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:

 

Finding the moments of inertia of a solid in two dimensions.

 Goal: Find the moment of inertia of a solid in two dimensions

Let's go back to the lamina considered in earlier post where we calculated its mass. In order to do that we considered the region R occupied by the lamina. We divided the lamina into tiny subrectangles. Our goal now is to find the different moments of inertia of the solid.

.

Let's find the moment of inertia about the x-axis.

 The moment of inertia of a solid about an axis is equal to its mass by the square of its distance from this axis.

The moment of inertia of the lamina about the x-axis is equal to the sum of the moments of the tiny subrectangles of the lamina.

The moment of inertia of a subrectangle about the x-axis is equal to the product of its mass by the square of its distance from the x-axis. This moment is calculated as: 

Let's add all the moments of the subrectangles. This comes to take the Rieman sum of the product and to determine its limit.                                                                                                                                         

           

The moment of inertia of a subrectangle about the y-axis is equal to the product of its mass by the square of its distance from the t-axis.  Hence, the moment is given by:                                                                      

                                                                                                                     

Adding all the moments together allows to find the moment of the lamina about the y-axis. This leads to use the Rieman sum of the product and take its limit.

Example

Solution

Practice