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:

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