# area of a hyperbola

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• Feb 4th 2009, 01:12 PM
chubigans
area of a hyperbola
hey... i've been trying this problem for like 30 minutes and I can't seem to get it. I keep ending up with a weird integrand with 12 (sec cubed x -sec x). Can someone explain how to go about this? (we're doing trigonomentric substitution in calc 2)

Find the area of the region bounded by the hyperbola 9x2 - 4y2 = 36 and the line x = 3.
• Feb 4th 2009, 07:17 PM
DeMath
Quote:

Originally Posted by chubigans
hey... i've been trying this problem for like 30 minutes and I can't seem to get it. I keep ending up with a weird integrand with 12 (sec cubed x -sec x). Can someone explain how to go about this? (we're doing trigonomentric substitution in calc 2)

Find the area of the region bounded by the hyperbola 9x2 - 4y2 = 36 and the line x = 3.

You should have received this integral $\int\limits_2^3 {\sqrt {9{x^2} - 36} dx}$, which you can easily calculate, using the hyperbolic substitution $x = 2\cosh t$.
• Feb 5th 2009, 12:50 AM
DeMath
View a graphical representation of your problem

http://s56.radikal.ru/i154/0902/78/4f919de3312c.jpg