Single Variable Calculus, Chapter 8, 8.3, Section 8.3, Problem 34

Determine the area of the region bounded by the hyperbola $9x^2 - 4y^2 = 36$ and the line $ x= 3$



By using vertical strips, Since the graph above the $x$-axis is the same as the graph below $x$-axis, we can multiply the area above the $x$-axis by 2 to get the area of the entire bounded region. So,


$
\begin{equation}
\begin{aligned}
A &= 2 \int^b_a \left( y_{\text{upper}} - u_{\text{lower}} \right) dx\\
\\
A &= 2 \int^3_2 \left( \sqrt{\frac{9x^2 - 36}{4}} - 0 \right) dx\\
\\
A &= 2 \int^3_2 \left( \frac{1}{2} \right) \sqrt{9x^2 - 36} dx\\
\\
A &= \int^3_2 \sqrt{9(x^2-4)} dx\\
\\
A &= 3 \int^3_2 \sqrt{x^2 - 4} dx
\end{aligned}
\end{equation}
$


From the triangle,



$
\begin{equation}
\begin{aligned}
\cos \theta &= \frac{2}{x} \qquad \text{and} \qquad \tan \theta = \frac{\sqrt{x^2 - 4}}{2}\\
\\
x &= \frac{2}{\cos \theta} = 2 \sec \theta\\
\\
dx &= 2 (\sec \theta \tan \theta) d \theta
\end{aligned}
\end{equation}
$


Thus,

$
\begin{equation}
\begin{aligned}
A = 3 \int \sqrt{x^2-4} dx &= 3 \int (2 \tan \theta) (2 \sec \theta \tan \theta) d \theta\\
\\
&= 12 \int \sec \theta \tan^2 \theta d \theta
\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int \sec \theta \tan \theta d \theta$,
Recall that $\tan^2 \theta = \sec^2 \theta - 1$

$
\begin{equation}
\begin{aligned}
\int \sec \theta \tan^2 \theta d \theta &= \int \sec \theta (\sec^2 \theta - 1) d \theta\\
\\
\int \sec \theta \tan^2 \theta d \theta &= \int \left( \sec^3 \theta - \sec \theta \right) d\theta\\
\\
\int \sec \theta \tan^2 \theta d \theta &= \sec^3 \theta d \theta - \int \sec \theta d \theta
\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int sec^3 \theta d \theta$, we will use $u = \sec \theta$ and $dv = \sec^2 \theta d \theta$, then
$du = \sec \theta \tan \theta d \theta$ and $\displaystyle v = \int \sec^2 \theta d \theta = \tan \theta$

So,
$\displaystyle \int \sec^3 \theta d \theta = uv - \int v du = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d \theta$

Going back from the previous equation,

$
\begin{equation}
\begin{aligned}
\int \sec \theta \tan^2 \theta d \theta &= \int \sec^3 \theta d \theta - \int \sec \theta d \theta\\
\\
\int \sec \theta \tan^2 \theta d \theta &= [\sec \theta \tan \theta - \int \sec \tan^2 \theta d \theta] - \ln (\sec \theta + \tan \theta)

\end{aligned}
\end{equation}
$


By combining like terms,

$
\begin{equation}
\begin{aligned}
2 \int \sec \theta \tan^2 \theta d \theta &= \sec \theta \tan \theta - \ln (\sec \theta + \tan \theta)\\
\\
\int \sec \theta \tan^2 \theta d \theta &= \frac{\sec \theta \tan \theta - \ln (\sec \theta + \tan \theta)}{2}
\end{aligned}
\end{equation}
$


Therefore,

$
\begin{equation}
\begin{aligned}
12 \int \sec \theta \tan^2 \theta d \theta &= 12 \left[ \frac{\sec \theta \tan \theta - \ln (\sec \theta + \tan \theta)}{2} \right]\\
\\
&= \left[ 6 \sec\theta \tan\theta - 6 \ln (\sec \theta + \tan \theta ) \right]^b_a
\end{aligned}
\end{equation}
$


From the triangle,

$
\begin{equation}
\begin{aligned}
12 \int \sec \theta \tan^2 \theta d \theta &= \left[6 \left( \frac{x}{2} \right)\left( \frac{\sqrt{x^2-4}}{2} \right) - 6 \ln \left( \frac{x}{2} + \frac{\sqrt{x^2-4}}{2} \right) \right]^3_2\\
\\
&= \left[ \frac{3}{2} x \sqrt{x^2 - 4} - 6 \ln \left( \frac{x+\sqrt{x^2-4}}{2} \right) \right]\\
\\
&= \frac{9}{2} \sqrt{5} - 6 \ln \left( \frac{3+\sqrt{5}}{2} \right) \text{ square units}
\end{aligned}
\end{equation}
$