Find the area "A" of the given graph
Answers should be in terms of a, b and h
My attempt:
(1) Look for an equation in terms of y (or maybe in terms of x since the region is bounded between $y=0$ and $y=h$?)
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \\ b^2x^2-a^2y^2 = a^2b^2 \\ a^2y^2 = b^2x^2-a^2b^2 \\ y^2 = (\frac{bx}{a})^2-b^2 \\ y = \sqrt{(\frac{bx}{a})^2-b^2} $
The expression inside the square root looks pretty similar to $\frac{d}{dx}arcsin(x)=\frac{1}{\sqrt{1-x^2}}$... I don't know if/how to make that substitution (if that's what I need to do). After that, I believe I just have to evaluate it from $y=0$ to $y=h$
Integrate over $y$, it will be easier to work out.
$$A = 2a\int_0^h \sqrt{1+\frac{y^2}{b^2}}dy = 2ab \int_0^{h/b} \sqrt{1+u^2}du $$
The primitive of $\sqrt{1+u^2}$ can be found by substitution. I suggest $u=\sinh t$, then
$$\int \sqrt{1+u^2}du = \int \cosh^2 t dt = \frac{1}{4} \sinh(2t) + \frac{t}{2} = \frac{1}{2} u\sqrt{1+u^2} + \frac{\text{arcsinh}(u)}{2} \; .$$