Answer:
To find the area under one arch of the cycloid defined by the parametric equations \(x = t - \sin(t)\) and \(y = 1 - \cos(t)\), we can use the formula:
\[A = \frac{1}{2} \oint_C (x \, dy - y \, dx)\]
Let's calculate this step by step:
1. Find \(dx\) and \(dy\):
\[dx = (1 - \cos(t)) \, dt\]
\[dy = (-\sin(t)) \, dt\]
2. Plug these values into the formula:
\[A = \frac{1}{2} \oint_C [(t - \sin(t))( -\sin(t) \, dt) - (1 - \cos(t))((1 - \cos(t)) \, dt)]\]
3. Simplify the expression:
\[A = \frac{1}{2} \int_0^{2\pi} [(-t\sin(t) + \sin^2(t) - (1 - \cos(t))^2) \, dt]\]
4. Expand and integrate:
\[A = \frac{1}{2} \int_0^{2\pi} [-t\sin(t) + \sin^2(t) - (1 - 2\cos(t) + \cos^2(t)) \, dt]\]
5. Now, integrate each term separately:
\[A = \frac{1}{2} \left[-\int_0^{2\pi} t\sin(t) \, dt + \int_0^{2\pi} \sin^2(t) \, dt - \int_0^{2\pi} (1 - 2\cos(t) + \cos^2(t)) \, dt\right]\]
6. Evaluate each integral:
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - \int_0^{2\pi} (1 - 2\cos(t) + \cos^2(t)) \, dt\right]\]
7. Now, we need to evaluate the last integral:
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - \int_0^{2\pi} (1 - 2\cos(t) + \cos^2(t)) \, dt\right]\]
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - \int_0^{2\pi} (1 - 2\cos(t) + \frac{1}{2}(1 + \cos(2t))) \, dt\right]\]
8. Continue to simplify and integrate:
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - \left(t - 2\sin(t) + \frac{1}{4}\sin(2t)\right)\Big|_0^{2\pi}\right]\]
9. Evaluate the definite integrals at the upper and lower limits:
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - (2\pi - 2\sin(2\pi) + \frac{1}{4}\sin(4\pi) - 0)\right]\]
10. Simplify further:
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - (2\pi - 0 + 0)\right]\]
11. Calculate the final result:
\[A = \frac{1}{2} \left[-2\pi + \frac{\pi}{2} - 2\pi\right] = \frac{1}{2} \left[-3\pi + \frac{\pi}{2}\right] = -\frac{5\pi}{4}\]
So, the area under one arch of the cycloid is \(-\frac{5\pi}{4}\) square units. Note that the negative sign indicates that the area is below the x-axis.
Explanation: