Question

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the curve \(r=\sqrt{\cos \theta}\) and outside the circle \(r=1 / \sqrt{2}\)

Short answer

Question: Find the area of the region that is inside the curve \(r = \sqrt{\cos \theta}\) and outside the circle \(r = \frac{1}{\sqrt{2}}\). Answer: The area of the region is \(-\frac{\pi}{2}\) square units.

Step-by-step solution

Identify the region

We need to find the area of the region that is inside the curve \(r = \sqrt{\cos \theta}\) and outside the circle \(r = \frac{1}{\sqrt{2}}\).

Sketch both the curves

First, plot the curve \(r = \sqrt{\cos \theta}\) and the circle \(r = \frac{1}{\sqrt{2}}\) on the same polar graph. Note that \(\theta\) varies from 0 to \(\pi\) for \(r = \sqrt{\cos \theta}\) as it becomes an undefined value past those angles. In the case of the circle, \(r = \frac{1}{\sqrt{2}}\), it is fixed at a distance of \(\frac{1}{\sqrt{2}}\) from the origin, resulting in a circle of radius \(\frac{1}{\sqrt{2}}\).

Identify the point of intersection

To find the point where both curves intersect, we need to find \(\theta\) for which both \(r\) values are equal. So, we have: \(\sqrt{\cos \theta} = \frac{1}{\sqrt{2}}\) Squaring both sides, \(\cos \theta = \frac{1}{2}\) Solving for \(\theta\), we get \(\theta = \frac{\pi}{3}\) or \(\theta = \frac{5\pi}{3}\).

Find the area of the region

Now that we know the points of intersection and the graphic representation, we can find the area of the region inside the curve and outside the circle. We will use the formula for the area in polar coordinates: \(A = \frac{1}{2}\displaystyle\int_\alpha^\beta r^2 d\theta\) To find the area of the region inside the curve \(r = \sqrt{\cos \theta}\), we integrate from \(0\) to \(\pi\): \(A_1 = \frac{1}{2}\displaystyle\int_0^\pi (\displaystyle\cos \theta) d\theta\) To find the area inside the circle \(r = \frac{1}{\sqrt{2}}\), we integrate from \(\theta = \frac{\pi}{3}\) to \(\theta = \frac{5\pi}{3}\): \(A_2 = \frac{1}{2}\displaystyle\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (\frac{1}{2}) d\theta\) Finally, to find the area of the region we're interested in, we take the difference between \(A_1\) and \(A_2\): \(A = A_1 - A_2\) Calculate and simplify the integrals to get the final area: \(A_1 = \frac{1}{2}\displaystyle\int_0^\pi (\displaystyle\cos \theta) d\theta = \frac{1}{2}[\sin(\pi) - \sin(0)] = 0\) \(A_2 = \frac{1}{2}\displaystyle\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (\frac{1}{2}) d\theta = \frac{1}{4}[\theta]_{\frac{\pi}{3}}^{\frac{5\pi}{3}} = \frac{1}{4}(\frac{5\pi}{3} - \frac{\pi}{3}) = \frac{\pi}{2}\) So, the area we're interested in is: \(A = A_1 - A_2 = 0 - \frac{\pi}{2} = - \frac{\pi}{2} (\in\ square\ units)\)

Key concepts

Area Calculation

Calculating the area of regions bounded by curves in polar coordinates involves understanding how polar equations work. In polar coordinates, a point is identified by its distance from the origin, denoted by \( r \), and the angle \( \theta \) it makes with the positive x-axis.
For our problem, we are tasked with finding the area of a region inside the curve \( r = \sqrt{\cos \theta} \) and outside the unit circle \( r = \frac{1}{\sqrt{2}} \).

Using Polar Area Formula

The area \( A \) of a region defined by polar curves from \( \theta = \alpha \) to \( \theta = \beta \) can be determined using the formula:\[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta\]To calculate the area effectively:

• The first integral \( A_1 \) calculates the area inside the curve \( r = \sqrt{\cos \theta} \) from \( \theta = 0 \) to \( \theta = \pi \).
• The second integral \( A_2 \) finds the area inside the circle \( r = \frac{1}{\sqrt{2}} \) from \( \theta = \frac{\pi}{3} \) to \( \theta = \frac{5\pi}{3} \).

Subtracting \( A_2 \) from \( A_1 \) gives the desired area. Understanding how these integrations work clarifies why polar area calculations can result in negative values, reminding us to carefully check our setup and limits.

Curve Sketching

Sketching polar curves gives a geometric insight into the problem and usually helps to visualize the intersection and region of interest vividly.
When sketching the curve \( r = \sqrt{\cos \theta} \), notice that it has some unique properties. It only remains valid for \( \theta \) values where \( \cos \theta \) is non-negative, specifically between \( 0 \) and \( \pi \).

Visualizing the Curves

• The curve \( r = \sqrt{\cos \theta} \) forms a shape that looks like a bean, lying on the positive x-axis with a gradual decrease to the origin as \( \theta \) approaches \( \pi \).
• The circle \( r = \frac{1}{\sqrt{2}} \) remains as a constant-distance circle centered at the origin.
  

By overlaying these two curves, you can better understand where they intersect and determine the bounded regions required for area calculations. This intersection marks the section of the curve inside which we calculate our area in polar coordinates.

Intersection Points

Intersection points of curves in polar coordinates play a critical role in determining limits for integration. These intersection points are where the two curves \( r = \sqrt{\cos(\theta)} \) and \( r = \frac{1}{\sqrt{2}} \) meet, and they define the main boundaries of our integration.
Finding intersection points involves solving:
\[ \sqrt{\cos \theta} = \frac{1}{\sqrt{2}}\]
By squaring both sides, we get:\[\cos \theta = \frac{1}{2}\]Then, solving for \( \theta \), we derive critical angle points, \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \).

Significance of Intersections

• These points, \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \), represent where the curves intersect, providing bounds for the necessary region of integration when computing the area outside the circle but inside the curve.
• These intersections help to establish the integration limits for \( A_2 \), confirming the corridor in the curve that is conditioned by the circle's reach.
  

Correctly identifying these points ensures that the area calculation is confined exactly to the desired regions.