounds:

• θ: The region is symmetric around the z-axis, so 0 ≤ θ ≤ 2π.
• z: At the tip of the cone, z = 0. At the top of the paraboloid, z = 2. So we have 0 ≤ z ≤ 2.
• r: For a fixed z and θ, r goes from the cone to the paraboloid:
  r = z (from z = r) to r = √(2 - z) (from z = 2 - r²).

 The region E is given by:

E = {(r, θ, z)/ z ≤ r≤⎷2 - z, 0≤z≤2, 0≤θ≤2ℼ}

The volume is given by:

.