91Ó°ÊÓ

The concept of the double integral is fundamental in calculus. It allows us to compute the accumulation of quantities over a region in the plane. When we refer to a double integral like \(\iint_{R} f(x, y) \, dA\), \(f(x, y)\) is a function evaluated over the region \(R\). The notation \(dA\) stands for a differential area element, which can be thought of as a tiny rectangle or section of the region where we're evaluating the function.A double integral computes the "volume under a surface" for real-valued functions defined over a two-dimensional region. This can visualize as the mass, area, or other cumulative quantities depending on the context. In our problem, this involves integrating an expression over a given region within the plane.The double integral allows us to consider accumulative effects across two dimensions:- **Volume of a solid**: when the function \(f(x, y)\) represents height, the integral computes the three-dimensional volume under a surface.- **Average value**: dividing the result of the integral by the area of the region gives an average value of \(f(x, y)\) over \(R\).- **Applications in physics**: for instance, integrating density functions to find total mass.In our particular exercise, the double integral is used to assess the function \(\arctan(y/x)\) over a specified region \(R\). In these types of exercises, transforming coordinates, as will be seen later, simplifies the process and possibly makes the computation more intuitive.

The region of integration \(R\) plays a crucial role in evaluating double integrals. It defines the boundary or area over which the integral is computed. In the current problem, the region is specified by the inequalities \(1 \leq x^2 + y^2 \leq 4\) and \(0 \leq y \leq x\). Understanding this region helps in visualizing the problem and correctly setting the limits for integration.This region, described by \(1 \leq x^2 + y^2 \leq 4\), represents an annular (ring-shaped) section bounded between two circles with radii 1 and 2 centered at the origin. To further narrow this down, the requirement \(0 \leq y \leq x\) restricts us to a specific portion of this annulus — specifically, the "pie-slice" region in the first quadrant.Key insights about the region of integration:- **Shape**: Understanding the geometric shape formed by the given inequalities helps in setting up the integration limits.- **Quadrant-based restrictions**: Knowing \(0 \leq y \leq x\) implies constraints regarding the permissible quadrant or directional restrictions.- **Changing limits**: These need to be adapted to fit a new coordinate system if the transformation is applied, as demonstrated in our exercise.Grasping the region of integration ensures accuracy when setting up double integrals and choosing the appropriate coordinate transformation, as we will discuss next.

Coordinate transformation is a powerful tool in multivariable calculus, often simplifying complex integrals. Converting from Cartesian \((x, y)\) coordinates to polar coordinates \((r, \theta)\) is particularly useful when the region of integration exhibits radial symmetry, as in our problem.In polar coordinates, the rectangular coordinates \(x\) and \(y\) are expressed in terms of \(r\) (the radial distance from the origin) and \(\theta\) (the angle from the positive \(x\)-axis). The transformations are defined as:- \(x = r \cos(\theta)\)- \(y = r \sin(\theta)\)One of the advantage of using polar coordinates is simplifying the integration process by aligning the integrals with the geometry of the region:- **Angular limits**: In our problem, \(\theta\) goes from \(0\) to \(\pi/4\), reflecting the region's angular extent.- **Radial symmetry**: The annular region translates nicely into radial bounds \(1 \leq r \leq 2\).- **Easier area element**: The area element \(dA\) becomes \(r \, dr \, d\theta\), capturing both radial and angular components of change.Transformations are an analytical strategy to solve integrals by reorienting the axis and limits to match the problem's natural symmetry. In our exercise, this drastically simplifies calculation, particularly in recognizing that \(\arctan(y/x) = \theta\) directly in polar coordinates.