Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\iint_{R} \sqrt{x^{2}+y^{2}} d A ; R=\left\\{(x, y): 1 \leq x^{2}+y^{2} \leq 4\right\\}$$

Short answer

Question: Evaluate the integral \(\iint_{R} \sqrt{x^{2}+y^{2}} d A\) over the region R defined by \(1 \leq x^2+y^2 \leq 4\). Answer: \(\frac{14\pi}{3}\)

Step-by-step solution

Sketch the Region R

The region R is defined as the set of all (x, y) points such that \(1 \leq x^2+y^2 \leq 4\). This describes the area between two circles centered at the origin. The first circle has a radius 1, and the second one has a radius 2. Sketching these circles will help us visualize R better.

Convert to Polar Coordinates

To change the integral to polar coordinates, we need to rewrite the function \(\sqrt{x^2+y^2}\) and the region R in terms of r and θ. In polar coordinates, the equation \(x^2+y^2= r^2\) holds, so the region R can be written as \(1 \leq r^2 \leq 4\), or equivalently, \(1 \leq r \leq 2\). The polar coordinate integral also involves multiplying by an extra r inside the integral, so we get: $$\iint_{R} \sqrt{x^{2}+y^{2}} d A = \int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2} r\cdot r dr d\theta$$ Here, \(\theta_1\) and \(\theta_2\) are the beginning and end angles of the region R. Since the whole region R is enclosed by both circles, \(\theta_1 = 0\) and \(\theta_2 = 2\pi\). Our integral becomes: $$\int_{0}^{2\pi}\int_{1}^{2} r^2 dr d\theta$$

Solve the Integral

Now we can solve the integral by calculating the inner integral first: $$\int_{1}^{2} r^2 dr = \left[\frac{1}{3}r^3\right]_{1}^{2} = \frac{1}{3}(2^3 - 1^3) = \frac{7}{3}$$ Next, we calculate the outer integral: $$\int_{0}^{2\pi} \frac{7}{3} d\theta = \left[\frac{7\theta}{3}\right]_{0}^{2\pi} = \frac{14\pi}{3}$$ So the final answer is: $$\iint_{R} \sqrt{x^{2}+y^{2}} d A = \frac{14\pi}{3}$$

Key concepts

Polar Coordinates

Polar coordinates offer an alternative way to locate points in the plane using a distance and an angle instead of x and y coordinates. This system is especially handy for dealing with problems involving circular or radial symmetry. In Polar coordinates, any point P can be represented as \( (r, \theta) \) where \( r \) is the radial distance from the origin (O) to the point, and \( \theta \) is the angle measured from the positive x-axis to the line segment OP. For conversion, the relationships \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \) are used.

When graphing in polar coordinates, circles are represented by simple equations like \( r = \text{constant} \) and spirals by equations involving \theta. This is much simpler than their Cartesian counterparts, highlighting one of the key advantages of using polar coordinates in certain integration problems.

Area Between Curves

In calculus, the concept of finding the area between curves can be visualized by considering the region bounded between two or more specific curves. This region is essentially the 'difference' in area of the regions under each curve. In Cartesian coordinates, the area is found by integrating the difference of the functions (curves) over the interval they bound the area.

In the context of polar coordinates, the 'curves' often end up being circles or functions of \( r \) with respect to \theta. The integral will run from one curve (the inner boundary) to the other (the outer boundary), often resulting in a need to integrate a function of \( r \) or \( r^2 \) due to the nature of the polar equations. The incremental area in polar coordinates is given by \( \frac{1}{2}r^2 d\theta \) which accounts for the 'slice of pie' shape rather than a rectangle.

Calculus

Calculus is the branch of mathematics that deals with the properties of derivatives and integrals of functions. This powerful tool allows us to understand changes and motion, and it is essential for solving problems in physics, engineering, economics, and beyond. In the realm of double integrals, calculus enables us to calculate the volume under a surface or the area of a region by integrating a function of two variables.

The process generally involves finding the anti-derivative, or integral, of a function, which provides the accumulated value, such as area under a curve in single-variable calculus, or volume under a surface in multivarable calculus. In the case of the exercise given, we're interested in finding the accumulated radial distance—a sort of 'weighted area'—of a region in the plane.

Multivariable Integration

Multivariable integration extends the idea of a single integral to multiple dimensions. While a single integral finds the area under a curve, a double integral like the one in our exercise can find the volume under a surface or the area of a region in more complex planes. In polar coordinates, when working with a double integral, the area element \( dA \) is represented as \( r dr d\theta \) rather than \( dx dy \) which acknowledges the fact that space is being partitioned into segments of circles rather than rectangles.

When performing a double integral in polar coordinates, as in the step-by-step solution, it is necessary to correctly set up the limits of integration for \( r \) and \theta. The process involves first integrating with respect to \( r \) and then integrating the resulting expression with respect to \theta. It is a powerful technique for evaluating areas of regions bounded by curves that are more naturally described in terms of their distance from the origin and angle from the horizontal.