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 \(x=2 u, y=4 v+2 u\)

Short answer

Question: Evaluate the integral \(\iint_{R} x^{2} y \; dA\), where R is the region in the xy-plane bounded by \(0 \leq x \leq 2\), \(x \leq y \leq x+4\), using the change of variables \(x=2u\) and \(y=4v+2u\). Answer: \(\iint_{R} x^{2} y \; dA = \frac{256}{3} + 64\)

Step-by-step solution

Sketch R and S

For R, we have the following inequalities: \(0 \leq x \leq 2\), \(x \leq y \leq x+4\). To sketch R, plot the lines \(x=0\), \(x=2\), \(y=x\) and \(y=x+4\). The region is enclosed by these lines, forming a trapezoid. Now, for S, apply the change of variables that are given: \(x=2u\) and \(y=4v+2u\). Solve these equations for \(u\) and \(v\) to get \(u = 0.5x\) and \(v = 0.25(y-x)\). Substituting the bounds of R, we find the bounds for S as \(0 \leq u \leq 1\) and \(0 \leq v \leq 1\). Hence, the region S is a square in the uv-plane.

Find limits of integration

Since we have found the bounds for S to be \(0 \leq u \leq 1\) and \(0 \leq v \leq 1\), we now know the limits of integration for the new integral with respect to \(u\) and \(v\).

Compute the Jacobian

The Jacobian is determined by the matrix: $\begin{bmatrix} \frac{d x}{d u} & \frac{d x}{d v}\\ \frac{d y}{d u} & \frac{d y}{d v} \end{bmatrix}$ Calculate the partial derivatives: \(\frac{d x}{d u} = 2\) \(\frac{d x}{d v} = 0\) \(\frac{d y}{d u} = 2\) \(\frac{d y}{d v} = 4\) Now, we can find the determinant of the matrix: \(J = (2)(4) - (0)(2) = 8\)

Change variables and evaluate the new integral

Now, we have all the information we need to change variables and evaluate the new integral. We need to change the integrand as well: \(x^{2} y \; dA = x^{2} y \; J\; dudv\) Using the given change of variables, substitute \(x = 2u\) and \(y = 4v + 2u\). Multiplying the integrand by the Jacobian, we get: \((2u)^{2} (4v+2u) \cdot 8 \; dudv = 64u^{2} (4v+2u) \; dudv\) Now, evaluate the new integral: \(\iint_{S} 64u^{2} (4v+2u) \; dudv = \int_{0}^{1}\int_{0}^{1} 64u^{2} (4v+2u) \; dudv \) Integrate first with respect to \(v\): \(= 64u^{2} \int_{0}^{1} (4v+2u) \; dv\) Integrating and evaluating gives: \(= 64u^{2} \left[ 2v^{2}+2uv \right]_{0}^{1} = 64u^{2}(2+2u)\) Now integrate with respect to \(u\): \(= \int_{0}^{1} 64u^{2}(4+4u) \; du\) Integrating and evaluating gives: \(= \left[ 64\left(\frac{4}{3}u^{3} + u^{4}\right) \right]_{0}^{1} =\frac{256}{3} + 64\) So, the value of the given integral is: \(\iint_{R} x^{2} y \; dA = \frac{256}{3} + 64\)

Key concepts

Jacobian transformation

When dealing with double integrals, especially when changing variables, the concept of the Jacobian transformation is crucial. The Jacobian allows us to relate two different sets of variables by providing a scaling factor that adjusts the area element appropriately. It ensures that our integral transformation maintains the correct value.

In our example, we have the transformation from variables \(x\) and \(y\) to \(u\) and \(v\), given by \(x = 2u\) and \(y = 4v + 2u\). To compute the Jacobian, we construct a 2x2 matrix consisting of partial derivatives:

• \(\frac{\partial x}{\partial u} = 2\)
• \(\frac{\partial x}{\partial v} = 0\)
• \(\frac{\partial y}{\partial u} = 2\)
• \(\frac{\partial y}{\partial v} = 4\)

The determinant of this matrix, which we refer to as the Jacobian \(J\), is crucial in understanding how the area scales: \[ J = \left| \begin{array}{cc} 2 & 0 \ 2 & 4 \end{array} \right| = (2)(4) - (0)(2) = 8 \]

This value tells us that each area in the \(uv\)-plane is scaled by a factor of 8 relative to the original \(xy\)-plane. When converting the integral to this new coordinate system, we multiply by this Jacobian to maintain the integrity of the integration.

Change of variables in integration

The change of variables is a powerful technique in calculus, especially when the region of integration is complex. By transforming variables, we can simplify the geometry of the region and the integrand itself.

In the given problem, we are initially working with the region \(R\) in the xy-plane, defined by inequalities: \(0 \leq x \leq 2\) and \(x \leq y \leq x+4\). By applying the change of variables \(x = 2u\) and \(y = 4v + 2u\), we transform this region into a new region \(S\) in the uv-plane. This transformation clearly redefines the boundaries into a simpler form: \(0 \leq u \leq 1\) and \(0 \leq v \leq 1\), forming a perfect square.

This change not only makes the limits of integration straightforward but also simplifies the function within the integral, allowing easier computation. An added benefit is it often turns the double integral into simpler bounded values, making the process of solving them much more accessible.

Limits of integration

When solving double integrals, setting your limits of integration correctly is vital. These limits define the region over which you are integrating and ensure you cover the desired area without overshooting or missing parts.

Initially, in the xy-plane, the limits are defined by the problem, with \(x\) going from 0 to 2 and \(y\) ranging from \(x\) to \(x+4\). These limits describe a trapezoidal shape.

After applying the transformation variables to convert into the uv-plane, these complex bounds convert to much simpler ones: both \(u\) and \(v\) range from 0 to 1. These new limits square off the region \(S\), making it easy to compute the integral as you process from double to a single iterated integral step-by-step.

Understanding and correctly applying these limits ensures that your computed integral accurately reflects the desired area and variables, which is crucial for acquiring the correct integral result.