Integral to Solve Area of Hyperbola

Given the standard hyperbola I need to find the area bounded by the curve and the line

Here is the visual representation

https://dl.dropboxusercontent.com/u/.../hyperbola.PNG

Solving for y gives me

I then set up the integral

I set , substituted and changed limits of integration accordingly

Using a trig identity I can simplify down to

And that is where I am stuck, I tried another u-substitution to no avail.

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 but what now?

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

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

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 , then you can make the substitution , noting that when and when and your integral becomes

Re: Integral to Solve Area of Hyperbola

That makes it a lot easier. Thanks

Re: Integral to Solve Area of Hyperbola

Quote:

Originally Posted by

**Prove It**

When making the substitution, where does that come from? shouldn't

1 Attachment(s)

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:

Your original trig substitution also works. See the attachment:

Attachment 28681