Find the volume inside of an ellipsoid and outside of a sphere

 We are going to use an example to do that:

Example:

Solution

First let's find the volume using a = 75ft b = 80 ft c = 90ft using the result from the example treated previously about the volume of the ellipsoid. Hence the volume of the ellipsoid is: 

From the result from the example about the volume of a sphere, we get:

Volume of an ellipsoid using spherical coordinates

 Let's use an example to calculate the volume of an ellipsoid using spherical coordinates.

Example

Solution

Let's use the change of variables that corresponds to an ellipsoid. We still change the variables from rectangular coordinates to spherical coordinates. 

The volume V of the ellipsoid is given by:

Let's change dV = dxdydz in spherical coordinates:

Let's apply the change-of-variables formula

$$dx\,dy\,dz=\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right|\;d\rho\,d\theta\,d\varphi.$$

$$quotient\; of\; the\; partial\; derivatives.$$

So, the task is to compute the Jacobian:

$$J(\rho,\theta,\varphi)=\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}.$$

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Compute the partial derivatives

Differentiate $x,y,\mathrm{z\; with\; respect\; to }$$\rho,\theta,\varphi$.

With respect to $\rho$:

$$\frac{\partial(x,y,z)}{\partial\rho} =\Big(a\cos\varphi\sin\theta,\; b\sin\varphi\sin\theta,\; c\cos\theta\Big).$$

With respect to $\theta$:

$$\frac{\partial(x,y,z)}{\partial\theta} =\Big(a\rho\cos\varphi\cos\theta,\; b\rho\sin\varphi\cos\theta,\; -c\rho\sin\theta\Big).$$

With respect to $\varphi$:

$$\frac{\partial(x,y,z)}{\partial\varphi} =\Big(-a\rho\sin\varphi\sin\theta,\; b\rho\cos\varphi\sin\theta,\; 0\Big).$$

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 Form the Jacobian determinant

Put those three vectors as columns (or rows—just be consistent). Using columns:

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = \left| \begin{matrix} a\cos\varphi\sin\theta & a\rho\cos\varphi\cos\theta & -a\rho\sin\varphi\sin\theta\\ b\sin\varphi\sin\theta & b\rho\sin\varphi\cos\theta & \ \ b\rho\cos\varphi\sin\theta\\ c\cos\theta & -c\rho\sin\theta & 0 \end{matrix} \right|.$$

 Factor constants (the key simplification)

• Factor $a$ from row 1, $b$ from row 2, $c$ from row 3 $<br>$

• Factor $\rho$ from column 2 and $\rho$ from column 3 $<br>$

So

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = abc\,\rho^2 \left| \begin{matrix} \cos\varphi\sin\theta & \cos\varphi\cos\theta & -\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \sin\varphi\cos\theta & \ \cos\varphi\sin\theta\\ \cos\theta & -\sin\theta & 0 \end{matrix} \right|.$$ 

Evaluating the determinant we obtain:

$$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| =abc\,\rho^2\sin\theta.$$ 

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Final conversion of dV

$$\boxed{dx\,dy\,dz=abc\,\rho^2\sin\theta\;d\rho\,d\theta\,d\varphi.}$$

The volume of the ellipsoid is calculated as follows:

​

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

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: