To find the area of the region enclosed by one loop of the polar curve $r = \sin(4\theta)$, we use the formula for the area enclosed by a polar curve:

$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta$

Step 1: Analyze the curve

The polar equation $r = \sin(4\theta)$ completes one loop when $\sin(4\theta)$ goes from 0 to 1 and back to 0. The period of $\sin(4\theta)$ is $\frac{\pi}{4}$, so each loop spans $\theta$ values from $0$ to $\frac{\pi}{4}$.

Step 2: Square $r$

$r^2 = (\sin(4\theta))^2$

Step 3: Set up the integral

The area enclosed by one loop is: $A = \frac{1}{2} \int_0^{\frac{\pi}{4}} (\sin(4\theta))^2 \, d\theta$

Step 4: Simplify the integral using a trigonometric identity

The identity $(\sin x)^2 = \frac{1 - \cos(2x)}{2}$ gives: $(\sin(4\theta))^2 = \frac{1 - \cos(8\theta)}{2}$ Thus, the integral becomes: $A = \frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{1 - \cos(8\theta)}{2} \, d\theta$ $A = \frac{1}{4} \int_0^{\frac{\pi}{4}} (1 - \cos(8\theta)) \, d\theta$

Step 5: Compute the integral

Split the integral: $A = \frac{1}{4} \left[ \int_0^{\frac{\pi}{4}} 1 \, d\theta - \int_0^{\frac{\pi}{4}} \cos(8\theta) \, d\theta \right]$

1. The first term: $\int_0^{\frac{\pi}{4}} 1 \, d\theta = \left[ \theta \right]_0^{\frac{\pi}{4}} = \frac{\pi}{4}$

2. The second term: $\int_0^{\frac{\pi}{4}} \cos(8\theta) \, d\theta = \frac{1}{8} \sin(8\theta) \Big|_0^{\frac{\pi}{4}} = \frac{1}{8} (\sin(2\pi) - \sin(0)) = 0$

Thus: $A = \frac{1}{4} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{16}$

Final Answer:

The area enclosed by one loop of the curve $r = \sin(4\theta)$ is: $\boxed{\frac{\pi}{16}}$

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Related Questions:

1. How do you calculate the area for more loops of $r = \sin(4\theta)$?
2. How does the formula for polar areas differ from Cartesian areas?
3. What if $r = \cos(4\theta)$ was the polar curve—how does this change the area calculation?
4. Can you explain why the integral of $\cos(8\theta)$ over a full loop is zero?
5. What is the total area enclosed by the entire curve $r = \sin(4\theta)$?

Tip:

In polar coordinates, symmetry can simplify calculations significantly; analyze the symmetry of the curve before integrating.