Problem 4 - Advanced Engineering Mathematics

Question

In Problems, use a double integral in polar coordinates to find the area of the region bounded by the graphs of the given polar equations. $$ r \quad 8 \sin 4 \theta, \text { one petal } $$

Step-by-Step Solution

Answer

The area of one petal is \( 4\pi \).

Step 1: Identify the range for \( \theta \)

The given polar equation is \( r = 8 \sin 4\theta \). Each petal of \( r = a \sin n\theta \) (where \( a > 0 \) and \( n \) is even) extends from 0 to \( \frac{\pi}{n} \). For \( n = 4 \), the range for one petal is \( 0 \leq \theta \leq \frac{\pi}{4} \).

Step 2: Set up the double integral

The area \( A \) in polar coordinates is given by the integral \( \int_{\alpha}^{\beta} \int_{0}^{r(\theta)} r \, dr \, d\theta \). For this case, \( \alpha = 0 \), \( \beta = \frac{\pi}{4} \), and the upper bound for \( r \) is \( 8 \sin 4\theta \). Thus, \( A = \int_{0}^{\frac{\pi}{4}} \int_{0}^{8 \sin 4\theta} r \, dr \, d\theta \).

Step 3: Evaluate the inner integral

Integrate \( r \) with respect to \( r \) from 0 to \( 8 \sin 4\theta \):\[\int_{0}^{8 \sin 4\theta} r \, dr = \left[ \frac{1}{2}r^2 \right]_{0}^{8 \sin 4\theta} = \frac{1}{2} (8 \sin 4\theta)^2 = 32 \sin^2 4\theta.\]

Step 4: Evaluate the outer integral

Substitute \( 32 \sin^2 4\theta \) into the outer integral:\[A = \int_{0}^{\frac{\pi}{4}} 32 \sin^2 4\theta \, d\theta.\]Use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) to make this integral solvable:\[32 \int_{0}^{\frac{\pi}{4}} \sin^2 4\theta \, d\theta = 32 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(8\theta)}{2} \, d\theta = 16 \int_{0}^{\frac{\pi}{4}} (1 - \cos(8\theta)) \, d\theta.\]

Step 5: Compute the definite integral

Evaluate the definite integral:\[16 \left[ \theta - \frac{1}{8} \sin(8\theta) \right]_{0}^{\frac{\pi}{4}} = 16 \left[ \frac{\pi}{4} - 0 \right] = 4\pi.\]The sine term evaluates to zero at both bounds due to periodicity, simplifying the calculation.

Key Concepts

Polar Equations

Polar equations describe curves in the polar coordinate system, where each point on the plane is specified by a distance from the origin (denoted as \( r \)) and an angle (denoted as \( \theta \)) from the positive x-axis. In our exercise, the polar equation is \( r = 8 \sin 4\theta \), which represents a rose curve. Rose curves have petal-like shapes, and the number of petals depends on the coefficient of \( \theta \). In a function \( r = a \sin n\theta \), if \( n \) is even, the curve has \( 2n \) petals; if \( n \) is odd, it has \( n \) petals. Therefore, for \( n = 4 \), there are 8 petals.Understanding the range of \( \theta \) for one petal is crucial when setting boundaries for integration. By analyzing the trigonometric component, we determine that each petal is drawn as \( \theta \) varies from 0 to \( \frac{\pi}{4} \). This insight is essential for solving problems involving the calculation of areas within polar coordinate systems.

Area of a Region

Finding the area of a region within a polar curve involves utilizing double integrals. The double integral in polar coordinates is set up as \( \int_{\alpha}^{\beta} \int_{0}^{r(\theta)} r \, dr \, d\theta \). This formula accounts for both radius \( r \) and angle \( \theta \), covering the entire area under the curve defined by the polar equation.In our particular case, \( \alpha = 0 \) and \( \beta = \frac{\pi}{4} \) are the limits for \( \theta \), corresponding to one petal of the rose curve. The inner integral \( \int_{0}^{8 \sin 4\theta} r \, dr \) calculates the area of thin radial strips emanating from the origin, while the outer integral adds up these strips over the range of \( \theta \). This careful combination results in the expression that captures the total area of one petal.

Trigonometric Integration

Solving the integral \( \int_{0}^{\frac{\pi}{4}} 32 \sin^2 4\theta \ d\theta \) requires understanding trigonometric identities. Directly integrating \( \sin^2 x \) can be tricky, but using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) simplifies the process. Substituting this identity transforms our problem into integrating more manageable components: constant terms and \( \cos \) functions.After the substitution, we have \( 16 \int_{0}^{\frac{\pi}{4}} (1 - \cos(8\theta)) \, d\theta \). The integration now becomes more straightforward: the integral of 1 with respect to \( \theta \) is simply \( \theta \), and the integral of \( \cos(8\theta) \) involves a \( \sin(8\theta) \) term, adjusted by a constant factor \( \frac{1}{8} \). Finally, evaluating this expression from 0 to \( \frac{\pi}{4} \), we find that the area evaluates to \( 4\pi \), marking the enclosed space covered by one petal of the rose in the exercise.