Chapter 11.1, Problem 59E

To determine

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

Answer to Problem 59E

The required surface area is 178.56 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: $x=t^2+2$ , $y=t+1$ , $0\le t\le 3$

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 $x=t^2+2$ , $y=t+1$ .

Differentiate the parametric equations with respect to t .

  $\frac{dx}{dt}=2t$ , $\frac{dy}{dt}=1$

Substitute 0 for a , 3 for b , $t+1$ for y , $2t$ for $\frac{dx}{dt}$ and 1 for $\frac{dy}{dt}$ in the given surface area.

  $\begin{array}{c}S={\displaystyle \underset{0}{\overset{3}{\int }}2\pi \left(t+1\right)}\sqrt{{\left(2t\right)}^2+{\left(1\right)}^2}dt\\ ={\displaystyle \underset{0}{\overset{3}{\int }}2\pi \left(t+1\right)}\sqrt{4t^2+1}dt\\ =2\pi {\displaystyle \underset{0}{\overset{3}{\int }}\left(t+1\right)}\sqrt{4t^2+1}dt\end{array}$

The value of the above expression can be found by using definite integral calculator.

So, $\begin{array}{c}S=2\pi \left(28.419\right)\\ =178.56\end{array}$

Hence, the required surface area is 178.56 square units.