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.

Change of variables for double integrals

 We have seen that under the change of variables T(u,v) = (x,y) where x = g(u,v) and y = h(u,v), a small region ΔA in the xy plane is related to the area formed by the product 𝜟u 𝜟v in the uv plane by the approximation: 𝜟A ≃J(u,v)𝜟u𝜟

Remember that the double integral is defined as:

In the following figure we divide the region S in small rectangles Sᵢⱼ and the region R in small rectangles Rᵢⱼ. A small rectangle Rᵢⱼ is the image of a small rectangle Sᵢⱼ under the transformation T.

 

Let's substitute f(xᵢⱼ, yᵢⱼ) and ΔA in the definition of the integral:

∫∫_R f(x,y) dA = ∫∫_S f(g(u,v),h(u,v)) |J(u,v)| du dv

Let's substitute J(u,v) in in the expression on the right side of the equality sign, we have:

Theorem

Example

Solution

Let's do the change of variables in the expression √x² + y² . By substituting x = rcosθ and y = sinθ, the expression becomes equal to r.

The expression dydx becomes J(r,θ) drdθ.  In a previous example J(r,θ) was equal to r. The integral becomes when we substitute everything:

Practice

What are Jacobians in multiple integrals?

 Let's consider a one-to-one C¹ transformation defined by: 

We want to see how it transforms a small rectangle with sides Δu and Δv in the (u,v) plane.

Let's call ΔA the area of the curved region R of sides Δurᵤ and Δvrᵥ. This area is approximately equal to  the norm of the product of the two sides. It can be written as:

ΔA≃ ॥Δurᵤ॥.॥Δvrᵥ॥ = ॥Δu॥॥rᵤ॥. ॥Δv॥rᵥ॥ = Δu॥rᵤ॥.Δv॥rᵥ॥ = ॥rᵤ॥॥rᵥ॥ ΔuΔv 

Let's find rᵤ and rᵥ in order to calculate the norm of their product.

Let's calculate the product of the vectors rᵤ and rᵥ. This is the tricky part. The surface is sitting in a tridimensional space, which is R³.  We can rewrite rᵤ and rᵥ as: 

rᵤ = ∂x/∂ui + ∂x/∂uj + 0k

rᵥ = ∂x/∂vi + ∂x/∂vj + 0k

So, the coordinates of the tangent vectors rᵤ and rᵥ at (u₀, v₀) are:

𝐫ᵤ = ( ∂x/∂u, ∂y/∂u, 0 )

𝐫ᵥ(u₀,v₀) = ( ∂x/∂v, ∂y/∂v, 0 )

Let's calculate the product of the vectors now:

Let's take the norm of the product:

॥rᵤ 🗙 rᵥ॥ =∥ ∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u) ) 𝐤 ) ∥

= ॥ (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)॥ ∥𝐤∥

=  (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)  since ∥𝐤∥ = 1

Let's substitute in the expression of ΔA. We have:

 . 

The jacobian can finally be written as:

J(u,v) = 

Example. Find the jacobian of the transformation given in the following example:

 

 

Solution

       

Finding the image of a triangle by a transformation

 The previous example consists in showing that a transformation is a one-to-one transformation. Now we are concerned about how to find the image of a triangle under a transformation. Let's use an example in this case.

Example

Solution

The triangle and its image are shown in the figure below. To understand how the sides of the triangle transform, call the side that joins (0,0) and (0,1) side A, the side that joins (0,0) and (1,1) side B and the side that joins (0,1) and (1,1) side C. 

Practice

   

Change of variables in multiple integrals. Planar Transformations

 Objective: Determine the image of a region under a given transformation of variables

Planar Transformation

A planar transformation is a function that transforms a region G in one plane into another region R into another plane by a change of variables. Both G and R are subsets of R². The figure bellow shows a region G in the (u,v) plane transformed into another region R in the (x,y) plane by the change of variables x = g(u,v) and y = h(u,v).

Definition

A transformation T: G → R defined as T(u,v) = (x,y) is said to be one- to- one transformation if no two points map the same image point.

Example

Solution

first quadrant of the xy plane. Hence R is a quarter circle bounded by x² + y² = 1 in the first quadrant.

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.