Question

For every two-dimensional set \(C\) contained in \(R^{2}\) for which the integral exists, let \(Q(C)=\int_{C} \int\left(x^{2}+y^{2}\right) d x d y\). If \(C_{1}=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\}\), \(C_{2}=\\{(x, y):-1 \leq x=y \leq 1\\}\), and \(C_{3}=\left\\{(x, y): x^{2}+y^{2} \leq 1\right\\}\), find \(Q\left(C_{1}\right), Q\left(C_{2}\right)\) and \(Q\left(C_{3}\right)\).

Short answer

Evaluate each double integral in order to compute \(Q(C_{1})\), \(Q(C_{2})\), and \(Q(C_{3})\). Each involves different patterns of integration due to the changing restrictions on \(x\) and \(y\). Polar coordinates are used for the computation of \(Q(C_{3})\), as the region \(C_{3}\) forms a unit circle.

Step-by-step solution

Computation of \(Q(C_{1})\)

For \(C_{1}\), \(x\) and \(y\) both vary from -1 to 1. Thus, the double integral can be written as follows: \[ Q(C_{1}) = \int_{-1}^{1} \int_{-1}^{1} (x^2 + y^2) \, dx \, dy \] Solve the inner integral first with respect to \(x\) and then solve the outer integral with respect to \(y\).

Computation of \(Q(C_{2})\)

Here, \(x = y\), the integral becomes: \[ Q(C_{2}) = \int_{-1}^{1} (x^2 + x^2) \, dx = 2 \int_{-1}^{1} x^2 \, dx \] Compute the integral to evaluate \(Q(C_{2})\).

Computation of \(Q(C_{3})\)

In this case, \(C_{3}\) represents a disc of radius 1 centered at the origin. It is easier to use polar coordinates here hence we use \(x = r \cos(\theta)\) and \(y= r \sin(\theta)\), where \(r \leq 1\) and \(\theta\) ranges from 0 to \(2\pi\). Convert the integral into polar coordinates, the integral for \(Q(C_{3})\) becomes: \[ Q(C_{3}) = \int_{0}^{2\pi} \int_{0}^{1} r^2 (r^2) \, dr \, d\theta \] Perform the inner integral with respect to \(r\) first followed by the outer integral with respect to \(\theta\).

Consolidation of Results

Through the process, evaluate the definite integrals after finding the primitive functions. Once completed, these results will yield the values for \(Q(C_{1})\), \(Q(C_{2})\), and \(Q(C_{3})\) respectively.

Key concepts

Polar Coordinates

Polar coordinates offer a different way of looking at points in a plane using distances and angles instead of the standard Cartesian coordinates. In this exercise, we use polar coordinates to simplify calculations for the set \( C_{3} \), which is a disc with radius 1 centered at the origin. By transforming to polar coordinates:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

In these expressions, \( r \) represents the radial distance from the origin, while \( \theta \) is the angle from the positive x-axis. This transformation allows us to convert double integrals in a circular region into simpler integrals. The area element \( dx \, dy \) becomes \( r \, dr \, d\theta \), due to the Jacobian of the transformation. This process can make integration over circular regions much more straightforward.

Two-Dimensional Set

A two-dimensional set refers to any collection of points within a two-plane coordinate system, like our common x-y plane. In the original exercise, we encountered three different two-dimensional sets:

• \( C_{1} \): A square region where both x and y range from -1 to 1.
• \( C_{2} \): A line segment where x equals y, running from -1 to 1.
• \( C_{3} \): A circular region, or disc, where \( x^2 + y^2 \leq 1 \).

Each of these sets corresponds to a spatial area where the integration takes place. For instance, \( C_{3} \) benefits from a switch to polar coordinates due to its circular nature. Understanding the geometry and boundaries of these sets is crucial to setting up integrals correctly.

Definite Integrals

Definite integrals are a fundamental concept in calculus that allow us to calculate the accumulation of quantities over a specified interval. In two dimensions, we use double integrals to find quantities such as area or mass over a region. In this exercise, definite integrals help us evaluate \( Q(C_{1}) \), \( Q(C_{2}) \), and \( Q(C_{3}) \).
For \( C_{1} \) and \( C_{2} \), integrations are performed straightforwardly in Cartesian coordinates. However, for \( C_{3} \), converting the integral into polar coordinates simplifies the computation:

• The radial function \( r^2 \) is integrated over the disc's area.
• The angles \( \theta \) go from 0 to \( 2\pi \), encompassing the entire circle.

Definite integrals find the total accumulated value (like area or volume) within specific bounds, wrapping up our calculations for each two-dimensional set.