Question

Let \(R\) be the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) where \(a>0\) and \(b>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v\) Evaluate \(\iint_{R}|x y| d A\)

Short answer

Based on the step-by-step solution provided, the double integral of \(|xy|\) over the region R enclosed by the ellipse can be calculated using a transformation, polar coordinates, and integration. After performing these steps, the final answer for the double integral of \(|xy|\) over the region R is \(\frac{(ab)^2 \pi}{4}\), where a and b are the semi-major and semi-minor axes of the ellipse, respectively.

Step-by-step solution

Set up the transformation

Let's express the transformation as a function, where T: \((x, y) \mapsto (u, v)\) with \(x = au\) and \(y = bv\). So, we have \(u = \frac{x}{a}\) and \(v = \frac{y}{b}\).

Find the Jacobian

The Jacobian \(J(u,v)\) of the transformation is the determinant of the matrix: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} a & 0\\ 0 & b \end{bmatrix} $$ And we find its determinant: $$J(u,v) = ab$$

Rewrite the double integral using the transformation

Now we need to change variables in the double integral. The given function \(|xy|\) becomes \(|abuv|\), and the domain R is transformed into \(u^2 + v^2 \leq 1\), which is a disk in the uv-plane. The double integral over the region R is then: $$\iint_{R}|xy| dA = \iint_{u^2+v^2\leq 1}|abuv|J(u,v) dudv$$ We substitute the Jacobian found in step 2, and we have: $$\iint_{R}|xy| dA = \iint_{u^2+v^2\leq 1}(ab)^2 uv \thinspace dudv$$

Convert to polar coordinates

Now we'll convert this integral to polar coordinates (\(u = r\cos\theta\), \(v = r\sin\theta\)): $$\iint_{r \leq 1} (ab)^2 (r\cos\theta)(r\sin\theta) rdrd\theta$$ Now we rewrite the integrand in terms of polar coordinates: $$(ab)^2 r^2 \cos\theta\sin\theta \cdot r$$ So, the integral becomes: $$(ab)^2 \int_0^1 \int_0^{2\pi} r^3 \cos\theta \sin\theta \thinspace d\theta dr$$

Evaluate the polar integral

Now we will separate the integral into two parts: the \(r\) integral and the \(\theta\) integral: $$ (ab)^2 \left(\int_0^1 r^3 dr\right)\left(\int_0^{2\pi} \cos\theta\sin\theta \thinspace d\theta\right) $$ We can now evaluate the two integrals separately. For the \(r\) integral: $$ \int_0^1 r^3 dr = \left[\frac{1}{4}r^4\right]_0^1 = \frac{1}{4} $$ For the \(\theta\) integral: $$ \int_0^{2\pi} \cos\theta\sin\theta d\theta $$ We can apply integration by parts, let \(u = \sin\theta, dv = \cos\theta d\theta\). We obtain: \(du = \cos\theta d\theta\) and \(v = \sin\theta\). Using integration by parts we have: $$ \int_0^{2\pi} \cos\theta\sin\theta d\theta = \left[\sin^2\theta\right]_0^{2\pi} - \int_0^{2\pi}\sin^2\theta d\theta $$ The first term evaluates to 0, so we have: $$ -\int_0^{2\pi}\sin^2\theta d\theta = -\frac{1}{2}\int_0^{2\pi}(1 - \cos(2\theta))d\theta $$ Evaluating the integral we get: $$ -\frac{1}{2}\int_0^{2\pi}(1 - \cos(2\theta))d\theta = -\pi $$ Now we multiply the two results and get: $$ \iint_{R}|xy| dA = (ab)^2 \cdot \frac{1}{4} \cdot (-\pi) = \frac{-(ab)^2 \pi}{4} $$ However, the original expression asks for the absolute value, so we take the absolute value of the result: $$ \iint_{R}|xy| dA = \abs{\frac{-(ab)^2 \pi}{4}} = \frac{(ab)^2 \pi}{4} $$ Therefore, the double integral of \(|xy|\) over the region R is \(\frac{(ab)^2 \pi}{4}\).

Key concepts

Ellipse Region

An ellipse is a geometric shape that can be thought of as an elongated circle. Mathematically, it is defined as the set of all points \(x, y\) that satisfy the equation of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. These parameters dictate the dimensions of the ellipse in the Cartesian plane.

In our context, the region bounded by this ellipse is represented as \(R\). This region describes the area inside the ellipse where the double integral will be evaluated. When dealing with transformations, especially in calculus, changing the perspective of this region into another coordinate system can simplify complex integrals. For instance, transforming an elliptical region into a circular region within a new coordinate system can make integration much more manageable.

• Semi-major axis \(a\): Length from the center to the farthest point horizontally.
• Semi-minor axis \(b\): Length from the center to the farthest point vertically.

Understanding these dimensions is crucial because they determine the scaling factors used in transformations, such as those used for converting elliptical regions into more computationally convenient forms, like a unit disk in the \(uv\) plane.

Polar Coordinates

Polar coordinates provide an alternative way to describe a point in the plane using a radius and an angle. This system is particularly useful for regions with circular symmetry, as it simplifies the integration process by aligning with the circular nature of the problem.

In the context of the Jacobian transformation, converting to polar coordinates involves transforming the variables \(u\) and \(v\) using formulas based on the radius \(r\) and angle \(\theta\):

• \(u = r\cos\theta\)
• \(v = r\sin\theta\)

By converting to polar coordinates, the double integral over the elliptical region can be re-evaluated over a simpler, circular region defined by \(r \leq 1\). This allows the complex variables in the integrand, such as \(uv\), to be represented in terms of simpler trigonometric functions, like \(\cos\theta\) and \(\sin\theta\), streamlining the integration process.

This approach is a common strategy in multi-variable calculus as it reduces an otherwise intricate integration task into more manageable components, leveraging the symmetry of polar angles and radial scaling to simplify the limits and computation of the integral.

Double Integral

A double integral is a way to accumulate values over a two-dimensional region. It is the 2D analogue of a single-variable integral, allowing us to calculate areas, volumes, and other quantities that depend on two variables.

The process of evaluating a double integral involves integrating a function of two variables across a specified region. In our context, this means calculating the accumulated 'weight' or value of \( |xy| \) over the ellipse region \( R \). When transformed via a Jacobian, the double integral is expressed as: \(\iint_{R}|xy| dA = \iint_{u^2+v^2\leq 1}(ab)^2 uv \thinspace dudv\), which can be solved via conversion to polar coordinates.

• Each infinitesimal portion \(dA\) is represented by \(rdrd\theta\) in polar form.
• The limits of integration change based on the geometry—here from a bounding ellipse to a disk.

Evaluating the resulting polar integral divides the task into more straightforward one-dimensional integrals over \(r\) and \(\theta\). This technique is beneficial for solving complex problems, reducing them into solvable steps by harnessing the power of coordinate transformations and symmetry in integration.