Problem 43 - Calculus Early Transcendentals

Question

In Exercises \(37-40,\) you found the intersection points of pairs of curves. Find the area of the entire region that lies within both of the following pairs of curves. $$r=1+\sin \theta \text { and } r=1+\cos \theta$$

Step-by-Step Solution

Answer

Answer: \(\pi\)

Step 1: Find Intersection Points

To find the intersection points of the given curves \(r=1+\sin\theta\) and \(r=1+\cos\theta\), set both equations equal and solve for \(\theta\). $$1 + \sin\theta = 1 + \cos\theta$$ $$\sin\theta = \cos\theta$$ Since \(\sin\theta = \cos\theta\) when \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{5\pi}{4}\), these are our intersection points.

Step 2: Set Up the Integrals for the Area

Now, we will set up two integrals to find the area enclosed by the curves using the following formula for the area of a region in polar coordinates: $$A = \frac{1}{2}\int_{\alpha}^{\beta}(r_1^2 - r_2^2)d\theta$$ We will split the region into two parts: from \(\theta = \frac{\pi}{4}\) to \(\theta = \frac{5\pi}{4}\), and from \(\theta = \frac{5\pi}{4}\) to \(\theta = \frac{9\pi}{4}\). For the first region, \(r_1 = 1+\sin\theta\) and \(r_2 = 1+\cos\theta\). But we need to find the values for \(\alpha\) and \(\beta\), the starting and ending angles of integration. From the intersection point \(\theta = \frac{\pi}{4}\), the curve \(r_1 = 1+\sin\theta\) is greater than the curve \(r_2 = 1+\cos\theta\). So, for the first region, $$\alpha = \frac{\pi}{4}, \beta = \frac{5\pi}{4}$$ For the second region, \(r_1 = 1+\cos\theta\) and \(r_2 = 1+\sin\theta\). From the intersection point \(\theta = \frac{5\pi}{4}\), the curve \(r_1 = 1+\cos\theta\) is greater than the curve \(r_2 = 1+\sin\theta\). So, for the second region, $$\alpha = \frac{5\pi}{4}, \beta = \frac{9\pi}{4}$$

Step 3: Compute the Area

Now, we will compute the area of both regions by evaluating the integrals and adding them together. Region 1: $$A_1 = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}[(1+\sin\theta)^2 - (1+\cos\theta)^2]d\theta$$ Region 2: $$A_2 = \frac{1}{2}\int_{\frac{5\pi}{4}}^{\frac{9\pi}{4}}[(1+\cos\theta)^2 - (1+\sin\theta)^2]d\theta$$ Now, compute both integrals and add them together: $$A = A_1 + A_2 = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}[(1+\sin\theta)^2 - (1+\cos\theta)^2]d\theta + \frac{1}{2}\int_{\frac{5\pi}{4}}^{\frac{9\pi}{4}}[(1+\cos\theta)^2 - (1+\sin\theta)^2]d\theta$$ $$A = \int_{\frac{\pi}{4}}^{\frac{9\pi}{4}}[(1+\sin\theta)^2 - (1+\cos\theta)^2]d\theta$$ Evaluating the integral yields: $$A = \pi$$ Thus, the area of the entire region that lies within the given pairs of curves is \(\pi\).

Key Concepts

Intersection Points

Finding intersection points of polar curves is a crucial step in understanding how these curves interact. For the given curves, \(r=1+\sin\theta\) and \(r=1+\cos\theta\), intersection points are where the curves coincide in the polar plane. To find them, equate the two expressions: \(1 + \sin\theta = 1 + \cos\theta\). Simplifying gives \(\sin\theta = \cos\theta\). This equation holds true when \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{5\pi}{4}\). These angles represent directions where both curves have the same radius, hence their intersection in polar coordinates.

Area of Polar Regions

Calculating the area of a region enclosed by polar curves involves integrating the difference of their squared radius functions. The area enclosed by polar curves like \(r=1+\sin\theta\) and \(r=1+\cos\theta\) is found using the polar area formula: \[A = \frac{1}{2}\int_{\alpha}^{\beta}(r_1^2 - r_2^2)\,d\theta\]For this problem, we split the area into two parts defined by the intersection points to fully enclose the region.

• First Region: Between \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{5\pi}{4}\), where \(r_1 = 1+\sin\theta\) is the outer curve.
• Second Region: Between \(\theta = \frac{5\pi}{4}\) and \(\theta = \frac{9\pi}{4}\), where \(r_1 = 1+\cos\theta\) is the outer curve.

Compute these integrals to find the total area of \(\pi\), reflecting the space shared by both curves.

Trigonometric Functions

Trigonometric functions like \(\sin\theta\) and \(\cos\theta\) are essential in polar coordinates as they determine the shape and behavior of curves. These functions repeat values at specific intervals, which is utilized in solving for intersection points or calculating areas. In our example, since \(\sin\theta = \cos\theta\), we utilize the periodic nature to identify exact angles of intersection such as \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{5\pi}{4}\). Trigonometric identities and the unit circle can be invaluable tools for visualizing and reasoning about these problems. These functions create the patterns and symmetries we often exploit when analyzing polar curves.