Question

Write an integral that represents the area of the surface generated by revolving the curve about the \(x\) -axis. Use a graphing utility to approximate the integral. $$ x=\theta+\sin \theta, \quad y=\theta+\cos \theta \quad 0 \leq \theta \leq \frac{\pi}{2} $$

Short answer

The integral that represents the area of the surface generated by revolving the curve around the \(x\)-axis is \[ \int_{0}^{\pi/2} 2\pi (\theta + \cos(\theta)) \sqrt{(1 + \cos(\theta))^2 + (1 - \sin(\theta))^2} d\theta \]

Step-by-step solution

Derivation

First calculate the derivatives of \(x\) and \(y\) with respect to \(\theta\), i.e., \(dx/d\theta\) and \(dy/d\theta\). \( x = \theta + \sin(\theta) \) thus \( dx/d\theta = 1 + \cos(\theta) \). Similarly, \( y = \theta + \cos(\theta) \) so \( dy/d\theta = 1 - \sin(\theta) \)

Calculation of the formula

Substitute everything into the formula for the surface area of rotation: \[ A = \int_{0}^{\pi/2} 2\pi (\theta + \cos(\theta)) \sqrt{(1 + \cos(\theta))^2 + (1 - \sin(\theta))^2} d\theta \] This is our integral for the surface area.

Approximation of the integral

Finally, the integral can be approximated using a numerical method or a graphing utility as asked in the problem. This final step usually requires a calculator or some kind of software and the precise result will depend on which method is being used. Without a specific method or tool stated, we can't provide a numerical approximation here.