Chapter 11.1, Problem 60E

To determine

To Find:The surface area when the parametric curve is revolved about the x-axis.

Answer to Problem 60E

The required surface area of the solid is $\pi$ .

Explanation of Solution

Given Information:

  $x=\ln \left(\sec t+\tan t\right)-\sin t$  ........ (1)

And

  $y=\cos t$  ........ (2)

Surface area of a solid is

  $S={\displaystyle \underset{a}{\overset{b}{\int }}2\pi y\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt}$  ........ (3)

Such that

  $0\le t\le \frac{\pi }{3}$

Calculation:

Differentiating equation (1) and (2)

  $\frac{dx}{dt}=\sec t-\cos t$

And

  $\frac{dy}{dt}=-\sin t$

Substitute the value $\sec t-\cos t$ for $\frac{dx}{dt}$ , $-\sin t$ for $\frac{dy}{dt}$ in equation (3)

  $\begin{array}{c}L={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}2\pi \left(\cos t\right)\sqrt{{\left(\sec t-\cos t\right)}^2+{\left(-\sin t\right)}^2}}dx\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}2\pi \left(\cos t\right)\sqrt{{\tan }^{2}t}}dt\\ =2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\sin t}dt\\ =2\pi {\left[-\cos t\right]}_0^{\frac{\pi }{3}}\\ =\pi \end{array}$

Conclusion:

Hence, the required surface area of the solid is $\pi$ .