5.7.1Determine the image of a region under a given transformation of variables.

5.7.2Compute the Jacobian of a given transformation.

5.7.3Evaluate a double integral using a change of variables.

5.7.4Evaluate a triple integral using a change of variables.

Recall from Substitution Rule the method of integration by substitution. When evaluating an integral such as we substitute Then or and the limits change to and Thus the integral becomes and this integral is much simpler to evaluate. In other words, when solving integration problems, we make appropriate substitutions to obtain an integral that becomes much simpler than the original integral.

We also used this idea when we transformed double integrals in rectangular coordinates to polar coordinates and transformed triple integrals in rectangular coordinates to cylindrical or spherical coordinates to make the computations simpler. More generally,

Where and and satisfy and

A similar result occurs in double integrals when we substitute and Then we get

where the domain is replaced by the domain in polar coordinates. Generally, the function that we use to change the variables to make the integration simpler is called a transformation or mapping.

Planar Transformations

A planar transformation is a function that transforms a region in one plane into a region in another plane by a change of variables. Both and are subsets of For example, Figure 5.71 shows a region in the transformed into a region in the by the change of variables and or sometimes we write and We shall typically assume that each of these functions has continuous first partial derivatives, which means and exist and are also continuous. The need for this requirement will become clear soon.

Figure 5.71The transformation of a region in the into a region in the

Definition

A transformation defined as is said to be a one-to-one transformation if no two points map to the same image point.

To show that is a one-to-one transformation, we assume and show that as a consequence we obtain If the transformation is one-to-one in the domain then the inverse exists with the domain such that and are identity functions.

Figure 5.71 shows the mapping where and are related to and by the equations and The region is the domain of and the region is the range of also known as the image of under the transformation

Example 5.65

Determining How the Transformation Works

Suppose a transformation is defined as where Find the image of the polar rectangle in the to a region in the Show that is a one-to-one transformation in and find

Solution

Since varies from 0 to 1 in the we have a circular disc of radius 0 to 1 in the Because varies from 0 to in the we end up getting a quarter circle of radius in the first quadrant of the (Figure 5.72). Hence is a quarter circle bounded by in the first quadrant.

Figure 5.72A rectangle in the is mapped into a quarter circle in the

In order to show that is a one-to-one transformation, assume and show as a consequence that In this case, we have

Dividing, we obtain

since the cotangent function is one-one function in the interval Also, since we have Therefore, and is a one-to-one transformation from into

To find solve for in terms of We already know that and Thus is defined as and

Example 5.66

Finding the Image under

Let the transformation be defined by where and Find the image of the triangle in the with vertices and

Solution

The triangle and its image are shown in Figure 5.73. To understand how the sides of the triangle transform, call the side that joins and side the side that joins and side and the side that joins and side

Figure 5.73A triangular region in the is transformed into an image in the

For the side transforms to so this is the side that joins and

For the side transforms to so this is the side that joins and

For the side transforms to (hence so this is the side that makes the upper half of the parabolic arc joining and

All the points in the entire region of the triangle in the are mapped inside the parabolic region in the

Checkpoint5.43

Let a transformation be defined as where Find the image of the rectangle from the after the transformation into a region in the Show that is a one-to-one transformation and find

Jacobians

Recall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that and exist and are also continuous. A transformation that has this property is called a transformation (here denotes continuous). Let where and be a one-to-one transformation. We want to see how it transforms a small rectangular region units by units, in the (see the following figure).

Figure 5.74A small rectangle in the is transformed into a region in the

Since and we have the position vector of the image of the point Suppose that is the coordinate of the point at the lower left corner that mapped to The line maps to the image curve with vector function and the tangent vector at to the image curve is

Similarly, the line maps to the image curve with vector function and the tangent vector at to the image curve is

Now, note that

Similarly,

This allows us to estimate the area of the image by finding the area of the parallelogram formed by the sides and By using the cross product of these two vectors by adding the k component as the area of the image (refer to The Cross Product) is approximately In determinant form, the cross product is

Since we have

Definition

The Jacobian of the transformation is denoted by and is defined by the determinant

Using the definition, we have

Note that the Jacobian is frequently denoted simply by

Note also that

Hence the notation suggests that we can write the Jacobian determinant with partials of in the first row and partials of in the second row.

Example 5.67

Finding the Jacobian

Find the Jacobian of the transformation given in Example 5.65.

Solution

The transformation in the example is where and Thus the Jacobian is

Example 5.68

Finding the Jacobian

Find the Jacobian of the transformation given in Example 5.66.

Solution

The transformation in the example is where and Thus the Jacobian is

Checkpoint5.44

Find the Jacobian of the transformation given in the previous checkpoint:

Change of Variables for Double Integrals

We have already seen that, under the change of variables where and a small region in the is related to the area formed by the product in the by the approximation

Now let’s go back to the definition of double integral for a minute:

Referring to Figure 5.75, observe that we divided the region in the into small subrectangles and we let the subrectangles in the be the images of under the transformation

Figure 5.75The subrectangles in the transform into subrectangles in the

Then the double integral becomes

Notice this is exactly the double Riemann sum for the integral

Theorem5.14

Change of Variables for Double Integrals

Let where and be a one-to-one transformation, with a nonzero Jacobian on the interior of the region in the it maps into the region in the If is continuous on then

With this theorem for double integrals, we can change the variables from to in a double integral simply by replacing

when we use the substitutions and and then change the limits of integration accordingly. This change of variables often makes any computations much simpler.

Example 5.69

Changing Variables from Rectangular to Polar Coordinates

Consider the integral

Use the change of variables and and find the resulting integral.

Solution

First we need to find the region of integration. This region is bounded below by and above by (see the following figure).

Figure 5.76Changing a region from rectangular to polar coordinates.

Squaring and collecting terms, we find that the region is the upper half of the circle that is, In polar coordinates, the circle is so the region of integration in polar coordinates is bounded by and

Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates

Derive the formula in triple integrals for

cylindrical and

spherical coordinates.

Solution

For cylindrical coordinates, the transformation is from the Cartesian to the Cartesian (Figure 5.81). Here and The Jacobian for the transformation is

We know that so Then the triple integral is

Figure 5.81The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region in to region in

For spherical coordinates, the transformation is from the Cartesian to the Cartesian (Figure 5.82). Here and The Jacobian for the transformation is

Expanding the determinant with respect to the third row:

Since we must have Thus

Figure 5.82The transformation from rectangular coordinates to spherical coordinates can be treated as a change of variables from region in to region in

Then the triple integral becomes

Let’s try another example with a different substitution.

Example 5.73

Evaluating a Triple Integral with a Change of Variables

Evaluate the triple integral

in by using the transformation

Then integrate over an appropriate region in

Solution

As before, some kind of sketch of the region in over which we have to perform the integration can help identify the region in (Figure 5.83). Clearly in is bounded by the planes We also know that we have to use for the transformations. We need to solve for Here we find that and

Using elementary algebra, we can find the corresponding surfaces for the region and the limits of integration in It is convenient to list these equations in a table.

Equations in for the region

Corresponding equations in for the region

Limits for the integration in

Figure 5.83The region in is transformed to region in

Now we can calculate the Jacobian for the transformation:

The function to be integrated becomes

We are now ready to put everything together and complete the problem.

Checkpoint5.48

Let be the region in defined by

Evaluate by using the transformation and

Section 5.7 Exercises

In the following exercises, the function on the region bounded by the unit square is given, where is the image of under

Justify that the function is a transformation.

Find the images of the vertices of the unit square through the function

Determine the image of the unit square and graph it.

[T] The transformations defined by and are called reflections about the origin, and the line respectively.

Find the image of the region in the through the transformation

Use a CAS to graph

Evaluate the integral by using a CAS. Round your answer to two decimal places.

410.

[T] The transformation of the form where is a positive real number, is called a stretch if and a compression if in the Use a CAS to evaluate the integral on the solid by considering the compression defined by and Round your answer to four decimal places.

[T] Lamé ovals have been consistently used by designers and architects. For instance, Gerald Robinson, a Canadian architect, has designed a parking garage in a shopping center in Peterborough, Ontario, in the shape of a superellipse of the equation with and Use a CAS to find an approximation of the area of the parking garage in the case yards, yards, and yards.