# Integral to Solve Area of Hyperbola

• Jun 26th 2013, 08:32 PM
ReneG
Integral to Solve Area of Hyperbola
Given the standard hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ I need to find the area bounded by the curve and the line $\displaystyle x = 2a$

Here is the visual representation
https://dl.dropboxusercontent.com/u/.../hyperbola.PNG

Solving for y gives me $\displaystyle y = b\sqrt{\frac{x^2}{a^2} - 1}$
I then set up the integral $\displaystyle A = 2b\int_{a}^{2a}\sqrt{\frac{x^2}{a^2} - 1} \,\,dx$
I set $\displaystyle x = a \sec{\theta} \Rightarrow \,\,dx = a\sec{\theta}\tan{\theta}\,\, d\theta$, substituted and changed limits of integration accordingly $\displaystyle A = 2b\int_{0}^{\frac{\pi}{3}} \sqrt{\sec^2{\theta} - 1}\, a\sec{\theta}\tan{\theta} \,\, d\theta$

Using a trig identity I can simplify down to $\displaystyle A = 2ab\int_{0}^{\frac{\pi}{3}} \tan^2\theta\sec\theta \,\,d\theta$

And that is where I am stuck, I tried another u-substitution to no avail.
• Jun 26th 2013, 10:45 PM
ReneG
Re: Integral to Solve Area of Hyperbola
Okay I now see that you can rewrite the two trig functions in terms of sine and cosine $\displaystyle A = 2ab\int_{0}^{\frac{\pi}{3}} \frac{\sin^2\theta}{\cos^3\theta}\,\, d\theta$ but what now?
• Jun 27th 2013, 12:23 AM
Shakarri
Re: Integral to Solve Area of Hyperbola
I think your integral wont work because half the area is above the x axis and have the area is below the x axis (negative) so the integral will be 0.
I would integrate with respect to y instead
Solving the equation of the hyperbola for x gives $\displaystyle x=a\sqrt{\frac{y^2}{b^2}-1}}$
To find the area between two lines you integrate the difference between them.
The limits of the integration have an x coordinate 2a and satisfy the equation of the hyperbola, they turn out to be $\displaystyle \pm b\sqrt{5}$
• Jun 27th 2013, 01:22 AM
Prove It
Re: Integral to Solve Area of Hyperbola
The OP's integral is correct, the integral of the function between a and 2a would give the area above the x-axis, doubling that gives the total area.

It would be easier to write the integral as \displaystyle \displaystyle \begin{align*} \frac{2b}{a}\int_a^{2a}{\sqrt{x^2 - a^2}\,dx} \end{align*}, then you can make the substitution \displaystyle \displaystyle \begin{align*} x = a \cosh{(t)} \implies dx = a\sinh{(t)}\,dt \end{align*}, noting that when \displaystyle \displaystyle \begin{align*} x = a, \, t = 0 \end{align*} and when \displaystyle \displaystyle \begin{align*} x = 2a, \, t = \cosh^{-1}{ (2) } \end{align*} and your integral becomes

\displaystyle \displaystyle \begin{align*} \frac{2b}{a}\int_a^{2a}{\sqrt{x^2 - a^2}\,dx} &= \frac{2b}{a}\int_{0 }^{ \cosh^{-1}{\left( 2 \right) } }{ \sqrt{ \left( a\cosh{(t)} \right) ^2 - a^2 } \, \sqrt{a}\,\sinh{(t)} \,dt} \\ &= \frac{2\sqrt{a}\,b}{a} \int_{0}^{\cosh^{-1}{ \left( 2 \right) }}{ \sqrt{ a^2 \left[ \cosh^2{(t)} - 1 \right] } \, \sinh{(t)}\,dt } \\ &= \frac{2\sqrt{a}\,b}{a} \int_0^{\cosh^{-1}{(2)}}{ a\sinh{(t)}\sinh{(t)}\,dt } \\ &= 2\sqrt{a}\,b \int_0^{\cosh^{-1}{(2)}}{\sinh^2{(t)}\,dt} \\ &= 2\sqrt{a}\,b \int_0^{\cosh^{-1}{(2)}}{\frac{1}{2}\left[\cosh{(2t)} - 1 \right] dt} \\ &= \sqrt{a}\,b \int_0^{ \cosh^{-1}{(2)} }{ \cosh{(2t)} - 1 \, dt } \\ &= \sqrt{a}\,b \left[ \frac{1}{2}\sinh{(2t)} - t \right]_0^{\cosh^{-1}{(2)}} \\ &= \sqrt{a}\,b \left[ \cosh{(t)} \sqrt{ \cosh^2{(t)} - 1 } - t \right]_0^{\cosh^{-1}{(2)}} \\ &= \sqrt{a}\,b \left\{ \left[ 2\sqrt{ 2^2 - 1 } - \cosh^{-1}{(2)} \right] - \left[ 1\sqrt{ 1^2 - 1 } - 0 \right] \right\} \end{align*}

\displaystyle \displaystyle \begin{align*} = \sqrt{a}\,b \left[ 2\sqrt{3} - \cosh^{-1}{(2)} \right] \end{align*}
• Jun 27th 2013, 08:40 AM
ReneG
Re: Integral to Solve Area of Hyperbola
That makes it a lot easier. Thanks
• Jun 27th 2013, 06:34 PM
ReneG
Re: Integral to Solve Area of Hyperbola
Quote:

Originally Posted by Prove It
\displaystyle \displaystyle \begin{align*} \frac{2b}{a}\int_a^{2a}{\sqrt{x^2 - a^2}\,dx} &= \frac{2b}{a}\int_{0 }^{ \cosh^{-1}{\left( 2 \right) } }{ \sqrt{ \left( a\cosh{(t)} \right) ^2 - a^2 } \, \sqrt{a}\,\sinh{(t)} \,dt} \end{align*}

When making the substitution, where does that $\displaystyle \sqrt{a}$ come from? shouldn't $\displaystyle dx = a \sinh{(t)} \,dt$
• Jun 27th 2013, 08:48 PM
johng
Re: Integral to Solve Area of Hyperbola
Hi ReneG,
ProveIt has a small error in the first line. It should not be square root of a, but a in the integrand. So the correct answer to to the problem is:
$\displaystyle ab(2\sqrt3-cosh^{-1}(2))$

Your original trig substitution also works. See the attachment:

Attachment 28681