Chapter 11.1, Problem 57E

To determine

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

Answer to Problem 57E

The required surface area is $8{\pi }^2$ square units.

Explanation of Solution

Given information:

The surface area of the solid: $S={\displaystyle \underset{a}{\overset{b}{\int }}2\pi y}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

The parametric equations: $x=\cos t$ , $y=2+\sin t$ , $0\le t\le 2\pi$

Calculation:

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

The parametric equations are $x=\cos t$ , $y=2+\sin t$ .

Differentiate the parametric equations with respect to t .

  $\frac{dx}{dt}=\sin t$ , $\frac{dy}{dt}=\cos t$

Substitute 0 for a , $2\pi$ for b , $2+\sin t$ for y , $-\sin t$ for $\frac{dx}{dt}$ and $\cos t$ for $\frac{dy}{dt}$ in the given surface area.

  $\begin{array}{c}S={\displaystyle \underset{0}{\overset{2\pi }{\int }}2\pi \left(2+\sin t\right)}\sqrt{{\left(-\sin t\right)}^2+{\left(\cos t\right)}^2}dt\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}2\pi \left(2+\sin t\right)}\sqrt{{\sin }^{2}t+{\cos }^{2}t}dt\\ =2\pi {\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(2+\sin t\right)}\left(1\right)dt\end{array}$

Simplify the above expression further.

  $\begin{array}{c}2\pi {\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(2+\sin t\right)}\left(1\right)dt={2\pi \left(2t-\cos t\right)|}_0^{2\pi }\\ =2\pi \left[2\left(2\pi \right)-\cos \left(2\pi \right)-0+\cos \left(0\right)\right]\\ =2\pi \left(4\pi -\cancel{1}-0+\cancel{1}\right)\\ =8{\pi }^2\end{array}$

Hence, the required surface area is $8{\pi }^2$ square units.