Question

For every two-dimensional set \(C\) contained in \(R^{2}\) for which the integral exists, let \(Q(C)=\iint_{C}\left(x^{2}+y^{2}\right) d x d y .\) 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

\(Q(C_1) = 4/3\), \(Q(C_2) = 4/3\), and \(Q(C_3) = \pi/2\).

Step-by-step solution

Calculating Q(C1)

First, let's find the volume under the surface \(f(x,y) = x^2 + y^2\) and bounded by the region \(C_1\). By inspecting the region \(C_1\), we can see that both x and y range from -1 to 1. Thus, \(Q(C_1) = \int_{-1}^{1} \int_{-1}^{1} (x^2 + y^2) dy dx\). We can split this into two separate integrals: \(Q(C_1) = \int_{-1}^{1} x^2 dx + \int_{-1}^{1} y^2 dy \). Solving these we get \(Q(C_1) = [x^3/3]_{-1}^{1} + [y^3/3]_{-1}^{1} = 2/3 + 2/3 = 4/3\).

Calculating Q(C2)

Next, let's find the volume under the same surface bounded by the region \(C_2\). The region \(C_2\) represents a straight line from (-1,-1) to (1,1), since \(x = y\). Since x=y, the function simplifies to \(2x^2\). Thus, \(Q(C_2) = \int_{-1}^{1} 2x^2 dx = [2x^3/3]_{-1}^{1} = 4/3\).

Calculating Q(C3)

Lastly, the region \(C_3\) represents a disk centered at the origin with radius 1. To make the integral easier, we can use polar coordinates. Thus the integral becomes \(Q(C_3) = \int_{0}^{2\pi} \int_{0}^{1} r^3 dr d\theta\). Solving this we get \(Q(C_3) = [2\pi r^4/4]_{0}^{1} = \pi/2\).

Final Answers

So, the volume under the surface \(f(x,y) = x^2 + y^2\) above \(C_1\), \(C_2\), and \(C_3\) are \(4/3\), \(4/3\), and \(\pi/2\), respectively.

Key concepts

Integral of Polynomial Functions

Understanding the integral of polynomial functions is foundational for solving various types of calculus problems. Polynomials are expressions that consist of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents.

For example, in the exercise, we integrate the function \(f(x, y) = x^2 + y^2\), which is a polynomial function of two variables. In a broader sense, when we integrate a polynomial function, we are essentially finding the area under its curve within a given range.

Single Variable Integration

With single-variable polynomials, the basic rule of thumb is to increase the exponent of each term by one and divide by the new exponent.

Multiple Variables

When dealing with functions of multiple variables, like in the exercise, we may integrate with respect to each variable separately, applying the same principle. It's crucial to understand the region over which we are integrating, as this defines the limits of integration.

Volume Under Surface

The volume under a surface is a key concept in multivariable calculus. It extends the idea of finding the area under a curve to three dimensions, where we now seek the volume beneath a surface over a given domain or region.

In the provided solution, we calculate the volume by setting up a double integral over the region \(C_1\). The limits of integration for both variables \(x\) and \(y\) are determined by the bounds of the region. In essence, the double integral accumulates all the infinitesimal volumes \(dx dy\) multiplied by the height of the function being integrated, \(x^2 + y^2\), over the entire region.

Interpreting the Results

The integral tells us how much space lies between the surface defined by the function and the \(xy\)-plane across the area of integration. This concept is crucial not only in mathematics but also in physics and engineering where such calculations are often made to determine physical quantities.

Polar Coordinates Integration

Polar coordinates integration allows us to calculate areas and volumes in problems where the setup is more naturally expressed in terms of radii and angles rather than \(x\) and \(y\) coordinates. This coordinate system can simplify the complexity of certain integrals, particularly when dealing with circular or sector-shaped regions.

In the case of \(C_3\), we have a circle centered at the origin, making polar coordinates an ideal choice. Here, \(r\) represents the distance from the origin, and \(\theta\) is the angle measured from the positive \(x\)-axis.

Conversion to Polar Coordinates

The function \(f(x, y) = x^2 + y^2\) conveniently converts to \(r^2\) in polar coordinates, and the integral simplifies as we're integrating \(r^3\) over a full circle (\(0\) to \(2\pi\)) and radius (\(0\) to \(1\)). The Jacobian determinant in polar coordinates (\(r\)) is automatically accounted for in the polar integration.

Mathematical Statistics

While not directly used in the solution of this exercise, mathematical statistics involves the use of integrals to summarize data, evaluate probabilities, and estimate parameters. Integrals are at the heart of probability density functions (PDFs) and cumulative distribution functions (CDFs) for continuous random variables.

By integrating a PDF over a range of values, we can find the probability that a random variable falls within that range. In the context of volume calculations, one could imagine using statistical distributions to model uncertainties in the boundaries or the surface shape, from which the volume under that 'random' surface could be evaluated.

Connection to Calculus

Calculus, and integrals specifically, provide the tools necessary to quantify and manage such uncertainties, underpinning the statistical methods applied in fields ranging from engineering to economics. Understanding the role of integrals in statistics is essential for students delving into data science and related areas.