Problem 6 - Thomas Calculus

Question

Find the areas of the regions Inside the six-leaved rose \(r^{2}=2 \sin 3 \theta\)

Step-by-Step Solution

Answer

The total area of the six-leaved rose is 2 square units.

Step 1: Understand the Polar Equation

The given equation is \( r^2 = 2 \sin 3\theta \). This equation describes a six-leaved rose in polar coordinates. The key is to find the area of one petal and then multiply it by the number of petals, which in this case is six.

Step 2: Set up the Integral for One Petal

One petal of this rose corresponds to a sector where \( \theta \) ranges from 0 to \( \pi/3 \), as \( \sin 3\theta \) represents one full sine wave cycle in the interval \(0 \leq \theta < 2\pi/3\), completing two petals. To find the area of one petal, integrate from 0 to \(\pi/3\).

Step 3: Convert to Area Formula

The formula for the area of a region defined by polar coordinates is \( A = \frac{1}{2} \int (r^2) \, d\theta \). For one petal, this becomes \( A = \frac{1}{2} \int_{0}^{\pi/3} (2\sin 3\theta) \, d\theta \).

Step 4: Evaluate the Integral

Evaluate the integral \( \frac{1}{2} \cdot 2 \cdot \int_{0}^{\pi/3} \sin 3\theta \, d\theta \). Simplify to \( \int_{0}^{\pi/3} \sin 3\theta \, d\theta = -\frac{1}{3} \cos 3\theta \bigg|_0^{\pi/3} \). Compute this as \( \left[-\frac{1}{3} \cos (\pi) + \frac{1}{3} \cos (0)\right] = \left[\frac{1}{3}(-(-1) + 1)\right] = \frac{2}{3} \).

Step 5: Calculate Total Area

Since we have found the area of one petal to be \( \frac{1}{3} \), multiply this by 6 for the total area of the rose: \(6 \times \frac{1}{3} = 2\).

Step 6: Conclusion: Final Area

The area of the entire six-leaved rose is 2 square units.

Key Concepts

Area of Polar Regions

In polar coordinates, the area of regions like those of a six-leaved rose can be a fascinating concept to explore. This involves understanding how polar equations describe curves and calculating the corresponding areas. Polar regions are defined by equations with radius (\( r \)), a function of angle (\( θ \)). For example, the six-leaved rose given by \( r^2 = 2 \sin 3θ \) describes a symmetrical figure with six identical petals.

To compute the area of a polar region, especially one that forms a pattern like the petals of a rose, you often calculate the area of one petal and multiply it by the number of petals to get the total area. This involves setting an interval for \(θ\) that covers one petal. For our six-leaved rose, one petal is formed as \( θ \) ranges from 0 to \( π/3 \). Each petal occupies a specific sector of the plane centered around the origin. Therefore, integrating within this interval gives the area of one petal.

Integral Calculus

Integral calculus is the branch of mathematics that deals with the concept of integration as a tool for finding areas under curves and surfaces. In solving problems involving polar coordinates, like our six-leaved rose, we use definite integrals to calculate areas.

The polar area formula is derived from integral calculus and given by:

• \( A = \frac{1}{2} \int (r^2) \, dθ \)

This formula is instrumental in calculating the area of regions bounded by polar curves.

In our example, the exact setup of the integral becomes vital. For the equation \( r^2 = 2 \sin 3θ \), converting it into an integral involves calculating:

• \( A = \frac{1}{2} \int_{0}^{π/3} (2 \sin 3θ) \, dθ \)

Here, we integrate with respect to \(θ\) over the interval corresponding to one petal. This precise setup of limits ensures that we correctly capture the area for one petal and thus, the total area when multiplied by six.

Trigonometric Integrals

Trigonometric integrals appear in our calculations when we solve for the area of polar regions like a six-leaved rose. These integrals involve trigonometric functions such as sine and cosine, requiring specific integration techniques.

In our case, we need to solve the integral:

• \( \int_{0}^{π/3} \sin 3θ \, dθ \)

This requires finding the antiderivative of \( \sin 3θ \), which can be tackled using basic integral rules related to trigonometric functions. To solve this, recognize that:

• The antiderivative of \( \sin 3θ \) is: \(-\frac{1}{3} \cos 3θ \)

Evaluating this expression between the limits \(0\) and \(π/3\) involves substituting these values into \( \cos 3θ \) and finding their difference:

• \( \left[-\frac{1}{3} \cos(π) + \frac{1}{3} \cos(0)\right] = \frac{2}{3} \)

Through these steps, the integral provides us the area of one petal, which after multiplying by the number of petals yields the desired area of the whole rose.