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