Volume of a sphere using spherical coordinates

Let's use an example to calculate the volume of a sphere in spherical coordinates. 

Example

Solution

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