Problem 194 - Calculus Volume 3

Question

Determine a definite integral that represents the area. Region in the first quadrant enclosed by \(r=2-\cos \theta\)

Step-by-Step Solution

Answer

The area is \(\frac{3\pi - 8}{4}.\)

Step 1: Understand the Polar Function

The polar function given is \(r = 2 - \cos \theta\). This function describes a limaçon curve. Since the function is defined in polar coordinates, the area calculation will be different from that in Cartesian coordinates.

Step 2: Set Up the Integral for Area in Polar Coordinates

The area \(A\) of a region defined in polar coordinates is given by the integral \(A = \frac{1}{2} \int (r^2) \, d\theta\). Here, it will be \(\frac{1}{2} \int (2 - \cos \theta)^2 \, d\theta\).

Step 3: Determine Limits of Integration

To find the area in the first quadrant, determine the limits of \(\theta\). In the first quadrant, \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\).

Step 4: Simplify the Integral Expression

Expand the expression within the integral: \((2 - \cos \theta)^2 = 4 - 4\cos \theta + \cos^2 \theta\). Use the identity \(\cos^2 \theta = \frac{1+\cos 2\theta}{2}\) for further simplification.

Step 5: Compute the Integral

Substitute the simplified expression into the integral: \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( 4 - 4\cos \theta + \frac{1+\cos 2\theta}{2} \right) \, d\theta. \]Simplify the integrand: \[ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( \frac{9}{2} - 4\cos \theta + \frac{1}{2}\cos 2\theta \right) \, d\theta. \]Integrate term by term.

Step 6: Evaluate the Definite Integral

Calculate each part:- \(\int 1 \, d\theta = \theta\).- \(\int \cos\theta \, d\theta = \sin\theta\).- \(\int \cos 2\theta \, d\theta = \frac{1}{2}\sin 2\theta\). Apply the limits from \(0\) to \(\frac{\pi}{2}\) to the integral solution.The final result simplifies to \(\frac{3\pi}{4} - 2 + 0\).

Step 7: Finalize the Area Calculation

Simplify the expression from the previous step: \[ \frac{3\pi}{4} - 2 = \frac{3\pi}{4} - \frac{8}{4} = \frac{3\pi - 8}{4}. \]Thus, the area of the region in the first quadrant enclosed by \(r = 2 - \cos \theta\) is \(\frac{3\pi - 8}{4}.\)

Key Concepts

Limaçon Curve

A limaçon is a fascinating type of curve, often introduced when studying polar coordinates. It has the general form of the polar equation \( r = a \, + \, b \cos\theta \) or \( r = a \, + \, b \sin\theta \). These curves can take on multiple shapes, depending on the values of \(a\) and \(b\).

In our exercise, the curve is defined by \( r = 2 - \cos \theta \), which gives it a distinctive limaçon shape. This equation subtly combines a circle with a loop, potentially creating a dimple or even a cardioid shape depending on where \( b = \cos \theta \) lines up with the constant term \(a = 2\).

Understanding the visual representation of this curve helps deeply. A limaçon may appear similar to other curves, like ellipses, but its unique lobe-like features distinguish it clearly and make it an interesting study subject in polar geometry.

Definite Integral

A definite integral is a tool used to calculate areas, volumes, and other quantities where you need to sum infinitesimal amounts. In this context, it gives a precise area under a curve, between set limits of \(\theta\) or \(x\).

The process of integration involves calculating the area, as mentioned above, by the sum of many tiny rectangles under the curve. Here, we deal with the integral to find the area of the limaçon within specific boundaries – in this case, the first quadrant between \(\theta = 0\) and \(\theta = \frac{\pi}{2}\).

• Boundaries: These are your limits of integration \([0, \frac{\pi}{2}]\).
• Expression: The expression \((2 - \cos \theta)^2\) needs expansion for integrating.

This integral gives the region's exact area by summing the values \( \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (2 - \cos \theta)^2 \, d\theta \).

Polar Coordinates

Polar coordinates provide an alternative way of representing a point in a plane, using a radius and an angle instead of the typical \(x, y\) Cartesian system. This coordinate system is particularly convenient for curves pertaining to rotational symmetry and those expressed more naturally in terms of angles such as circles, spirals, and cardioids.

**Key Features:**

• Radius \(r\): This represents the distance from the origin to the point.
• Angle \(\theta\): This measures the counterclockwise direction from the positive x-axis to the point.

In our situation, polar coordinates help describe the limaçon curve's shape and determine the area enclosed by this curve. The challenge shifts toward mastering the unique characteristics of polar coordinates like radial lines and rotational intervals, rather than typical up-down or left-right movements.

Area Calculation in Polar Coordinates

Finding the area in polar coordinates may seem daunting compared to Cartesian systems, but it follows clear steps. The key formula for area calculation is:

\[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]

Where \(r\) is the radius from the polar equation, and \(\alpha\) and \(\beta\) are the lower and upper limits of \(\theta\).
**Important Hints:**

• Expand \( (r(\theta))^2 \) if necessary for simplification.
• Use trig identities like \(\cos^2\theta = \frac{1+\cos2\theta}{2}\) for easier integration.
• Plug in the limits after integrating to find the definite area.

For example, by expanding and simplifying \((2 - \cos \theta)^2\) and integrating over \([0, \frac{\pi}{2}]\), one finds the enclosed area in the first quadrant, expressing it neatly as the definite integral's solution.