Problem 31 - University Calculus: Early Transcendentals

Question

Find the area of the region cut from the first quadrant by the cardioid \(r=1+\sin \theta.\)

Step-by-Step Solution

Answer

The area is \( \frac{3\pi}{8} + 1 \).

Step 1: Understand the Cardioid Equation

The given equation is \( r = 1 + \sin \theta \), which is a cardioid in polar coordinates. In the first quadrant, we limit \( \theta \) to values from \( 0 \) to \( \frac{\pi}{2} \).

Step 2: Use the Polar Area Formula

The area \( A \) of a region in polar coordinates is calculated using the formula: \[ A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta \] where \( r = f(\theta) \). For our problem, \( r = 1 + \sin \theta \) and \( \theta \) varies from \( 0 \) to \( \frac{\pi}{2} \).

Step 3: Set Up the Integral for Area Calculation

Substitute \( r = 1 + \sin \theta \) into the area formula: \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 + \sin \theta)^2 \, d\theta \].

Step 4: Simplify the Expression Inside the Integral

Expand the term \((1 + \sin \theta)^2\):\[ (1 + \sin \theta)^2 = 1 + 2\sin \theta + \sin^2 \theta \].

Step 5: Solve the Integral

Evaluate \[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 + 2\sin \theta + \sin^2 \theta) \, d\theta \]. Using the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \), split and integrate each part: 1. \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \).2. \( \int_{0}^{\frac{\pi}{2}} 2\sin \theta \, d\theta = \left[ -2\cos \theta \right]_{0}^{\frac{\pi}{2}} = 2 \).3. \( \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2\theta}{2} \, d\theta = \left[ \frac{\theta}{2} - \frac{\sin 2\theta}{4} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4} \).

Step 6: Combine the Results

Add the results of the integrals:\[ A = \frac{1}{2} \left( \frac{\pi}{2} + 2 + \frac{\pi}{4} \right) = \frac{1}{2} \left( \frac{3\pi}{4} + 2 \right) \].

Step 7: Final Area Result

Calculate the expression to get the final area: \[ A = \frac{3\pi}{8} + 1 \].

Key Concepts

Cardioid

A cardioid is a unique heart-shaped curve represented in polar coordinates. It gets its name from the Greek word "kardia," meaning heart.
In polar coordinates, it often takes the form \( r = 1 + \sin \theta \) or \( r = 1 + \cos \theta \).
This type of curve has valuable applications in both mathematics and engineering.
It's important to recognize that cardioids are a special form of limaçon, a broader family of curves.
The cardioid manifests fascinating properties such as being tangent to the axes and going through the origin.

Area Calculation

Calculating the area of a region enclosed by polar curves like a cardioid involves a distinct method.
We apply the formula for the area in polar coordinates:
\[A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta\]
Here, \( r \) represents a function dependent on \( \theta \), and \([a, b]\) represents the interval over which the curve extends.
For a cardioid: \( r = 1 + \sin \theta \), you determine \( a \) and \( b \) based on the part of the curve you want to calculate, like focusing on the first quadrant.
By setting up this integral correctly, you can determine the precise area within this defined section of the polar plane.

Integral Calculus

Integral calculus plays a vital role in computing areas under curves, especially in polar coordinates.
The integral showcases its power by accumulating the tiny, infinitesimal segments of area under a curve to find a total area.
In our cardioid example, we evaluate:
\[\int_{0}^{\frac{\pi}{2}} (1 + 2\sin \theta + \sin^2 \theta) \, d\theta\]
Breaking down the function inside the integral into manageable parts facilitates this evaluation:

• \( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta \)
• \( \int_{0}^{\frac{\pi}{2}} 2\sin \theta \, d\theta \)
• \( \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2\theta}{2} \, d\theta \)

These calculations, using known trigonometric identities, help ascertain the area accurately.
Don't be intimidated by the calculus, though! It's just about adding up areas over and over until you have a complete picture.

First Quadrant

The first quadrant of the coordinate system is where both \( x \) and \( y \) values are positive.
In polar coordinates, it corresponds to angles \( \theta \) ranging from \( 0 \) to \( \frac{\pi}{2} \).
When solving problems, especially with polar curves like cardioids, the first quadrant is crucial because it simplifies calculations.
By restricting our focus to this quadrant, we limit the angle \( \theta \), making integral calculations more straightforward.
Understanding quadrant boundaries ensures that we don't double-count areas or compute beyond the required region, keeping it simple and easy to manage.