# How do you calculate the area under hyperbola?is this correct? what do I do next?

• Jan 7th 2010, 09:34 AM
Aerospank
How do you calculate the area under hyperbola?is this correct? what do I do next?
Attachment 14707

I found $\displaystyle y = \sqrt{\frac{b^2x^2}{a^2} - b^2}$

then Area = $\displaystyle b \int \sqrt{\frac{x^2}{a^2} - 1}dx$ with upper limit c and lower a.

I then let x = acosht, is this correct?

then did the method of substitution.

then what do I do? My end answer looks very weird...so i just want to make sure.
• Jan 7th 2010, 12:14 PM
shawsend
Solve it parametrically if we know that the parametric representation of the right side of a hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $\displaystyle x=a\cosh(t), y=b\sinh(t)$. So then the area is:

$\displaystyle \int_a^c ydx=\int_{t_0}^{t_1} y(t) d(x(t))=\int_{t_0}^{t_1} b\sinh(t) a\sinh(t)dt$.

You can find the limits in terms of t right?
• Jan 7th 2010, 01:52 PM
Aerospank
Quote:

Originally Posted by shawsend
Solve it parametrically if we know that the parametric representation of the right side of a hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $\displaystyle x=a\cosh(t), y=b\sinh(t)$. So then the area is:

$\displaystyle \int_a^c ydx=\int_{t_0}^{t_1} y(t) d(x(t))=\int_{t_0}^{t_1} b\sinh(t) a\sinh(t)dt$.

You can find the limits in terms of t right?

i only submitted the x one....is that ok?
• Jan 7th 2010, 03:08 PM
shawsend
Yes, that's fine what you did. Sorry, I didn't notice that initially. And you know of course to express the limits of integration in t which I get:

$\displaystyle ab\int_0^{x_0} \sinh^2(t)dt$