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.

Center of mass and moments

 The expressions of mass, center of mass, moments of inertia expressed in double integrals can be modified by replacing the double integrals with triple integrals.

Example

Solution

The region Q is a tetrahedron meeting the axes at the point (6,0,0), (0,3,0) and (0,0,2) (see figure below). To find the limits of integration, let z = 0 in the slanted plane z = 1/3(6-x-2y). Then for x and y find the projection of Q on the the xy plane which is bounded by the axes and the line x_+ 2y = 6. The mass is calculated as follow: