Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the lemniscate \(r^{2}=2 \sin 2 \theta\) and outside the circle \(r=1\)

Short answer

To find the area of the region inside the lemniscate \(r^2 = 2\sin{2\theta}\) and outside the circle \(r = 1\), we first identify the points of intersection as \((\frac{\pi}{12}, 1)\) and \((\frac{5\pi}{12}, 1)\). Then, we subtract the area enclosed by the circle from the area enclosed by the lemniscate between these points of intersection. The final area of the region is \(\sqrt{6} - \frac{\pi}{3}\).

Step-by-step solution

Sketch the lemniscate and the circle

Begin by sketching the graphs of the two functions: 1. Lemniscate: \(r^{2} = 2 \sin 2 \theta\) 2. Circle: \(r = 1\) The lemniscate's equation can be written as \(r = \sqrt{2 \sin 2 \theta}\), this curve is symmetric about the origin and has points of intersection with the polar axes at \((\frac{\pi}{4}, 0)\), \((\frac{3\pi}{4}, 0)\), \((\frac{5\pi}{4}, 0)\), and \((\frac{7\pi}{4}, 0)\). The circle is centered at the origin with radius 1.

Find points of intersection

We need to find the points where the lemniscate and the circle intersect. To do this, set their equations equal to each other: 1. Lemniscate: \(r^{2} = 2 \sin 2 \theta\) 2. Circle: \(r = 1\) Setting \(r^{2} = 1\), we get: $$1 = 2 \sin 2 \theta$$ $$\frac{1}{2} =\sin 2 \theta$$ From here, we find the angles \(\theta\) for which this is true: $$2 \theta = \frac{\pi}{6}, \frac{5\pi}{6}$$ $$\theta = \frac{\pi}{12}, \frac{5\pi}{12}$$ The points of intersection are \((\frac{\pi}{12}, 1)\) and \((\frac{5\pi}{12}, 1)\).

Find the area

Now that we have the points of intersection, we can find the area of the region inside the lemniscate and outside the circle. The area can be found by subtracting the area enclosed by the circle from the area enclosed by the lemniscate between the points of intersection. First, find the area enclosed by the lemniscate between the points of intersection using the formula for polar areas: $$A_{lem} = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} [r(\theta)]^2 d\theta =\frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 2\sin 2\theta d\theta$$ Now, we find the area enclosed by the circle between the points of intersection: $$A_{cir} =\frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} [1]^2 d\theta = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} d\theta$$ The area of the region we are interested in is given by the difference between these two areas: $$A = A_{lem} - A_{cir}$$ Now compute the integrals: $$A_{lem} = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 2\sin 2\theta d\theta =\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \sin 2\theta\ d\theta = -\frac{1}{2}[\cos(2\theta)]\Big|_{\frac{\pi}{12}}^{\frac{5\pi}{12}} = \frac{1}{2}[\frac{\sqrt{6}}{2}] = \frac{\sqrt{6}}{4}$$ $$A_{cir} =\frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} d\theta = \frac{1}{2}[\theta]\Big|_{\frac{\pi}{12}}^{\frac{5\pi}{12}} = \frac{\pi}{6}$$ Now subtract the areas: $$A = A_{lem} - A_{cir} = \frac{\sqrt{6}}{4} - \frac{\pi}{6}$$ Since the region of interest is symmetric with respect to the \(x\)-axis, we need to multiply this area by 2: $$A_{total} = 2A = \sqrt{6} - \frac{\pi}{3}$$ Therefore, the area of the region inside the lemniscate and outside the circle is \(\sqrt{6} - \frac{\pi}{3}\).

Key concepts

Lemniscate

A lemniscate is a fascinating curve that resembles the shape of a figure-eight or the infinity symbol (∞). In polar coordinates, a lemniscate is often expressed in equations like \(r^2 = a^2 \sin 2\theta\) or \(r^2 = a^2 \cos 2\theta\).
For the specific lemniscate in this problem, the equation is given by \(r^2 = 2 \sin 2\theta\).
This curve is symmetric about the origin and intersects the polar axes at specific angles, making it a unique plotting challenge in polar graphs.

• It has a loop on either side of the origin.
• Each loop reaches its farthest point away from the origin along lines at \(\theta = \frac{\pi}{4}\), \(\theta = \frac{3\pi}{4}\), \(\theta = \frac{5\pi}{4}\), and \(\theta = \frac{7\pi}{4}\).

Understanding how these key points form this elegant curve is essential for visualizing lemniscates in polar plots.

Circle

In polar coordinates, a circle centered at the origin with a constant radius can be simply represented as \(r = c\), where \(c\) is the radius.
In the current problem, the circle described by \(r = 1\) is easy to understand: it creates a perfect circle centered at the origin with a radius of 1.

• This circle encompasses all points that are exactly one unit away from the origin regardless of the angle \(\theta\).

Such simplicity of representation makes circles one of the easiest shapes to handle in polar coordinates.

Area Between Curves in Polar Coordinates

The area between two curves in polar coordinates involves subtracting the area of the inner curve from the area of the outer curve.
To solve these problems, the formula for the area enclosed by a polar curve \(r(\theta)\) between two angles \(\theta = a\) and \(\theta = b\) is used: \[ A = \frac{1}{2} \int_{a}^{b} [r(\theta)]^2 \,d\theta\]In this exercise, the area inside the lemniscate but outside the circle is found by:

• Calculating the area under the lemniscate \(r^2 = 2 \sin 2\theta\) between specific angles related to the points of intersection.
• Subtracting the area under the circle \(r = 1\) between the same angles.
• Doubling the result since the region of interest is symmetric about the x-axis.

This approach allows us to determine the precise region of overlap of these curves in a polar plot, providing a clear sense of dimension within their confines.

Intersection of Polar Curves

Finding the intersection points of polar curves is crucial for determining regions enclosed by them.
For intersection, set the equations of the curves equal: this means solving \(r^2 = 2 \sin 2\theta\) and \(r = 1\) together. This leads to:
\[1 = 2 \sin 2\theta\]
From this, angles \(\theta\) that satisfy this equation indicate where both curves cross each other.

• The process involves solving \(\sin 2\theta = \frac{1}{2}\), resulting in angles \(2\theta = \frac{\pi}{6}\) and \(2\theta = \frac{5\pi}{6}\), leading to \(\theta = \frac{\pi}{12}\) and \(\theta = \frac{5\pi}{12}\).
• These angles help pinpoint the polar coordinates of intersection: \((\frac{\pi}{12}, 1)\) and \((\frac{5\pi}{12}, 1)\).

Understanding and calculating such intersections is key for evaluating the specific segments of curves that contribute to the area calculations.