Question

Find the area of the regions bounded by the following curves. The complete three-leaf rose \(r=2 \cos 3 \theta\)

Short answer

Answer: The area of the regions bounded by the curve \(r = 2 \cos 3\theta\) is \(2\pi\) square units.

Step-by-step solution

Identify the range of \(\theta\) for one leaf of the rose

Observe that when \(\theta=0\), the equation \(r = 2 \cos 3 \theta\) describes one petal as it starts at the origin. As we increase \(\theta\), the function traces out one petal until it returns back to the origin. To find the range of \(\theta\) for which the curve traces out one petal, find when the equation becomes \(0\) (i.e., when it returns back to the origin). \(r=2\cos 3\theta=0\) \(\cos 3\theta = 0\) \(3\theta = \pm \frac{\pi}{2}\), since the cosine of \(\frac{\pi}{2}\) and \(-\frac{\pi}{2}\) is zero. One leaf covers the region between \(\theta = -\frac{\pi}{6}\) and \(\theta = \frac{\pi}{6}\). Thus, one leaf of the rose covers the range of \(\theta\) from \(-\frac{\pi}{6}\) to \(\frac{\pi}{6}\).

Set up the integral

Since we now know the range of \(\theta\) for which the curve explains one petal, we can set up the double integral to find the area. The area of the region bounded by the curve is given by the integral: \(A = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (2 \cos 3\theta)^2 d\theta\)

Simplify and solve the integral

Now we can evaluate the integral to find the area of one petal of the rose. \(A = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (2 \cos 3\theta)^2 d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (4\cos^2 3\theta) d\theta\) Using the double-angle formula, we know that \(\cos^2x = \frac{1+\cos 2x}{2}\) So, \(A = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (4 \frac{1+\cos 6\theta}{2}) d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (2+2\cos 6\theta) d\theta\) \(A = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 2 d\theta + \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 2 \cos 6\theta d\theta\) We can now solve the two remaining integrals: \(A = \left[ 2\theta \right]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} + \left[ \frac{1}{3} \sin 6\theta \right]_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\) \(A = 2 \left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right) + \frac{1}{3}(\sin \pi - \sin(-\pi))\) \(A = \frac{2\pi}{3}\) The area of one leaf of the rose is \(\frac{2\pi}{3}\) square units. Since there are three leaves in the rose, we need to multiply this value by \(3\) to find the area enclosed by the entire curve. \(A_{total} = 3 \cdot \frac{2\pi}{3}\) \(A_{total} = 2\pi\) The area of the regions bounded by the curve \(r = 2 \cos 3\theta\) is \(2\pi\) square units.

Key concepts

Area of Regions

To find the area enclosed by a curve in polar coordinates, we focus on the region bounded by the polar curve. With polar coordinates, calculating the area isn't quite the same as in Cartesian coordinates. This is because area in polar coordinates is found using an integral that accounts for the circular nature of the system.

For the petals of a polar rose like the one from the exercise, defined by a polar equation like \(r = 2 \cos 3\theta\), the task is to determine the space covered as the angle \(\theta\) changes. We look for the exact range of \(\theta\), where the curve completes one leaf of the rose before repeating. Once this is determined, the area of one complete leaf can be calculated by setting up the appropriate integral over this range.

The general formula for the area \(A\) of a region in polar coordinates is given by \( A = \frac{1}{2} \int_\alpha^\beta r^2 \,d\theta \), where \(\alpha\) and \(\beta\) are the limits of \(\theta\). For the complete curve, we multiply the result for one leaf by the total number of leaves.

Polar Equations

Polar equations describe curves using a radius \(r\) and an angle \(\theta\), forming a polar coordinate system. Unlike Cartesian coordinates, which use the x-y grid system, polar coordinates position a point based on its angle and distance from the origin.

In the exercise, the polar equation used is \(r = 2 \cos 3\theta\). This particular equation means as \(\theta\) varies, \(r\) changes in a way that traces a three-leave rose. The rose pattern emerges due to the oscillation of the \(\cos\) function and the multiplication by \(3\) inside the cosine function, which affects the frequency and number of petals. Such sinusoidal polar equations are commonly used to describe roses or star-like shapes in polar graphs.

Understanding how to interpret and manipulate polar equations is crucial for visualizing and determining the layout of these interesting curves.

Integral Calculus

Integral calculus is the branch of mathematics focused on the accumulation of quantities and the areas under curves. When dealing with polar coordinates, integral calculus helps us find areas that aren't simply rectangular or based on straight edges.

In the context of the exercise, integral calculus is used to compute the area enclosed by the curve \(r = 2 \cos 3\theta\). By converting the polar equation into an integral form, \(A = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (2 \cos 3\theta)^2 \,d\theta\), we are able to calculate the area of one leaf of the rose. This formula finds its basis in the way areas under curves can be summed up by integration, a core principle of integral calculus.

• The factor of \(\frac{1}{2}\) accounts for the radial nature of the area sweep.
• The integral bounds determine the specific section of the curve.
• Through integration, all the infinitesimally small radial areas are aggregated to find the whole area of the shape.

Breaking down the integral by substituting values and applying trigonometric identities results in an accurate area calculation for complex shapes like the polar rose.