Let's solve both parts step by step: --- ## (a) **Length of One Arch of the Cycloid** Given parametric equations: \[ x = \theta - \sin\theta, \quad y = 1 - \cos\theta \] for \(0 \leq \theta \leq 2\pi\). The length \(L\) of a curve in parametric form is: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} d\theta \] Find the derivatives: \[ \frac{dx}{d\theta} = 1 - \cos\theta \] \[ \frac{dy}{d\theta} = \sin\theta \] Plug into the length formula: \[ L = \int_{0}^{2\pi} \sqrt{(1 - \cos\theta)^2 + (\sin\theta)^2} d\theta \] Expand and simplify inside the square root: \[ (1 - \cos\theta)^2 + \sin^2\theta = 1 - 2\cos\theta + \cos^2\theta + \sin^2\theta \] But \(\cos^2\theta + \sin^2\theta = 1\), so: \[ = 1 - 2\cos\theta + 1 = 2 - 2\cos\theta \] \[ = 2(1 - \cos\theta) \] So, \[ L = \int_{0}^{2\pi} \sqrt{2(1 - \cos\theta)} d\theta \] Using the trigonometric identity: \[ 1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right) \] Therefore, \[ \sqrt{2(1 - \cos\theta)} = \sqrt{2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} = 2|\sin\left(\frac{\theta}{2}\right)| \] For \(0 \leq \theta \leq 2\pi\), \(\sin\left(\frac{\theta}{2}\right)\) is always non-negative, so: \[ L = \int_{0}^{2\pi} 2\sin\left(\frac{\theta}{2}\right) d\theta \] Let \(u = \frac{\theta}{2} \implies d\theta = 2 du\) When \(\theta = 0, u = 0\), when \(\theta = 2\pi, u = \pi\) So, \[ L = \int_{u=0}^{u=\pi} 2\sin(u) \cdot 2 du = 4 \int_{0}^{\pi} \sin u \, du \] \[ = 4 [-\cos u]_{0}^{\pi} \] \[ = 4 \left( -\cos\pi + \cos 0 \right) \] \[ = 4 \left( -(-1) + 1 \right) \] \[ = 4 (1 + 1) = 8 \] **Final Answer for (a):** \[ \boxed{8} \] --- ## (b) **Area of Surface Generated by Rotating the Arch Around the x-Axis** The formula for the surface area \(S\) of a surface of revolution about the x-axis (for parametric equations) is: \[ S = 2\pi \int_{a}^{b} y \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} d\theta \] We've already computed \(\sqrt{(dx/d\theta)^2 + (dy/d\theta)^2} = 2\sin\left(\frac{\theta}{2}\right)\). So, \[ S = 2\pi \int_{0}^{2\pi} (1 - \cos\theta) \cdot 2\sin\left(\frac{\theta}{2}\right) d\theta \] \[ = 4\pi \int_{0}^{2\pi} (1 - \cos\theta)\sin\left(\frac{\theta}{2}\right) d\theta \] Expand the integrand: \[ = 4\pi \left[ \int_{0}^{2\pi} \sin\left(\frac{\theta}{2}\right) d\theta - \int_{0}^{2\pi} \cos\theta \sin\left(\frac{\theta}{2}\right) d\theta \right] \] ### Compute Each Integral: #### First Integral: Let \(u = \frac{\theta}{2}\), \(d\theta = 2 du\), \(\theta = 0 \rightarrow u = 0\), \(\theta = 2\pi \rightarrow u = \pi\) \[ \int_{0}^{2\pi} \sin\left(\frac{\theta}{2}\right) d\theta = 2 \int_{0}^{\pi} \sin u du = 2 [-\cos u]_{0}^{\pi} = 2 [-( -1 - 1)] = 2 \times 2 = 4 \] #### Second Integral: Recall: \(\cos\theta \sin\left(\frac{\theta}{2}\right)\) Use the product-to-sum identity: \[ \cos\theta \sin\left(\frac{\theta}{2}\right) = \frac{1}{2} \left[ \sin\left(\theta + \frac{\theta}{2}\right) + \sin\left(\frac{\theta}{2} - \theta\right) \right] \] \[ = \frac{1}{2} \left[ \sin\left(\frac{3\theta}{2}\right) + \sin\left(-\frac{\theta}{2}\right) \right] \] \[ = \frac{1}{2} \left[ \sin\left(\frac{3\theta}{2}\right) - \sin\left(\frac{\theta}{2}\right) \right] \] So, \[ \int_{0}^{2\pi} \cos\theta \sin\left(\frac{\theta}{2}\right)d\theta = \frac{1}{2} \int_{0}^{2\pi} \sin\left(\frac{3\theta}{2}\right) d\theta - \frac{1}{2} \int_{0}^{2\pi} \sin\left(\frac{\theta}{2}\right) d\theta \] We already computed \(\int_{0}^{2\pi} \sin\left(\frac{\theta}{2}\right) d\theta = 4\). Now, compute \(\int_{0}^{2\pi} \sin\left(\frac{3\theta}{2}\right) d\theta\): Let \(v = \frac{3\theta}{2}\), \(d\theta = \frac{2}{3} dv\), when \(\theta = 0, v = 0\), \(\theta = 2\pi, v = 3\pi\) \[ \int_{0}^{2\pi} \sin\left(\frac{3\theta}{2}\right) d\theta = \frac{2}{3} \int_{0}^{3\pi} \sin v dv = \frac{2}{3} [-\cos v]_{0}^{3\pi} = \frac{2}{3} [-\cos(3\pi) + \cos(0)] \] \(\cos(3\pi) = -1, \cos(0) = 1\) \[ = \frac{2}{3} [ -(-1) + 1 ] = \frac{2}{3} [1 + 1] = \frac{2}{3} \times 2 = \frac{4}{3} \] So, the second integral: \[ \int_{0}^{2\pi} \cos\theta \sin\left(\frac{\theta}{2}\right)d\theta = \frac{1}{2} \left( \frac{4}{3} - 4 \right) = \frac{1}{2} \left( \frac{4 - 12}{3} \right) = \frac{1}{2} \left( -\frac{8}{3} \right) = -\frac{4}{3} \] ### Combine Results: \[ S = 4\pi \left( 4 - \left(-\frac{4}{3}\right) \right) = 4\pi \left( 4 + \frac{4}{3} \right) \] \[ = 4\pi \left( \frac{12}{3} + \frac{4}{3} \right) = 4\pi \left( \frac{16}{3} \right) = \frac{64\pi}{3} \] **Final Answer for (b):** \[ \boxed{\frac{64\pi}{3}} \] --- ## **Summary of Answers:** **(a) Length of one arch:** \(\boxed{8}\) **(b) Surface area when rotated:** \(\boxed{\dfrac{64\pi}{3}}\)