- #1
sporff
- 24
- 0
This seems like it should be easy but I can't seem to wrap my brain around it right now. I'm integrating to find the area of a hyperbola cut off by the line x=4 (I assume just the nose of the hyperbola cut off by the line on the positive side)
hyperbola:
[tex]\frac{x^2}{9} - \frac{y^2}{4} = 1[/tex]
Since it's cut off by the line x=4 I rearrange the equation in terms of x.
[tex]y = \frac{2}{3}\sqrt{x^2-9}[/tex]
I should just need [tex]\int{\sqrt{x^2-9}} dx[/tex] (mulltiplied by 2/3 and then solve for the area, of course) then, yes? The only way I've tried is substituting in x = 3sec(u) in which leaves you with (including the 2/3 in this):
[tex]6\int\tan^2{u} sec{u} du[/tex]
Which doesn't seem any easier. Any hints? This is not homework ask I'm self-teaching in my spare time. Hints rule but I'm not turning down a full answer either as this has been bugging me for a couple of days now.
hyperbola:
[tex]\frac{x^2}{9} - \frac{y^2}{4} = 1[/tex]
Since it's cut off by the line x=4 I rearrange the equation in terms of x.
[tex]y = \frac{2}{3}\sqrt{x^2-9}[/tex]
I should just need [tex]\int{\sqrt{x^2-9}} dx[/tex] (mulltiplied by 2/3 and then solve for the area, of course) then, yes? The only way I've tried is substituting in x = 3sec(u) in which leaves you with (including the 2/3 in this):
[tex]6\int\tan^2{u} sec{u} du[/tex]
Which doesn't seem any easier. Any hints? This is not homework ask I'm self-teaching in my spare time. Hints rule but I'm not turning down a full answer either as this has been bugging me for a couple of days now.