# Thread: area of a hyperbola

1. ## 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.

2. 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 $x = 2\cosh t$.

3. View a graphical representation of your problem

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# Find The area of the region bounded by the curve 9x^2 4y^2-36=0 is?

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