Problem 30 - 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. \(\iint_{R} x y d A,\) where \(R\) is bounded by the ellipse \(9 x^{2}+4 y^{2}=36\) use \(x=2 u, y=3 v\).

Step-by-Step Solution

Answer

Answer: The value of the double integral after changing variables and evaluation is 18.

Step 1: Sketch the region of integration R in the xy-plane and the new region S in the uv-plane

The region R is bounded by the ellipse \(9x^2 + 4y^2 = 36\). Divide both sides of the equation by 36, we get: \(\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1\) This is an ellipse with semi-major axes of 2 and 3 in the x and y-directions respectively. The change of variables given is \(x=2u, y=3v\). We can solve for \(u\) and \(v\): \(u=\frac{x}{2}, v=\frac{y}{3}\) To find the region S in the uv-plane, we substitute the new variables \(u\) and \(v\) into the ellipse equation: \(\frac{(2u)^2}{2^2} + \frac{(3v)^2}{3^2} = 1\) This simplifies to: \(u^2 + v^2 = 1\) So, the region \(S\) in the uv-plane is a circle with the radius 1 centered at the origin.

Step 2: Find the limits of integration for the new integral with respect to u and v

Since the new region is a circle with radius 1 centered at the origin (0,0) in uv-plane, the limits of the integration are: u: \(-1 \leq u \leq 1\) Next, we'll find the limits for v. We know that \(v^2 = 1 - u^2\) from the equation of the circle. Taking the square root of both sides, we get: \(v = \pm \sqrt{1-u^2}\) Hence, the limits for v are: v: \(-\sqrt{1-u^2} \leq v \leq \sqrt{1-u^2}\)

Step 3: Compute the Jacobian

We are given the transformation \(x=2u, y=3v\). To compute the Jacobian, we need to find the determinant of the matrix of the partial derivatives: $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} $ Taking the partial derivatives, we get: $\begin{vmatrix} 2 & 0\\ 0 & 3 \end{vmatrix}$ The determinant is: \(J(u, v) = (2)(3) - (0)(0) = 6\)

Step 4: Change variables and evaluate the new integral

Now we can rewrite the given integral using the new variables and their limits, as well as the Jacobian: \(\iint_{R} xy dA = \int_{-1}^{1} \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} (2u)(3v)(6) dvdu\) Now, we can evaluate this integral. First, integrate with respect to \(v\): \(\int_{-1}^{1} [18uv^2]_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} du = 18u\int_{-1}^{1} [\sqrt{1-u^2}^2] du = 18u\int_{-1}^{1} (1-u^2) du\) Next, integrate with respect to \(u\): \(18\left[\frac{1}{2}u^2 - \frac{1}{4}u^4\right]_{-1}^{1} = 18(2\left(\frac{1}{2}-\frac{1}{4}\right)-(2\left(\frac{1}{2}-\frac{1}{4}\right))) = 18(1-0)=18\) So, the value of the given integral is 18.

Key Concepts

Change of Variables

In multivariable calculus, the change of variables technique is an invaluable tool for simplifying integrals. By transforming the original variables to new variables, it allows us to integrate over a more straightforward region. This technique is essential when dealing with complex-shaped regions like ellipses, which can be transformed into simpler shapes, such as circles, in a different coordinate system.

In the provided exercise, the transformation from \[ x = 2u, \, y = 3v \]is used to convert the ellipse \[ 9x^2 + 4y^2 = 36 \]into a unit circle \[ u^2 + v^2 = 1 \]in the uv-plane. This kind of transformation simplifies both the region over which we integrate and the calculation involved. It does so by making the limits of integration more manageable, usually aligning them along machinery axis of the new plane.

This technique often requires finding the relationship between the old and new variables. The relationship helps in setting up the bounds of integration and rewriting the function to be integrated in terms of the new variables. The next step is to calculate the Jacobian determinant, which is crucial to ensure the volume or area represented in different coordinate systems is maintained.

Jacobian Determinant

The Jacobian determinant is a critical factor when applying change of variables in multivariable calculus. It serves as a scaling factor that accounts for how the area or volume elements change during the transformation. In simple terms, it tells us how much the change in variables distorts the space over which we are integrating.

To compute the Jacobian, we form a matrix of partial derivatives that represent how each of the original coordinates changes relative to the new coordinates. For the transformation\[ x = 2u, \, y = 3v \]the Jacobian matrix is:\[ J(u, v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}\]Substituting the partial derivatives, we get:\[ \begin{vmatrix} 2 & 0 \ 0 & 3 \end{vmatrix} \]The determinant, a straightforward multiplication in this case, results in: \[ J(u, v) = 6 \]This determinant is used to scale the integrand accordingly when changing variables in the integral, ensuring that the region's measurement in the new coordinate system is correct.

Double Integrals

Double integrals are a fundamental tool in calculus for finding areas and volumes. They allow us to integrate a function over a two-dimensional region. This is often done using either Cartesian or polar coordinates to compute solutions specific to the region's shape.

In the original exercise, we start with a double integral over an ellipse in the xy-plane. Through change of variables, this ellipse was transformed into a simpler circle in the uv-plane:\[ \iint_{R} xy \, dA = \int_{-1}^{1} \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} (2u)(3v)(6) \, dvdv \]

Changing the region of integration by transforming ellipses into circles simplifies the calculation. Each step involves integrating with respect to one variable while considering it constant with respect to the other. This approach systematically reduces the dimensions one at a time.

After setting up the double integral in the transformed coordinates, the integration with respect to \(v\) is performed first, followed by the integration with respect to \(u\). This nested integration can significantly simplify resulting calculations, especially with symmetric regions like circles. Ultimately, the double integral evaluates the desired quantity over the specific region - in this case, yielding a final value of 18.