Problem 31 - Calculus Early Transcendentals

Question

Evaluate the following integrals using a change of variables. Sketch the original and new regions of integration, \(R\) and \(S\) $$\int_{0}^{1} \int_{y}^{y+2} \sqrt{x-y} d x d y$$

Step-by-Step Solution

Answer

#Answer# $$\frac{2}{3}$$

Step 1: Choose a change of variables

We can use the change of variables technique by defining new variables \(u\) and \(v\) such that: $$u = x - y$$ $$v = y$$ These new variables will help simplify the integrand and change the limits of integration.

Step 2: Determine the transformation equations for x and y

Solve the equations above for x and y: $$x = u + v$$ $$y = v$$

Step 3: Find the Jacobian of the transformation

The next step is to find the Jacobian of the transformation, which is the determinant of the matrix formed by the partial derivatives of the transformation equations: $$J(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 1\\ 0 & 1 \end{vmatrix} = 1$$

Step 4: Change the limits of integration

We need to change the limits of integration to match the new variables. From our transformation equations, the original region of integration \(R\) can be translated into a new region \(S\) with the following limits: $$x = u+v$$ $$u = x-y$$ When \(y = 0\), we have \(u = x\), and when \(y = 1\), we have \(u = x-1\). Also, we have \(x = y+2\) and \(x = y\). So, the new limits of integration are \(0 \leq v \leq 1\) and \(0 \leq u \leq 1\).

Step 5: Rewrite and evaluate the integral

Now we can rewrite our integral in terms of \(u\) and \(v\) with the new limits of integration: $$\int_{0}^{1} \int_{0}^{1} \sqrt{u}\, J(u,v)\, du\, dv = \int_{0}^{1} \int_{0}^{1} \sqrt{u}\, du\, dv$$ Let's first evaluate the inner integral with respect to \(u\): $$\int_{0}^{1} \frac{2}{3}(u^{3/2} - 0^{3/2}) dv$$ Now, integrate with respect to \(v\): $$\int_{0}^{1} \frac{2}{3}(1^{3/2} - 0^{3/2}) dv = \frac{2}{3} \int_{0}^{1} 1\, dv = \frac{2}{3}(v\Big|_0^1) = \frac{2}{3}(1-0)$$ Finally, the result is: $$\int_{0}^{1} \int_{y}^{y+2} \sqrt{x-y} d x d y = \frac{2}{3}$$

Key Concepts

Multiple Integration

Multiple integration refers to the process of integrating a function of two or more variables. For instance, in a double integral, the function depends on two variables, often denoted as \(x\) and \(y\). The goal is to find the accumulated value over a specified region.

Imagine spreading a thin layer of some material over a surface, and you wish to calculate the total amount of the material. Multiple integration allows you to add up the infinitesimally small quantities across the entire surface. In our textbook exercise, the double integral numerically captures the volume under a three-dimensional surface over the region \(R\). With the appropriate change of variables, as shown in the solutions, not only can you simplify the integrand by making it a function of new variables \(u\) and \(v\), you also streamline the evaluation of the integral.

Jacobian Determinant

The Jacobian determinant is a critical concept when performing a change of variables in multiple integrals. It's essentially a scaling factor that provides the necessary adjustment for area or volume when switching from the original coordinates to the new ones. The determinant is found by creating a matrix of partial derivatives and calculating its determinant.

The Jacobian for our exercise was calculated by taking the partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\). Despite its simple value of 1 in this context, the Jacobian can have far-reaching implications on the integral's evaluation, especially when it comes to more complex transformations where it's not unity. It compensates for distortions in the shape and size of infinitesimal regions during the transformation.

Integration Limits Transformation

When you change variables in an integral, you must also transform the integration limits to correspond to the new variables. This part requires understanding how the original limits, which are in terms of the initial variables, map onto the new variables.

In our textbook example, the original limits for \(x\) and \(y\) are shifted to new limits for \(u\) and \(v\), where \(u = x - y\) and \(v = y\). Correctly translating these limits ensures that the new integral is taken over the correct region \(S\) and that the result is equivalent to the original integral over the region \(R\). Mistakes in this step can lead to incorrect answers, so it's essential to approach this systematically, as shown in the step-by-step solution.

Region of Integration Sketching

Sketching the region of integration is a visual aid that can significantly enhance comprehension of multiple integrals. By drawing the domain over which the integration is to be performed, students gain insight into the geometry of the problem.

For our exercise, the region \(R\) could be visualized as a section on the \(xy\)-plane that ranges from \(y\) to \(y+2\) for \(x\), and from 0 to 1 for \(y\). After changing to the new variables \(u\) and \(v\), a new region \(S\) emerges, bounded by the transformed limits. These sketches serve as a conceptual map and can be used to verify the transformations of the integration limits. This practice can prevent errors and deepens the understanding of how the change of variables affects the integral.