Problem 28 - Calculus Early Transcendentals

Question

To evaluate the following integrals, carry out these steps. a. Sketch the original region of integration \(R\) in the xy-plane and the new region \(S\) in the uv-plane using the given change of variables. b. Find the limits of integration for the new integral with respect to \(u\) and \(v\) c. Compute the Jacobian. d. Change variables and evaluate the new integral. \(\iint_{R} x^{2} y d A,\) where \(R=\\{(x, y): 0 \leq x \leq 2, x \leq y \leq x+4\\}\) use \(x=2 u, y=4 v+2 u\).

Step-by-Step Solution

Answer

The value of the given integral after the change of variables and evaluation is \(\frac{64}{3}\).

Step 1: Sketch regions R and S

First, we need to sketch the regions of integration R in the xy-plane and S in the uv-plane. The region R is given by the inequalities: 0 ≤ x ≤ 2, x ≤ y ≤ x + 4 In the uv-plane, the region S, we need to find the corresponding inequalities using the change of variables: x = 2u, y = 4v + 2u We solve these inequalities for u and v, which gives us the new inequalities for the region S.

Step 2: Find the limits of integration for u and v

Now that we have the inequalities, we need to find the limits of integration for u and v. To do this, we find the range of u and v that satisfy the inequalities for the region S. From the transformation equations (x = 2u, y = 4v + 2u), we can see that: 0 ≤ 2u ≤ 2 ⟹ 0 ≤ u ≤ 1, 2u ≤ 4v + 2u ≤ 2u + 4 ⟹ 0 ≤ v ≤ 1 So the limits of integration for u and v are: 0 ≤ u ≤ 1, 0 ≤ v ≤ 1

Step 3: Compute the Jacobian

Next, we compute the Jacobian of the transformation (denoted by J). The Jacobian is given by the determinant of the matrix of partial derivatives: \[J = \left|\begin{matrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix}\right|\] Calculating the partial derivatives, we get: \[\frac{\partial x}{\partial u} = 2, \frac{\partial x}{\partial v} = 0, \frac{\partial y}{\partial u} = 2, \frac{\partial y}{\partial v} = 4\] Thus, the Jacobian is: \[J = \left|\begin{matrix}2 & 0 \\2 & 4\end{matrix}\right| = (2)(4) - (0)(2) = 8\]

Step 4: Change variables and evaluate the new integral

Finally, we change variables and evaluate the new integral. Since we have the Jacobian, the new integral becomes: \[\iint_{S} (8u^2)(4v+2u)(8)dudv\] Now, we can calculate the double integral using the limits of integration found in Step 2: \[\int_{0}^{1} \int_{0}^{1} (8u^2)(4v+2u)(8)dudv\] First, integrate with respect to u: \[\int_{0}^{1}\left[\frac{32}{3}u^3(4v+2u)\right]_{0}^{1}dv\] \[= \int_{0}^{1}\frac{32}{3}(4v+2)dv\] Next, integrate with respect to v: \[\left[\frac{32}{3}(2v^2+2v^3)\right]_{0}^{1}\] \[= \frac{32}{3}(2+2) = \boxed{\frac{64}{3}}\] So, the value of the given integral is \(\frac{64}{3}\).

Key Concepts

Double Integral

A double integral is a form of multiple integration that allows us to compute the volume under a surface over a given region. It's written as \(\textstyle \(\iint\)_{R} f(x, y) dA\), where \(R\) is the region of integration in the xy-plane, and \(f(x, y)\) is a function that gives the height of the surface above point \((x, y)\). To evaluate a double integral, we typically integrate with respect to one variable while keeping the other constant, and then integrate the resulting expression with respect to the second variable.

When faced with more complex regions or functions, sometimes it's necessary to change the order of integration or use a more convenient coordinate system to simplify the calculation process. This is where the concept of changing variables could come into play, which is also displayed in the exercise provided.

Change of Variables in Integrals

Changing variables in double integrals can greatly simplify an otherwise complicated integration problem. This process involves introducing a new pair of variables, often denoted as \(u\) and \(v\), which transforms the original region of integration into a new one that's easier to work with. The new variables and the original ones are connected through transformation equations, like \(x = g(u, v)\) and \(y = h(u, v)\).

For example, in our exercise, the transformation is given by \(x = 2u\) and \(y = 4v + 2u\). Such a change of variables isn't arbitrary; it should simplify the region of integration or the function to be integrated, or ideally, both. After changing variables, it's also essential to update the limits of integration accordingly which reflect the new variables' region.

Jacobian Determinant

The Jacobian determinant, often simply called 'the Jacobian,' is crucial when changing variables in double integrals. It arises from the need to preserve the area (or volume in higher dimensions) under transformation and is represented by the determinant of a matrix of partial derivatives associated with the transformation. Mathematically, for a change of variables from \((x, y)\) to \((u, v)\), the Jacobian \(J\) is defined as:\[J = \left| \begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right|\]

The value of the Jacobian tells us how a small area element changes under the transformation. In the given exercise, after calculating the Jacobian, we multiply the function being integrated by it. This step compensates for the changes in scale introduced by the change of variables.

Integration Limits

When performing a double integral, the integration limits define the region over which we are integrating. In the case of Cartesian coordinates, these would be the x and y bounds that contain the area. If the region is bounded by curves or conditions involving the variables, finding these limits can require solving inequalities or equations.

After changing variables, as we have done in the exercise by introducing \(u\) and \(v\), it's necessary to re-establish these limits for the new variables. This often simplifies the region to be a rectangle or box in the new variable space, with constant numerical limits, though this isn't always the case. Determining the correct new limits is a crucial step in the solution process since incorrect limits can lead to wrong answers or integrals that are impossible to evaluate.