A curve is given by the following parametric equations. x(t)=3cos(t)+cos(3t), y=3sin(t)-sin(3t), 0 is less than or equal to t, is less than or equal to pi. Find the area of the region bounded above by the curve and below by the x axis.

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.

Solve the problem of finding the area bounded by the parametric curve given by x(t)=3cos(t)+cos(3t) and y=3sin(t)-sin(3t) for t in [0, pi]. This solution uses trigonometric identities and integration techniques to calculate the area. Suitable for college-level calculus students.

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