Problem 47 - Thomas Calculus

Question

Evaluate the integral \(\iint_{R} \sqrt{x^{2}+y^{2}} d A,\) where \(R\) is the region inside the upper semicircle of radius 2 centered at the origin, but outside the circle \(x^{2}+(y-1)^{2}=1\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{7\pi}{3} \).

Step 1: Understand the Region R

The region, \( R \), is defined as the area inside the upper semicircle of radius 2 centered at the origin, but outside the circle \( x^2 + (y-1)^2 = 1 \). This describes an annular sector that lies above the x-axis from \( x = -2 \) to \( x = 2 \), and excludes a circle centered at \( (0, 1) \) with radius 1.

Step 2: Convert to Polar Coordinates

To simplify the evaluation of the integral, we convert to polar coordinates. Recall the transformations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The differential area element transforms as \( dA = r \, dr \, d\theta \). The innermost circle becomes \( (r \sin\theta - 1)^2 = 1 - r^2 \cos^2 \theta \).

Step 3: Determine the Limits for r and θ

In polar coordinates, the outer radius of the semicircle is 2, so \( r \) ranges from 1 to 2. The angle \( \theta \) sweeps from \( 0 \) to \( \pi \) since we're considering the upper semicircle. The region described is an annular sector in polar coordinates between these limits.

Step 4: Integrate with Respect to r

The integrand in polar coordinates becomes \( \sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta} = r \). Therefore, the integral is \( \int_{1}^{2} \int_{0}^{\pi} r^2 \, dr \, d\theta \). First, integrate \( r^2 \) with respect to \( r \):\[ \int_{1}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}. \]

Step 5: Integrate with Respect to θ

Now integrate the result with respect to \( \theta \):\[ \int_{0}^{\pi} \frac{7}{3} \, d\theta = \frac{7}{3} \times \pi = \frac{7\pi}{3}. \]

Step 6: Final Step: Evaluate the Result

The final result of the integration is \( \frac{7\pi}{3} \). This is the value of the double integral over the specified region \( R \).

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system that represent a point in the plane by an angle and a distance. Instead of using

• x and y, like in Cartesian coordinates,
• polar coordinates use r and \(\theta\).

The distance from the origin, \(r\), and the angle \(\theta\), are key concepts.
These measurements provide a natural way to describe circular or rotational systems. In polar coordinates, any point \((x, y)\) is converted using the formulas:

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

This is useful whenever dealing with problem regions that are circular or rotational in nature, like semicircles or annuli. The angle \(\theta\) generally spans from \(0\) to \(2\pi\), making it versatile for covering the entire circle. Hence, switching to polar coordinates simplifies calculations involving symmetrically circular shapes.

Annular Region

An annular region in two-dimensional space is a ring-shaped area between two circles. Imagine a flat donut shape, where you have an inner circle that is cut out and an outer circle that defines the boundary. In the context of double integrals, especially those involving polar coordinates, annular regions are common, as they naturally appear when circles are defined by different radii of the same center, or slightly offset centers.
For example, in our exercise, the region is an annular sector. The outer boundary is defined by the semicircle of radius 2, centered at the origin.
The inner boundary is another circle, centered at (0, 1) with radius 1. By identifying these boundaries, one can set up appropriate limits for integration, like extending from radius 1 to 2 and angle \(0\) to \(\pi\).
This understanding helps in determining the area you need to integrate over.

Area Integrals

Area integrals involve calculating two-dimensional integrals to find areas, masses, or other quantities distributed over a plane. This is done by integrating a function over a specified region. In cases where the region is circular or annular, as in our example, changes to polar coordinates facilitate the determination of these integrals.
In polar coordinates, the area element \(dA\) is expressed as \(r \, dr \, d\theta\), which naturalizes integration over circular domains. When evaluating the function over an annular region like the one in our example, the integration is split into two parts:

• The integral with respect to \(r\) covers the radial distances from the inner to the outer boundary.
• The integral with respect to \(\theta\) spreads across the angular width of the region, which in the problem is \(0\) to \(\pi\).

Using these constructs, the double integral gives a comprehensive solution over the desired area.
The calculation follows through by first solving the inner integral, then the outer, employing fundamental calculus techniques to achieve the final value.