Problem 32 - Calculus Early Transcendentals

Question

Evaluate the following integrals using a change of variables. Sketch the original and new regions of integration, \(R\) and \(S\) $$\begin{aligned} &\iint_{R} \sqrt{y^{2}-x^{2}} d A, \text { where } R \text { is the diamond bounded by }\\\ &y-x=0, y-x=2, y+x=0, \text { and } y+x=2 \end{aligned}$$

Step-by-Step Solution

Answer

The given double integral can be evaluated using the change of variables method. After identifying a suitable transformation of \(u=y-x\) and \(v=y+x\), and evaluating the Jacobian, the transformed integral is found to be \(\frac{1}{2}\iint_S u du dv\). Evaluating this integral over the new region \(S = \{ (u,v) \in [0,2] \times [0,2] \}\), we get the final result, which is 2.

Step 1: Identify the transformation

We need to find a transformation that simplifies the region for the double integral. We start off by observing that the region R is a parallelogram enclosed by the lines \(y-x=0, y-x=2, y+x=0,\) and \(y+x=2\). To make our life easier, we will try the following transformation: $$u=y-x, v=y+x.$$

Step 2: Determine the region boundaries in terms of new variables

Now, the region defined by the lines becomes simple: $$0 \leq u \leq 2, \hspace{1cm} 0 \leq v \leq 2.$$ So, now the region is bounded by \(S = \{ (u,v) \in [0,2] \times [0,2] \}\).

Step 3: Evaluate the Jacobian

Now, we need to compute the Jacobian of our transformation. It is given by: $$J(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{vmatrix},$$ where $$x(u,v) = \frac{v-u}{2}, \hspace{1cm} y(u,v) = \frac{u+v}{2}.$$ Now compute the partial derivatives and insert them into the matrix: $$J(u,v) = \begin{vmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{vmatrix} = \frac{1}{4} - \frac{(-1)}{4} = \frac{1}{2}.$$

Step 4: Write the integral in terms of transformed variables

Now that we have the Jacobian, we can write the double integral in terms of the new variables u and v: $$\begin{aligned} \iint_R \sqrt{y^2 - x^2} dA &= \iint_S \sqrt{\left(\frac{u+v}{2}\right)^2 - \left(\frac{v-u}{2}\right)^2} \cdot \frac{1}{2} du dv\\ &= \frac{1}{4}\iint_S (u+v-v+u) du dv \\ &= \frac{1}{4}\iint_S 2u du dv \\ &= \frac{1}{2}\iint_S u du dv. \end{aligned}$$

Step 5: Evaluate the transformed integral

Finally, evaluate the transformed double integral: $$\begin{aligned} \frac{1}{2}\iint_S u du dv &= \frac{1}{2} \int_0^2 \left( \int_0^2 u du\right) dv \\ &= \frac{1}{2} \int_0^2 \left[\frac{u^2}{2}\right]_0^2 dv \\ &= \frac{1}{2} \int_0^2 2 dv \\ &= \int_0^2 dv \\ &= [v]_0^2 \\ &= 2. \end{aligned}$$ So the integral evaluates to 2.

Key Concepts

Double Integral

When dealing with the concept of a double integral, we are essentially extending the idea of integration to two dimensions. A double integral, denoted by \(\iint_R f(x,y) dA\), allows us to calculate the 'volume' under a surface defined by the function \(f(x,y)\) over a certain region \(R\). This region is typically within the xy-plane and can have different shapes such as rectangles, circles, or more complex polygons. The value of the double integral is equivalent to summing up infinitely small pieces of area \(dA\), each multiplied by the function's value at that area, to find the aggregate of these products over the entire region. In the context of our textbook exercise, we work with a double integral over a diamond-shaped region \(R\), specifically attempting to simplify the process of evaluation through a change of variables.

Region of Integration

The region of integration, \(R\), plays a crucial role when computing a double integral. It is the domain over which we are integrating our function. A clear understanding of this region is vital for setting up the limits of integration. In simple cases, we can define \(R\) by a pair of inequalities for each variable. However, with complex shapes like the diamond shape in our exercise, identifying \(R\) and integrating over it directly can be cumbersome. That's why we often convert the region into a simpler one, \(S\), using a suitable coordinate transformation, making the integration process more manageable. In the described exercise, the region \(R\) bounded by lines \(y-x=0\), \(y-x=2\), \(y+x=0\), and \(y+x=2\), forms a diamond shape that's not straightforward to integrate over directly. Hence, we use a transformation to redefine this region in a way that the limits of integration become constants, greatly simplifying the double integral computation.

Jacobian

The Jacobian is akin to a conversion factor between coordinate systems when performing integrations. Mathematically, the Jacobian determinant of a coordinate transformation provides information on how a function is stretched or compressed as a result of the transformation. When changing variables in a double integral, the Jacobian adjustment is necessary to accurately reflect the new limits of integration. Derived from the partial derivatives of the transformed variables, the Jacobian accounts for the area distortion caused by the transformation. In the exercise, we convert coordinates from \(x, y\) to \(u, v\). The computed Jacobian, \(J(u,v)=\frac{1}{2}\), is used to adjust the integral. It ensures that the final integral indeed represents the same physical or geometric quantity, such as area or volume, as the original integral. The application of the Jacobian is indispensable to the change of variables technique in double integrals.

Coordinate Transformation

Coordinate transformation is a powerful method used to simplify the evaluation of integrals, especially when faced with complex regions of integration like in our example. It involves changing the variables of integration to new variables that can describe the region more conveniently. This change leads to an equivalent integral with potentially simpler limits of integration. In our textbook solution, the transformation is defined by \(u=y-x\) and \(v=y+x\), effectively converting the parallelogram \(R\) to a square \(S\) with side length 2. This makes the process of finding the new limits of integration much simpler, as they are now constants. Students should pay close attention to how this transformation is set up and how it simplifies the original problem, as this technique is widely applicable in various branches of mathematics and physics where integrals over complex regions are common.