Converting a triple integral in rectangular coordinates to spherical coordinates

 Converting a triple integral from rectangular coordinates to spherical coordinates. Let's do that through an example.

Example

Solution

Let's start by finding the ranges for θ, ⍴, 𝛗.

1) Look only at the outer two integrals (the x,y bounds):

0 ≤ y ≤ 3, 0 ≤ x ≤ √(9 − y²)

Rewrite the x-bound as an inequality:

0 ≤ x ≤ √(9 − y²)
⇔ x² ≤ 9 − y²
⇔ x² + y² ≤ 9

Combine this with x ≥ 0 and y ≥ 0.
This describes the first-quadrant portion of the disk x² + y² ≤ 9 in the xy-plane.

In polar (or spherical) coordinates, θ is the angle in the xy-plane measured from the positive x-axis.
The first quadrant therefore gives:

0 ≤ θ ≤ π⁄2

2) Let's find the ranges for ⍴:

The top z-surface is given by:

z = √(18 − x² − y²)

Square the equation (this is valid here since z ≥ 0):

z² = 18 − x² − y²
⇔ x² + y² + z² = 18

In spherical coordinates, the relation between rectangular and spherical variables is:

x² + y² + z² = ρ²

Therefore, this surface becomes:

ρ² = 18
ρ = 3√2

The range for ⍴ is then: 0 ≤ ⍴ ≤ 3⎷2

3) The bottom z-surface is given by:

z = √(x² + y²)

Let r = √(x² + y²).
Then the surface can be written as:

z = r

This represents a cone opening upward with vertex at the origin.

In spherical coordinates, the relationships are:

z = ρ cos φ
r = ρ sin φ

Substitute these into z = r:

ρ cos φ = ρ sin φ

For ρ > 0, divide both sides by ρ:

cos φ = sin φ

This implies:

tan φ = 1
φ = π/4

The original bounds satisfy z ≥ √(x² + y²), which means the region lies above the cone.
In spherical coordinates, this corresponds to angles smaller than π/4.

Therefore, the φ-range is:

0 ≤ φ ≤ π/4

From the coordinate transformation, we have:

x² + y² + z² = ρ²
dV = ρ² sin φ dρ dφ dθ

The integrand becomes:

x² + y² + z² = ρ²

Finally, the triple integral becomes:

 ​

Converting triple integrals from rectangular coordinates to cylindrical coordinates

 Converting a triple integral from rectangular coordinates to cylindrical coordinates require to change the function f(x,y,z) in cylindrical form i.e f(r,θ,z). Let's model this through an example.

Example

Solution

We have to transform the given triple integral in cylindrical coordinates form. Let's use the Fubini's theorem:

In polar form, we have x = rcosθ y = rsinθ.

Let's find for the limits of integration for r, θ and z. The limits of integration of x and y from the given integral are:

Let's solve the system of inequalities:

−1 ≤ y ≤ 1
0 ≤ x ≤ √(1 − y²)

Since x ≥ 0 and √(1 − y²) ≥ 0, we can square the inequality x ≤ √(1 − y²):

x ≤ √(1 − y²)  ⟺  x² ≤ 1 − y²  ⟺  x² + y² ≤ 1.

Therefore the solution set is

{ (x, y) ∈ ℝ² : x ≥ 0 and x² + y² ≤ 1 }.

Geometrically, this is the right half of the closed unit disk (including the boundary).

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In polar form we have:

0 ≤ r ≤ 1,   −π/2 ≤ θ ≤ π/2.

Interchanging order of integration in spherical coordinates

 As we saw before the order of integration can be changed. We use an example to show that.

Example

Let E be the region bounded below by the cone z = ⎷x² + y² and above by the sphere z = x² + y² + z².Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration.

a. dρdഴdθ

b. dഴd𝜌dθ

Solution

Evaluating a triple integral in spherical coordinates

 Integration in spherical coordinates

Let f (⍴, θ, ψ) be a function continuous over a bounded spherical box defined by:

Let's divide each interval in l, m. n subintervals such that 

Let's consider any sample point (ρᵢⱼₖ, θᵢⱼₖ, ψᵢⱼₖ) in the subbox Bᵢⱼₖ. The volume element ΔV of the subbox B can be written in spherical coordinates by:

as shown in the following figure:

Let's take the Rieman sum of the expression:

The limit of this expression when l, m, n approach infinity is the triple integral of the function in spherical coordinates as defined above

Definition of a triple integral in spherical coordinates

The properties already examined for previous integrals work for triple integrals in cylindrical coordinates as well as iterated integrals. As always, Fubini's theorem allows us to evaluate a triple a integral by setting it up as an iterated integral. The theorem is stated below:

Theorem

Example

Solution

The variables being independent of each other, we can integrate each piece and multiply:

Integration in spherical coordinates

After spending some time setting up and evaluating triple integrals in cylindrical coordinates, we are going to work with triple integrals in spherical coordinates. Before doing this, let's get an overview of spherical coordinates 

Overview of spherical coordinates 

In a three-dimensional system of coordinates, a point P (x, y, z) is defined by:

ρ : the distance from the origin to the point P

θ: the angle from the positive direction of the x-axis as in the cylindrical coordinates system

ѱ: the angle from the positive z axis and the line OP.

Relationships between rectangular coordinates and spherical coordinates

Other important relationships for conversion

The following figures show a few regions that are useful to express in spherical coordinates

Finding the volume with triple integrals in three ways

 In a previous example I showed how to set up a triple integral in three ways. Calculating a volume in three ways comes to using the same procedure. The following example shows how to calculate the volume with triple integrals in three ways.

Example

Let E be the region bounded below by the rθ plane, above by the sphere x² + y² + z² = 4 and on the sides by the cylinder x² + y² = .1. Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and in each case find the volume and check that the answers are the same.

a. dzdrdθ

b.drdzdθ

Solution

In the expression of E₁ below, r is a function of z. Therefore, we have 0 ≤ r ≤ ⎷4-z² and not 0 ≤ r ≤ ⎷4-r²

Practice

Redo the previous example with the following order dθdzdr,

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:

.