To find the area of the region bounded by the parametric curve given by:

$x(t) = 3\cos(t) + \cos(3t), \quad y(t) = 3\sin(t) - \sin(3t),$ where $0 \leq t \leq \pi$, and the area is bounded above by the curve and below by the x-axis, we use the following formula for the area enclosed by a parametric curve:

$A = \int_{t_1}^{t_2} y(t) \cdot x'(t) \, dt.$

Here:

• $x(t) = 3\cos(t) + \cos(3t)$,
• $y(t) = 3\sin(t) - \sin(3t)$,
• The limits of integration are $t_1 = 0$ and $t_2 = \pi$.

Step 1: Compute $x'(t)$

Differentiate $x(t)$ with respect to $t$:

$x'(t) = \frac{d}{dt}[3\cos(t) + \cos(3t)] = -3\sin(t) - 3\sin(3t).$

Step 2: Set up the integral

Now, the area $A$ is given by:

$A = \int_0^{\pi} \left(3\sin(t) - \sin(3t)\right) \cdot \left(-3\sin(t) - 3\sin(3t)\right) dt.$

Step 3: Simplify the expression

Multiply the terms inside the integrand:

$A = \int_0^{\pi} \left[ -9\sin^2(t) - 9\sin(t)\sin(3t) + 3\sin(3t)\sin(t) + 3\sin^2(3t) \right] dt.$

This simplifies to:

$A = \int_0^{\pi} \left[-9\sin^2(t) + 3\sin^2(3t)\right] dt,$ because the terms involving $\sin(t)\sin(3t)$ cancel out.

Step 4: Use trigonometric identities

We now use the trigonometric identity:

$\sin^2(t) = \frac{1}{2}(1 - \cos(2t)).$

Applying this identity to both $\sin^2(t)$ and $\sin^2(3t)$, we get:

$A = \int_0^{\pi} \left[-9 \cdot \frac{1}{2}(1 - \cos(2t)) + 3 \cdot \frac{1}{2}(1 - \cos(6t))\right] dt.$

Simplifying:

$A = \int_0^{\pi} \left[-\frac{9}{2}(1 - \cos(2t)) + \frac{3}{2}(1 - \cos(6t))\right] dt.$

Step 5: Evaluate the integral

Now, we can split the integral into two parts:

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

Evaluating each term:

$\int_0^{\pi} 1 \, dt = \pi, \quad \int_0^{\pi} \cos(2t) \, dt = 0, \quad \int_0^{\pi} \cos(6t) \, dt = 0.$

Thus:

$A = \left(-\frac{9}{2}\right)(\pi) + \frac{3}{2}(\pi) = -\frac{9\pi}{2} + \frac{3\pi}{2} = -\frac{6\pi}{2} = -3\pi.$

Step 6: Take the absolute value

Since the area is always positive:

$A = 3\pi.$

Thus, the area of the region bounded by the curve and the x-axis is $\boxed{3\pi}$.

Would you like to explore more details or have any questions?

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Here are 5 related questions:

1. How would you find the arc length of this parametric curve?
2. What is the geometric interpretation of the parametric curve for different intervals of $t$?
3. How would the area change if the curve were symmetric with respect to the y-axis as well?
4. Can you determine the points where the curve intersects the x-axis?
5. What happens to the area if the bounds of $t$ are extended beyond $\pi$?

Tip: Parametric curves are often best visualized graphically to understand their behavior and symmetries.