Problem 31 - Thomas Calculus

Question

Cardioid in the first quadrant 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 Problem

We are given a polar equation of a cardioid: \(r = 1 + \sin \theta\). Our task is to find the area of this cardioid only within the first quadrant.

Step 2: Set up the Polar Area Formula

Recall that the area \(A\) of a region bounded by a polar curve \(r(\theta)\) from \(\theta = a\) to \(\theta = b\) is given by \[ A = \frac{1}{2} \int_{a}^{b} r(\theta)^2 \, d\theta \] For the first quadrant, \(\theta\) will range from \(0\) to \(\frac{\pi}{2}\).

Step 3: Substitute the Function in the Formula

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: Expand \((1 + \sin \theta)^2\)

Expand the expression \((1 + \sin \theta)^2\) to simplify the integral: \((1 + \sin \theta)^2 = 1 + 2\sin \theta + \sin^2 \theta\).

Step 5: Integrate each Term Separately

Separate the integral into individual terms: \[ A = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, d\theta + 2\int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta + \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \right) \]

Step 6: Solve Each Integral

Calculate each integral:1. \(\int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}\)2. \(2\int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta = -2\left[ \cos \theta \right]_{0}^{\frac{\pi}{2}} = -2(0 - 1) = 2\)3. \(\int_{0}^{\frac{\pi}{2}} \sin^2 \theta \, d\theta = \frac{\pi}{4}\) (using the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \)).

Step 7: Combine and Simplify the Results

Substitute the results back and simplify: \[ A = \frac{1}{2} \left( \frac{\pi}{2} + 2 + \frac{\pi}{4} \right) = \frac{1}{2} \times \left( \frac{2\pi}{4} + 2 + \frac{\pi}{4} \right) = \frac{1}{2} \times \left( \frac{3\pi}{4} + 2 \right) \]\[ A = \frac{3\pi}{8} + 1 \]

Step 8: Final Step: Calculate Result

Thus, the area of the region in the first quadrant bounded by the cardioid is \(\boxed{\frac{3\pi}{8} + 1}\).

Key Concepts

Polar Coordinates

Polar coordinates are a way to represent points in a plane using a distance and an angle, rather than the traditional x-y coordinates of the Cartesian system. In the polar coordinate system, any point is represented as \( (r, \theta) \), where:

• \( r \) is the radial distance from the origin to the point.
• \( \theta \) is the angle measured from the positive x-axis to the line connecting the origin and the point.

This system is particularly useful when dealing with curves that are functions of angles, such as the cardioid given by \( r = 1 + \sin \theta \). Polar graphs can depict curves that would be complicated to express with Cartesian equations. For instance, a cardioid is a curve that resembles a heart and can succinctly be described with a simple polar equation.

Integral Calculus

Integral calculus involves finding the integral of functions, which basically sums up infinite small quantities to get an accumulated value. This is especially useful in finding areas, volumes, and other quantities that require summation.
Mathematically, the integral of a function \( f(x) \) is denoted \[ \int f(x) \, dx \], where \( dx \) indicates we're adding up infinitesimally small slices of \( f(x) \) along the x-axis.
In the problem at hand, integral calculus helps in calculating the area bounded by the cardioid in the first quadrant. Each part of the area involves integrating the function \( (1 + \sin \theta)^2 \) over the angle \( \theta \). By breaking down this function and integrating term by term, we achieve the final desired area for the region in focus.

First Quadrant

The first quadrant refers to the upper-right section of a Cartesian plane where both x and y coordinates are positive. In the context of polar coordinates, it corresponds to the angular range from \( 0 \) to \( \frac{\pi}{2} \) radians. This means that when considering a polar curve like a cardioid, we focus only on this sector and disregard the rest of the curve that lies outside of it.
To solve the exercise, we consider this specific section defined by the cardioid equation and integrate only over this portion. This restricts our integration limits to \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \), ensuring that only the area of the curve lying within the first quadrant is calculated.

Polar Area Formula

The polar area formula is a specific integral calculus method to find the area enclosed by a polar curve \( r(\theta) \) from \( \theta = a \) to \( \theta = b \). It is given by:\[ A = \frac{1}{2} \int_{a}^{b} r(\theta)^2 \, d\theta \] This formula provides a powerful tool to calculate areas that are difficult to handle with traditional Cartesian methods.
When we apply it, each part of the curve from \( \theta = 0 \) to \( \frac{\pi}{2} \) contributes to the total area. This technique involves squaring the radius function, multiplying by half, and integrating over the desired angle range. For the cardioid in our problem, \( r = 1 + \sin \theta \), substituting this into the polar area formula gives us the area calculation specific to that unique shape and section.