Question

Finding the Area between Two Polar Curves Find the area outside the cardioid \(r=2+2 \sin \theta\) and inside the circle \(r=6 \sin \theta\).

Short answer

The area is calculated by integrating the difference of the polar curves from \(\theta = \frac{\pi}{6}\) to \(\theta = \frac{5\pi}{6}\).

Step-by-step solution

Understand the Problem

We need to find the area that lies outside the cardioid defined by \(r = 2 + 2 \sin \theta\) and inside the circle given by \(r = 6 \sin \theta\). This is a region within the polar coordinate system.

Determine Intersection Points

To find where the curves intersect, set the two equations equal: \(2 + 2 \sin \theta = 6 \sin \theta\).Solving, subtract \(2 \sin \theta\) from both sides:\[ 2 = 4 \sin \theta \]Divide both sides by 4:\[ \sin \theta = \frac{1}{2} \]This corresponds to angles \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\). These are the points where the curves intersect in the first and second quadrants.

Define Area Region

The area we need to find lies between \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\). Within these limits, we calculate the area inside the circle and subtract the area inside the cardioid to find the desired area.

Set Up Area Integrals

The formula for the area \(A\) in polar coordinates is \(A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta\) where \(\alpha\) and \(\beta\) are the bounds.- For the circle: \(A_{\text{circle}} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (6 \sin \theta)^2 \, d\theta\).- For the cardioid: \(A_{\text{cardioid}} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (2 + 2 \sin \theta)^2 \, d\theta\).

Evaluate Integrals for Circle and Cardioid

Evaluate each integral separately:For the circle:\[ A_{\text{circle}} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 36 \sin^2 \theta \, d\theta \]Use \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\) identity:\[ = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 18 (1 - \cos(2\theta)) \, d\theta\]Carry out the integral:\[ = \frac{1}{2} [18 \theta - 9 \sin(2\theta)]\bigg|_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\]For the cardioid:\[ A_{\text{cardioid}} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (4 + 8\sin \theta + 4 \sin^2 \theta) \, d\theta\]Solve for each term (use \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\) similarly) and evaluate the integral in the given bounds.

Calculate the Final Area

Subtract the area of the cardioid from the area of the circle:\[ A_{\text{desired}} = A_{\text{circle}} - A_{\text{cardioid}} \]From the evaluated integrals, plug in the values and solve for the final numerical result. This area represents the region outside the cardioid and inside the circle between the angles \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Key concepts

Area Between Curves

Finding the area between two curves in polar coordinates involves analyzing the region where each polar curve overlaps or encloses space between them. In this instance, we aim to find the space that lies outside one given curve, the cardioid, and inside another curve, the circle.

When working with polar coordinates, regions can take on irregular shapes, making the calculation unique compared to Cartesian coordinates. Instead of using vertical slices, like in rectangular coordinates, we use angular sections measured in radians.

• First, identify the intersection points of the polar curves as these will determine the limits of integration.
• Then, set up integrals to represent the areas under each curve from one intersection point to the other.
• Lastly, the desired area between curves is the difference between these two integrals within the specified limits.

Integrals in Polar Coordinates

In polar coordinates, calculating area involves integrating the square of the radius function. The basic formula used is \[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \ d\theta \] where \( r \) is the function of \( \theta \), and \( \alpha \) and \( \beta \) are the bounds.

This differs from Cartesian integrals, where you integrate over the x-axis typically. Here, your variable is \( \theta \) representing angles in radians, and \( r^2 \) reflects the idea of accumulating "angular slices" of area, fanning out from the origin.

• Evaluate the integral separately for each curve: from the intersection angle \( \alpha \) to \( \beta \) .
• Transform trigonometric identities where necessary, especially if integrating functions like \( \sin^2 \theta \).
• Subtract the integral of the cardioid from the one for the circle to find the area outside the cardioid.

This method provides an elegant way to compute areas that are circular or curved, allowing us to handle more complex forms than in Cartesian systems.

Intersection Points

The intersection points of polar curves are crucial in determining the bounds for integration. They represent where the curves' radii coincide at certain angles \( \theta \).

To find intersection points, set the equations of the two curves equal. In our example:

• Equate the expressions for the radii: \( 2 + 2 \sin \theta = 6 \sin \theta \).
• Solve the equation to find \( \theta \). This will yield angles where the radial distances to the origin are the same for both curves.

We solve, inferring that \( \sin \theta = \frac{1}{2} \), giving us \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). These angles now serve as integration bounds.

Polar Curves

Polar curves describe loci in the plane based on an angle and distance from a central point, the pole. Unlike Cartesian graphs, they rely on \( (r, \theta) \) pairs, where \( r \) is the distance from the origin, and \( \theta \) is the angle from the polar axis.

This problem involves a cardioid and a circle:

• The cardioid \( r = 2 + 2 \sin \theta \) is a heart-shaped curve that touches the pole.
• The circle \( r = 6 \sin \theta \) is a symmetric circular shape centered at the origin.

When dealing with different polar curves, understanding their shapes and how they can interact is critical. The circular path creates constant radial measures for varying \( \theta \), while the cardioid's radius changes with \( \sin \theta \)—affecting the enclosed area between these two when graphed. Understanding these interactions is essential for visualizing the curves and setting up the correct expressions for integration.