To calculate the surface area generated when the cycloid $\{x = a(t - \sin(t)), y = a(1 - \cos(t))\}$ (where $0 \leq t \leq 2\pi$) is revolved about the $x$-axis, we use the formula for the surface area of a curve revolved about the $x$-axis:

$A = 2\pi \int_{a}^{b} y \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \, dt$

Here, $y = a(1 - \cos(t))$, and $x = a(t - \sin(t))$. The interval for $t$ is from $0$ to $2\pi$.

First, we need to find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

$\frac{dx}{dt} = a(1 - \cos(t))$ $\frac{dy}{dt} = a\sin(t)$

Next, we compute the integrand:

[ \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } = \sqrt{ \left[ a(1 - \cos(t)) \right]^2 + \left[ a\sin(t) \right]^2 } ]

Simplify inside the square root:

$\sqrt{a^2 (1 - \cos(t))^2 + a^2 \sin^2(t)} = a \sqrt{ (1 - \cos(t))^2 + \sin^2(t) }$

Using the Pythagorean identity, $(1 - \cos(t))^2 + \sin^2(t) = 2(1 - \cos(t))$:

$a \sqrt{ 2(1 - \cos(t)) } = a \sqrt{ 2(1 - \cos(t)) }$

Therefore, the integrand simplifies to:

$a \sqrt{ 2(1 - \cos(t)) }$

The surface area integral becomes:

$A = 2\pi \int_{0}^{2\pi} a(1 - \cos(t)) \cdot a \sqrt{2(1 - \cos(t))} \, dt$

$A = 2\pi a^2 \int_{0}^{2\pi} (1 - \cos(t)) \sqrt{2(1 - \cos(t))} \, dt$

Let $u = 1 - \cos(t)$, then $du = \sin(t) \, dt$, and when $t$ ranges from $0$ to $2\pi$, $u$ ranges from $0$ to $2$. Also, note that $\sin^2(t) = 1 - \cos^2(t)$:

$A = 2\pi a^2 \int_{0}^{2\pi} (1 - \cos(t)) \sqrt{2(1 - \cos(t))} \, dt$

Changing the variable:

$= 2\pi a^2 \int_{0}^{2} u \sqrt{2u} \, \frac{du}{\sin(t)}$

Now, integrating by simplifying the integral:

$= 2\pi a^2 \int_{0}^{2} \sqrt{2} u^{3/2} \, du$

$= 2\pi a^2 \sqrt{2} \int_{0}^{2} u^{3/2} \, du$

$= 2\pi a^2 \sqrt{2} \left[ \frac{2}{5} u^{5/2} \right]_{0}^{2}$

Evaluating the integral at the bounds:

$= 2\pi a^2 \sqrt{2} \left[ \frac{2}{5} (2)^{5/2} - 0 \right]$

$= 2\pi a^2 \sqrt{2} \left[ \frac{2}{5} \cdot 4\sqrt{2} \right]$

$= 2\pi a^2 \sqrt{2} \left[ \frac{8\sqrt{2}}{5} \right]$

$= 2\pi a^2 \frac{8 \cdot 2}{5}$

$= 2\pi a^2 \frac{16}{5}$

$= \frac{32\pi a^2}{5}$

Therefore, the surface area $A$ generated when the cycloid is revolved about the $x$-axis is:

$A = \frac{32\pi a^2}{5}$

Do you have any questions or need further details?

Additional Questions

1. What is the significance of the cycloid in mathematics?
2. Can you derive the length of one arch of the cycloid?
3. How does the integral change if we revolve the cycloid about the $y$-axis?
4. What happens to the surface area if $a$ is doubled?
5. How can this method be generalized for other parametric curves?
6. Can we use numerical integration methods to approximate the surface area for more complex curves?
7. What are some real-world applications of cycloid curves?
8. How does the surface area change if the curve is revolved around a different axis?

Tip

When working with parametric curves, always check for possible simplifications using trigonometric identities or substitutions to make the integrals more manageable.