Please help me with this problem. I have been trying 2 do it 4 half an hour and still can figure out. Thank You!
Find the area of the region bounded by the hyperbola 25x2 - 4y2 = 100 and the line x = 3.
So the area looks a little like a slice of a circle right?
the easiest way to do it is turn y in terms of x.
$\displaystyle 25x^2-4y^2=100$
$\displaystyle 25x^2-100-4y^2=0$
$\displaystyle 25x^2-100=-4y^2$
$\displaystyle \frac{25}{4}x^2-25=y^2$
$\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$
Since you need both the positive and negative portion instead of doing 2 integrals you can just take the integral to get the region above the x axis and multiply it's area by 2.
can you take it from there?
$\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$
$\displaystyle \int\sqrt{\frac{25}{4}x^2-25}dx$
Substituting when you take the integral
$\displaystyle x = 5sec(\theta)$
well what's $\displaystyle x^2$?
$\displaystyle x^2=25sec^2(\theta)$ right?
what if you let u = 2x?
and $\displaystyle dx=5sec\theta tan\theta \text{ }d\theta$ right?
when you substitute that in it gets messy.
$\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$
$\displaystyle =\sqrt{(25)*(\frac{1}{4}x^2-1)}$
$\displaystyle =5\sqrt{\frac{x^2}{4}-1}$
what if you let u = 2x?
Seems like you should do a trig substitution after that, though your initial idea to substitute secant isn't wrong, just needs somewhat different coefficients.