Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 24

Sketch the region enclosed by the curves $\displaystyle y = \cos x, y = 1 - \cos x$ and $0 \leq x \leq \pi$. Then find the area of the region.







By using a vertical strip,

$\displaystyle A = \int^{x_2}_{x_1} (y_{\text{upper}}, y_{\text{lower}}) dx$

In order to get the value of the upper and lower limits, we equate the two functions to get its points of intersection. So..


$
\begin{equation}
\begin{aligned}

& \cos x = 1 - \cos x
\\
\\
& 2 \cos x = 1
\\
\\
& x = \cos^{-1} \left[ \frac{1}{2} \right]
\\
\\
& x = \frac{\pi}{3} + 2 \pi n; \text{ where $n$ is any integer}

\end{aligned}
\end{equation}
$


For interval $0 \leq x \leq \pi$, we have point of intersection at $\displaystyle x = \frac{\pi}{3}$

Notice that orientation of the curve changes at the point of intersection. Let $A_1$ and $A_2$ be the area in left part respectively. Thus,


$
\begin{equation}
\begin{aligned}

A_1 =& \int^{\frac{\pi}{3}}_0 [\cos x - (1 - \cos x)] dx
\\
\\
A_1 =& \int^{\frac{\pi}{3}}_0 [2 \cos x - 1] dx
\\
\\
A_1 =& [2 \sin x - x]^{\frac{\pi}{3}}_0
\\
\\
A_1 =& 0.6849 \text{ square units}

\end{aligned}
\end{equation}
$


For the other,


$
\begin{equation}
\begin{aligned}

A_2 =& \int^{\pi}_{\frac{\pi}{3}} [1 - \cos x - (\cos x)] dx
\\
\\
A_2 =& \int^{\pi}_{\frac{\pi}{3}} [1 - 2 \cos x] dx
\\
\\
A_2 =& [x - 2 \sin x]^{\pi}_{\frac{\pi}{3}}
\\
\\
A_2 =& 3.8264 \text{ square units}

\end{aligned}
\end{equation}
$


Therefore, the total area is $A_1 + A_2 = 4.5113$ square units