Use a double integral to find the area of the region. one loop of the rose $r = 3 \cos(3\theta)$

Neetesh Kumar | November 28, 2024

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This is the solution to Math 1D
Assignment: 15.3 Question Number 7
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Step-by-step solution:

Step 1: Understanding the Problem

The equation $r = 3 \cos(3\theta)$ describes a rose curve with $3$ petals (since the number of petals is given by the absolute value of the coefficient of $\theta$ in the cosine function). To find the area of one loop (one petal), we need to compute the area enclosed by the curve for one full cycle of the curve, which corresponds to one petal.

The region enclosed by the curve can be computed using a double integral in polar coordinates. In polar coordinates, the area of a region is given by:

$A = \int_{\theta_1}^{\theta_2} \int_{r=0}^{r(\theta)} r \, dr \, d\theta.$

Here, $r = 3 \cos(3\theta)$ defines the boundary of the region, and $\theta$ will range from $0$ to $\frac{\pi}{3}$ to cover one petal (since the rose curve has $3$ petals, and one petal corresponds to $\frac{2\pi}{3}$ radians, but due to symmetry, we calculate one petal from $0$ to $\frac{\pi}{3}$).

Step 2: Set up the Integral

We set up the double integral for the area of one loop of the rose curve:

$A = \int_0^{\frac{\pi}{3}} \int_0^{3 \cos(3\theta)} r \, dr \, d\theta.$

Step 3: Evaluate the Inner Integral

We first evaluate the inner integral with respect to $r$:

$\int_0^{3 \cos(3\theta)} r \, dr = \frac{r^2}{2} \Big|_0^{3 \cos(3\theta)} = \frac{(3 \cos(3\theta))^2}{2} = \frac{9 \cos^2(3\theta)}{2}.$

Thus, the integral becomes:

$A = \int_0^{\frac{\pi}{3}} \frac{9 \cos^2(3\theta)}{2} \, d\theta.$

Step 4: Use a Trigonometric Identity

To simplify the integral, we use the double angle identity for cosine:

$\cos^2 x = \frac{1 + \cos(2x)}{2}.$

Applying this identity to $\cos^2(3\theta)$, we get:

$\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}.$

Thus, the integral becomes:

$A = \int_0^{\frac{\pi}{3}} \frac{9}{2} \left( \frac{1 + \cos(6\theta)}{2} \right) \, d\theta.$

Simplify the constants:

$A = \frac{9}{4} \int_0^{\frac{\pi}{3}} \left( 1 + \cos(6\theta) \right) \, d\theta.$

Step 5: Evaluate the Integral

Now, evaluate the integral term by term:

1. The integral of $1$ with respect to $\theta$ is:

$\int_0^{\frac{\pi}{3}} 1 \, d\theta = \frac{\pi}{3}.$

2. The integral of $\cos(6\theta)$ with respect to $\theta$ is:

$\int_0^{\frac{\pi}{3}} \cos(6\theta) \, d\theta = \frac{\sin(6\theta)}{6} \Big|_0^{\frac{\pi}{3}} = \frac{\sin(2\pi)}{6} - \frac{\sin(0)}{6} = 0.$

Thus, the integral simplifies to:

$A = \frac{9}{4} \left( \frac{\pi}{3} + 0 \right) = \frac{9\pi}{12} = \frac{3\pi}{4}.$

Final Answer:

The area enclosed by one loop of the rose curve $r = 3 \cos(3\theta)$ is: $\boxed{\frac{3\pi}{4}}$

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