Question

Using the transformation \(T: x=u+v, y=u-v,\) the image of the unit square \(S=\\{(u, v): 0 \leq u \leq 1,0 \leq v \leq 1\\}\) is a region \(R\) in the \(x y\) -plane. Explain how to change variables in the integral \(\iint_{R} f(x, y) d A\) to find a new integral over \(S\).

Short answer

To change variables in the integral and evaluate it over region \(S\), we need to substitute the transformation equations \(x = u+v \text{ and } y = u-v\) into the given integral and multiply by the absolute value of the Jacobian, which we found to be |-2|. After doing this, the new integral over \(S\) is given by: $$ \iint_{S} 2f(u+v, u-v) \,du \, dv $$ This integral should be evaluated over the range \(0 \leq u \leq 1\) and \(0 \leq v \leq 1\).

Step-by-step solution

Find the Transformation

First, we are given the transformation \(T: x=u+v, y=u-v,\) which will map points from the uv-plane to the xy-plane.

Determine Region R

Given the region \(S\) in the uv-plane as: \(S=\\{(u, v): 0 \leq u \leq 1,0 \leq v \leq 1\\}\), we can find the corresponding region R in the xy-plane by plugging the limits of \(S\) into the transformation equations: 1. When \(u=0\) and \(0 \leq v \leq 1\), we have \(x = v\) and \(y=-v\). So, the points \((0,0)\), and \((1,-1)\) are included in R. 2. When \(u=1\) and \(0 \leq v \leq 1\), we have \(x = 1+v\) and \(y=1-v\). So, the points \((1,1)\), and \((2,0)\) are included in R. Thus, the region R in the xy-plane is the parallelogram formed by the vertices \((0,0)\), \((1,1)\), \((1,-1)\), and \((2,0)\).

Calculate the Jacobian

To change variables in the integral, we will need to calculate the Jacobian of the transformation \(T\), given by: $$ J(u,v) = \det \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} $$ Calculating the partial derivatives: $$ \frac{\partial x}{\partial u} = 1, \; \frac{\partial x}{\partial v} = 1 \\ \frac{\partial y}{\partial u} = 1, \; \frac{\partial y}{\partial v} = -1 $$ Now, we can find the Jacobian: $$ J(u,v) = \det \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = (1)(-1) - (1)(1) = -2 $$

Change Variables in the Integral

To change variables in the integral, we will substitute the transformation equations into the given integral and multiply by the absolute value of the Jacobian: $$ \iint_{R} f(x, y) dA = \iint_{S} f(u+v, u-v) |J(u,v)| \, du \, dv = \iint_{S} f(u+v, u-v) |\!|-2|\!| \, du \, dv $$

Final Result

Therefore, the new integral over \(S\) is: $$ \iint_{S} 2f(u+v, u-v) du \, dv $$ Integrating over the range \(0 \leq u \leq 1\) and \(0 \leq v \leq 1\).

Key concepts

Transformation of Variables

Changing variables in calculus is a powerful method, especially useful when solving complex integrals. The transformation of variables involves substituting one set of variables with another, often to simplify calculations.

For example, we transform from \(x, y\) to \(u, v\). Given the transformation \(x = u + v\) and \(y = u - v\), the points in the \(u, v\)-plane are mapped onto the \(x, y\)-plane. This helps manage integrals over complex regions by simplifying the area or function description.

Through transformation, the area under consideration may become more regular, like changing a parallelogram into a square. This technique is beneficial as it allows us to apply integration over simpler boundaries.

Jacobian Determinant

The Jacobian Determinant is crucial for transformations as it measures how the transformation manipulates area or volume. It appears in the context of multiple integrals when changing variables.

For a transformation \((x, y) = (u+v, u-v)\), the Jacobian determinant is the determinant of the matrix of partial derivatives:

$$\begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}.$$

In our case, it becomes:

\( J(u, v) = \begin{vmatrix} 1 & 1 \ 1 & -1 \end{vmatrix} = -2 \).

The sign of the Jacobian can indicate the nature of the transformation's orientation, while its magnitude indicates how much the area is stretched or compressed. In integrals, we use the absolute value because we're interested in the area change magnitude.

Multiple Integrals

Multiple integrals extend the concept of integration to functions of multiple variables.

They are used to compute volumes under surfaces in two or more dimensions. The transformation of variables in multiple integrals simplifies integration over complex regions by changing the domain of integration to a more manageable shape.

For instance, converting an integral over a parallelogram in \(xy\)-coordinates to a simpler square in \(uv\)-coordinates makes calculations straightforward. The integral transformation process involves substituting variables and multiplying by the Jacobian's absolute value to maintain equivalence between the \(xy\) and \(uv\) planes.

Coordinate Transformation

Coordinate transformation involves changing the coordinates used to describe equations or functions. This change can simplify both the region of integration and the function expression.

Consider the transformation \(x = u+v, y = u-v\). This shifts any point \( (u, v) \) in a unit square to points in a parallelogram in the \(xy\)-plane.

Through transformation, problems that are difficult to solve in one coordinate system become easier in another. Here, by transforming the integral over region \(R\), which is a parallelogram, to the unit square \(S\), computation becomes simpler, with both the boundaries and integrand being easier to handle.

• Transforms complex regions into regular ones.
• Facilitates easier calculations.
• Helps in visualizing and solving integrations over unconventional shapes.