Area Under Unit Hyperbola?

Question

Going through Strang's Calculus right now and don't understand a seemingly basic homework question. It asks to integrate under the unit hyperbola, from $(1,0)$ to $(\cosh t, \sinh t)$.

The answer in the book simply gives this: $\int y \text{d}x = \int \sinh t (\sinh t\ \text{d}t)$

I initially thought the integral would be $\int\sqrt {\cosh^2t -1}$ because $y = \sinh t = \sqrt {\cosh^2t -1} $.

Is the answer in the book because $y = \sinh t$, $\text{d}x = \cosh t$ and the equating that to $\text{d}t$ would be its derivative $\sinh t\ \text{d}t$?

Answer 1

If I didn't know about parametric equations, would there have been some other way of answering this question?

Going through Strang's Calculus right now... It asks to integrate under the unit hyperbola, from $(1,0)$ to $(\cosh t, \sinh t)$.

The author has specified the right half of the unit hyperbola $$x^2-y^2=1,$$ and introduced a parameter $t$ on $\mathbb R,$ where nonnegative values of $t$ correspond to points in the first quadrant.

So—avoiding explicit parameterisation—the required integral is \begin{align}&\frac t{|t|}\int_1^{\cosh t}\sqrt{x^2-1}\,\mathrm dx\\= &\frac t{|t|}\int_0^{|t|}\sinh^2\theta\,\mathrm d\theta\end{align} (substituting $x=\cosh \theta,\;\theta\geq0)$ \begin{align}=&\int_0^t\sinh^2\theta\,\mathrm d\theta\end{align} (since $\sinh^2\theta$ is an even function) \begin{align}=&\sinh2t-2t.\end{align}

The answer in the book simply gives this: $\int y \text{d}x = \int \sinh t (\sinh t\ \text{d}t)$

As above, but do note that the lower and upper limits of integration are $0$ and $t\;(t\in\mathbb R)$ respectively.

Answer 2

Your curve is parametrised by $$\begin{cases} x = \cosh(t) \\ y = \sinh(t) \end{cases} $$

A fast recap from the theory. Say if your curve is parametrised by $$x = f(t) ~~~~~~~~~~~~~ y = g(t)$$ Now to find the area under a curve $y = F(x)$ in the interval $[a, b]$ we simply calculate the integral $$A = \int_a^b F(x)\ \text{d}x$$

We will now think of the parametric equation $x = f(t)$ as a substitution in the integral. We will also assume that $\alpha = f(a)$ and $\beta = f(b)$. There is actually no reason to assume that this will always be the case and so I’ll give a corresponding formula later.

So, if this is going to be a substitution we’ll need, $$\text{d}x = f'(t)\ \text{d}t$$

Plugging this into the area formula above and making sure to change the limits to their corresponding $t$ values gives us

$$A = \int_{\alpha}^{\beta}F(f(t))f'(t)\ \text{d}t$$

Since we don’t know what $F(x)$ is we’ll use the fact that $$y = F(x) = F(f(t)) = g(t)$$

and we arrive at the formula that we want:

$$A = \int_{\alpha}^{\beta}g(t) f'(t)\ \text{d}t$$

In your case then $g(t) = y = \sinh(t)$, $f'(t)\ \text{d}t = \text{d}x = \sinh(t)\ \text{d}t$ whence

$$A = \int_1^0 \sinh(t)\sinh(t)\ \text{d}t = \int_1^0 \sinh^2(t)\ \text{d}t = \frac{1}{4} (2-\sinh (2))~~~~~ \text{u}^2$$

Where $u^2$ means unit square since we are calculating an area, whatever the unit be.