Problem 47 - University Calculus: Early Transcendentals

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 value of the integral is \( \pi \).

Step 1: Define the Region

Firstly, identify the region \( R \). The region is inside the upper semicircle of radius 2 centered at the origin, which has the inequality \( x^2 + y^2 \leq 4 \) and \( y \geq 0 \). Additionally, it's outside the circle \( x^2 + (y-1)^2 = 1 \). This circle is centered at \((0, 1)\) with radius 1.

Step 2: Convert to Polar Coordinates

To evaluate the integral \( \iint_{R} \sqrt{x^2+y^2} \, dA \), convert to polar coordinates. In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \). The Jacobian of the transformation is \( r \), so \( dA = r \, dr \, d\theta \). The integrand becomes \( r \sqrt{r^2} = r^2 \).

Step 3: Set Up Integration Limits

The semicircle \( x^2 + y^2 \leq 4 \) corresponds to \( 0 \leq r \leq 2 \) and the circle \( x^2 + (y-1)^2 = 1 \) can be parameterized in polar form, \( x = r\cos\theta \) and \( y - 1 = r\sin\theta \). Solving, \( r^2\cos^2\theta + r^2\sin^2\theta = 1 - 2r\sin\theta + r^2 \sin^2\theta \) simplifies to \( r \geq 2\sin\theta \). As such, \( r \) runs from \( 2\sin\theta \) to 2.

Step 4: Determine Angular Limits

Since the region is the upper semicircle, \( \theta \) ranges from \( 0 \) to \( \pi \). Furthermore, to satisfy both circle conditions, the area of influence from the inner circle means \( \theta \) runs only where it's possible\( \sin\theta \) is not zero, thus the integration takes place fully from \( \theta = 0 \) to \( \pi \).

Step 5: Evaluate the Integral

Integrate the function in polar coordinates. The given integral is now \( \int_0^{\pi} \int_{2\sin\theta}^{2} r^2 \, r \, dr \, d\theta \). This gives:\[\int_0^{\pi} \int_{2\sin\theta}^2 r^3 \, dr \, d\theta = \int_0^{\pi}\left[\frac{1}{4}r^4\right]_{2\sin\theta}^{2} \, d\theta\]Evaluate the \( r \) integral to get:\[\int_0^{\pi}\left(\frac{16}{4} - \frac{16\sin^4\theta}{4} \right) \, d\theta = \int_0^{\pi} (4 - \frac{4\sin^4\theta}{4}) \, d\theta\]Which simplifies to \( \int_0^{\pi} 4(1 - \sin^4\theta) \, d\theta \).

Step 6: Compute Final Integral

Finish computing the integral:\[\int_0^{\pi} 4 - 4\sin^4\theta \, d\theta = 4\left[\theta - \frac{3\theta}{4} \right]_0^{\pi} = 4\left[\frac{\pi}{4} \right] = \pi\]So the value of the integral is \( \pi \).

Key Concepts

Polar Coordinates

Polar coordinates provide a different way to represent points in a plane using distance and angle, rather than the standard Cartesian coordinates of \(x\) and \(y\). In polar coordinates, any point is defined by \(r\), the distance from the origin, and \(\theta\), the angle measured from the positive x-axis. Here’s how you relate the two systems:

• The relation for \(x\) is given by \(x = r \cos \theta\)
• For \(y\), it is \(y = r \sin \theta\)

In many mathematical problems, especially those involving circles or sectors, using polar coordinates simplifies the calculations.
This is what we've done in our exercise as we are dealing with circular regions.

Region of Integration

The region of integration is the area over which we are attempting to evaluate the integral. In this specific problem, we consider a particular shape:

• The region inside the upper semicircle of radius 2 centered at the origin
• The region outside a smaller circle of radius 1, centered at \( (0, 1)\)

Combining these defines the area we need to focus on, and understanding these constraints helps determine the boundaries of our integral when we set the limits of \(r\) and \(\theta\).
Visualizing the region can greatly aid in solving such integrals, as it simplifies determining the limits.

Conversion to Polar Coordinates

Conversion to polar coordinates is crucial when dealing with circular regions. First, we change the variables using polar representations:

• Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\)
• The differential area element \(dA\) in Cartesian becomes \(r dr d\theta\) in polar coordinates

The given function in Cartesian form \(\sqrt{x^2 + y^2}\) simplifies to \(r\) in polar coordinates.
Thus, this conversion effectively modifies the integral into a form that is often easier to solve.

Upper Semicircle

An upper semicircle is half of a circle that lies above a line, typically the x-axis. For this problem:

• The upper semicircle has a radius of 2 and is centered at the origin \( (0, 0)\)
• This implies \(x^2 + y^2 \leq 4\) and \(y \geq 0\)

Recognizing this segment helps us define the limits of \(\theta\) from 0 to \(\pi\) since we are dealing with the top half.
Understanding the structure allows us to adjust our region of integration accurately.

Evaluate Integrals

The final objective is to evaluate the integral over the specified region. In our problem, the steps to achieve this begin with setting up the integral in polar coordinates:

• The function \(r \sqrt{r^2}\) simplifies to \(r^2\)
• We integrate from \(r = 2 \sin \theta\) to \(r = 2\) and \(\theta\) from \(0\) to \(\pi\)

The integral thus becomes \(\int_0^\pi \int_{2\sin\theta}^{2} r^3 \, dr \, d\theta\), making it easier to compute.
Step-by-step evaluation provides the final solution, \(\pi\), rendering a clear conclusion to the problem.