To solve the problem involving the curve $y = 3\sin(x) - 3x\cos(x)$ over the interval $0 \leq x \leq \pi$, we need to complete the following steps:

(a) Set up an integral for the length of the curve

The formula for the arc length $L$ of a curve $y = f(x)$ over the interval $a \leq x \leq b$ is given by:

$L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$

First, we need to compute the derivative $\frac{dy}{dx}$ of the given function $y = 3\sin(x) - 3x\cos(x)$.

$\frac{dy}{dx} = 3\cos(x) - 3\cos(x) + 3x\sin(x)$

Simplifying:

$\frac{dy}{dx} = 3x\sin(x)$

Now, square the derivative:

$\left( \frac{dy}{dx} \right)^2 = (3x\sin(x))^2 = 9x^2\sin^2(x)$

Thus, the arc length integral becomes:

$L = \int_0^\pi \sqrt{1 + 9x^2\sin^2(x)} \, dx$

(b) Graph the curve to see what it looks like

I will plot the graph of the curve $y = 3\sin(x) - 3x\cos(x)$ over the interval $[0, \pi]$ to visualize the shape.

(c) Use a grapher's or computer's integral evaluator to find the curve's length numerically

To calculate the length numerically, I'll compute the integral $\int_0^\pi \sqrt{1 + 9x^2\sin^2(x)} \, dx$. Let me do this calculation now.The numerical value of the curve's length, calculated using the integral $\int_0^\pi \sqrt{1 + 9x^2\sin^2(x)} \, dx$, is approximately $10.32$ units.

Would you like a detailed explanation of any of the steps? Here are five related questions you might consider:

1. How is the formula for the arc length of a curve derived?
2. Can you explain why we need the derivative of $y$ to compute the arc length?
3. How can the arc length formula be applied to curves in parametric form?
4. What are common methods for numerically solving integrals like this?
5. How would the arc length formula change for a curve in polar coordinates?

Tip: When calculating arc lengths, always check if the function allows for simplifications in the integrand to make the integral easier to evaluate.