Solving a triple integral using the change of variables

 In the previous posts we spent time establishing formulas to solve a triple integral using a change of variables. Let's now solve an example.

Example

Solution

Let's apply the formula:

Let's take the following steps to apply the formula:

1) Let's find x, y, z from the given values of u, v, w. This leads to solve the following system of equations:

u = 2x-y/2 (1)

v = y/2 (2)

w = z/3 (3)

Solving this system, we find x = (u + v)/2 y = 2v z = 3w

2) Let's find the new function H(u,v,w) to be integrated by substituting x, y, z in the function F(x,y,z) = x +z/3. 

3) Let's find the limits of integration of the new function i.e u.v.w:

Let's isolate x, y, z as limits of integration of the given integral. We have:

x = y/2 x = y/2 + 1

y = 0 y = 4

z = 0 z = 3

These equations represent the planes that bound the surface G in the space xyz.

From the equation x=y/2 we have 2x = y 2x-y = 0 ⇒2x-y/2 = 0 ⇒ u = 0 since u = 2x-y/2 (1)

From the equation x = y/2 + 1 we have 2x = y+2 ⇒ 2x-y = 2 ⇒ 2x-y/2 = 1  ⇒ u = 1.

Let's use the equation v = y/2:

For y = 0 we have v = 0. For y = 4 v =2 

Let's use the equation w = z/3

For z = 0 w = 0

For z =3 w = 1

Let's calculate the jacobian:

Change of variables for triple integrals

 Objective: Set the formula for the change of variables in triple integrals.

Planar transformation in tridimensional space

a.The change of variables in triple integrals works the same way as in double integrals. We consider a transformation from a tridimensional space (u,v.w) to a tridimensional space (x,y,z). We define the Jacobian in the (u,v,w) space. Then we establish the formula for the triple integral defined in (u,v,w).

 Suppose that G i s a region in the uvww space transformed in another region D in the xyz space by a C¹ transformation by the transformation T so that T(u,v,w) = (x,y,z) where x = g(u,v,w) y = h(u,v.w) z = k(u,v,w).

Any function F(x.y,z) defined on D can be thought of another function H(u,v. w) defined G such that:

F(x.y.z) = F(g(u,v,w), h(u,v,w), k(u,v,w)) = H(u,v,w), Now let's define the Jacobian in a tridimensional space (u,v,w

Jacobian for three variables. Definition

The jacobian J(u,v,w) for three variables is defined as follows:

This is the same as:

Theorem

Change of Variables for Triple Integrals

Let T(u, v, w) = (x, y, z) where x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w), be a one-to-one C¹ transformation, with a nonzero Jacobian, that maps the region G in the uvw-space into the region R in the xyz-space. As in the two-dimensional case, if F is continuous on R, then

∫∫∫_R F(x,y,z) dV = ∫∫∫_G F(g(u,v,w), h(u,v,w), k(u,v,w))   | ∂(x,y,z) ∂(u,v,w) |  du dv dw
= ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw.

Example 

  Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates

Derive the formula in triple integrals for

1. cylindrical and
2. spherical coordinates.

Solution

Let's apply the formula for change of variables in triple integrals:

∫∫∫_R F(x,y,z) dV = ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw

In cylindrical coordinates it becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cosθ, r sinθ, z)  |J(r,θ,z)|  dr dθ dz

J(r, θ, z) =  |

∂x/∂r	∂x/∂θ	∂x/∂z
∂y/∂r	∂y/∂θ	∂y/∂z
∂z/∂r	∂z/∂θ	∂z/∂z

| = |

cos θ	−r sin θ	0
sin θ	r cos θ	0
0	0	1

| = r cos²θ + r sin²θ = r (cos²θ + sin²θ) = r.

We know that r ≥ 0, so |J(r,θ,z)| = r. Then the triple integral becomes by substituting the value of the jacobian: 
∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cos θ, r sin θ, z) r dr dθ dz.

b. Let's find the formula for the triple integral in spherical coordinates by applying the same process.

For spherical coordinates, the transformation is T(ρ, θ, φ) = (x, y, z) from the Cartesian pθφ-plane to the Cartesian xyz-plane . Here x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. The expression of the triple integral can be written as:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)  |J(ρ,θ,φ)|  dρ dφ dθ. Let's calculate the jacobian.

First write it as a quotient of partial derivatives:

J(ρ,θ,φ) = ∂(x,y,z) ∂(ρ,θ,φ)

Now write it as a determinant:

J(ρ,θ,φ) = |

∂x/∂ρ	∂x/∂θ	∂x/∂φ
∂y/∂ρ	∂y/∂θ	∂y/∂φ
∂z/∂ρ	∂z/∂θ	∂z/∂φ

|

Substitute the partial derivatives:

J(ρ,θ,φ) = |

sinφ cosθ	-ρ sinφ sinθ	ρ cosφ cosθ
sinφ sinθ	ρ sinφ cosθ	ρ cosφ sinθ
cosφ	0	-ρ sinφ

|

Expand along the third row:

J = cosφ |

-ρ sinφ sinθ	ρ cosφ cosθ
ρ sinφ cosθ	ρ cosφ sinθ

| - ρ sinφ |

sinφ cosθ	-ρ sinφ sinθ
sinφ sinθ	ρ sinφ cosθ

|

Calculate the first determinant:

(-ρ sinφ sinθ)(ρ cosφ sinθ) - (ρ cosφ cosθ)(ρ sinφ cosθ)

= -ρ² sinφ cosφ sin²θ - ρ² sinφ cosφ cos²θ

= -ρ² sinφ cosφ

Calculate the second determinant:

(sinφ cosθ)(ρ sinφ cosθ) - (-ρ sinφ sinθ)(sinφ sinθ)

= ρ sin²φ cos²θ + ρ sin²φ sin²θ

= ρ sin²φ

Therefore,

J = cosφ(-ρ² sinφ cosφ) - ρ sinφ(ρ sin²φ)

J = -ρ² sinφ cos²φ - ρ² sin³φ

J = -ρ² sinφ(cos²φ + sin²φ)

J(ρ,θ,φ) = -ρ² sinφ

So,

|J(ρ,θ,φ)| = ρ² sinφ

Let's substitute the jacobian in the expression of the integral. Then the triple integral becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)  ρ² sinφ dρ dφ dθ.

Finding the moments of inertia of a solid in three dimensions

 The formulas to calculate the moments of inertia being known, let's solve a problem to apply them.

Example

Suppose the region Q is bounded by the plane x+2y+3z = 0 and the coordinates planes with density ⍴ = x²yz (see figure in this example). Find the moments of inertia about the yz plane, the xz plane, the xy plane.

Solution

Let's use the formulas already established

Practice

 

Consider the same region Q with density function ⍴(x,y,z) = xy²z. Find the moments of inertia about the three coordinate planes.

Finding the center of mass of a solid in 3 dimensions

 Goal: Find the center of mass of a solid in 3 dimensions

We already stated the formulas to calculate the center of mass of a solid in 3 dimensions. Let's solve an example.

Example

Suppose Q is a solid region bounded by the plane x + 2y + 3z = 0, the coordinates planes with density ϼ(x,y,z) = x²yz (see figure in the example in the previous post). Find the center of mass using decimal approximation. Use the mass found in the previous example.

Solution

Practice

Consider the same region Q and the density function ρ(x,y,z) = xy²z. Find the center of mass using the following figure used in this example.

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:

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

---

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 $\varphi$:

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

---

 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:

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

---

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:

​

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).

---

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,