Chapter 8.2, Problem 29E

To determine

To find the area of the region enclosed by the lines and curves.

Answer to Problem 29E

The area of regions enclosed by the lines and curves is $6\sqrt{3}$ squared units.

Explanation of Solution

Concept used: Area between curves.

If f and g are continuous with $f\left(x\right)\ge g\left(x\right)$ throughout [a,b], then the area between

The curves $y=f\left(x\right)\text{ and }y=g\left(x\right)$ from a to b is the integral of $\left[f-g\right]$ from ato b.

  $\begin{array}{l}A={\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]}dx\\ \end{array}$

Given information:

The given functions is, $y=8\cos x\text{ , }y={\sec }^{2}x,-\frac{\pi }{3}\le x\le \frac{\pi }{3}$ .

Calculation: The area of the region enclosed by the lines and curves is obtained as,

  $\begin{array}{l}y=8\cos x\\ y={\sec }^{2}x\end{array}$

Set both equations equal to each other,

  $\begin{array}{l}8\cos x={\sec }^{2}x\left(\text{solve for the limits}\right)\\ 8\cos x=\frac{1}{{\cos }^{2}x}\\ {\cos }^{3}x=\frac{1}{8}\\ \cos x=\frac{1}{2}\\ x=\frac{\pi }{3},-\frac{\pi }{3}\end{array}$

The curves $y=8\cos x$ and $y={\sec }^{2}x$ intersect at $x=\frac{\pi }{3},-\frac{\pi }{3}$ .

Now, area bounded by these region is:

  $\begin{array}{l}\text{ A}={\displaystyle {\int }_a^b\left[f\left(t\right)-g\left(t\right)\right]}dt\\ ={\displaystyle {\int }_{-\pi /3}^{\pi /3}\left(8\cos x-{\sec }^{2}x\right)dx}\\ =2{\displaystyle {\int }_0^{\pi /3}\left(8\cos x-{\sec }^{2}x\right)dx}\\ =2{\left[8\sin x-\tan x\right]}_0^{\pi /3}\\ =2\left[8\sin \left(\frac{\pi }{3}\right)-\tan \left(\frac{\pi }{3}\right)-\left(8\sin 0-\tan 0\right)\right]\\ =2\left[8.\frac{\sqrt{3}}{2}-\sqrt{3}-\left(0-0\right)\right]\\ =6\sqrt{3}\end{array}$

Hence, the area of regions enclosed by the lines and curves is $6\sqrt{3}$ squared units.