Area of a Hyperbolic Sector via the Parametrization (cosh t, sinh t)

The parametric sector-area formula

When a curve is given parametrically by $\bigl(x(t),y(t)\bigr)$ and you want the area of the region swept out from the origin as $t$ runs from $t_0$ to $t_1$ (the two radial segments to the endpoints plus the arc between them), use the sector form of the shoelace integral:

$A=\frac{1}{2}\int_{t_0}^{t_1}\bigl(x\,y'-y\,x'\bigr)\,dt.$

The integrand $x\,y'-y\,x'$ is the cross product $\vec r\times\vec r{}'$; geometrically $\tfrac12(x\,y'-y\,x')\,dt$ is the tiny triangle the radius vector sweeps in time $dt$. (In polar form this is just $A=\tfrac12\int r^{2}\,d\theta$.)

Why $(\cosh t,\sinh t)$ is special

The point $(\cosh t,\sinh t)$ lies on the right branch of the unit hyperbola $x^{2}-y^{2}=1$, because $\cosh^{2}t-\sinh^{2}t=1.$ This is the exact analogue of $(\cos t,\sin t)$ on the unit circle $x^{2}+y^{2}=1$. Differentiating gives $x'=\sinh t$ and $y'=\cosh t$, so the integrand telescopes beautifully: $x\,y'-y\,x'=\cosh t\cdot\cosh t-\sinh t\cdot\sinh t=\cosh^{2}t-\sinh^{2}t=1.$

Hence the hyperbolic sector from $t=0$ to $t=t_1$ has area $A=\frac{1}{2}\int_{0}^{t_1}1\,dt=\frac{t_1}{2}.$

This is the defining property of the hyperbolic functions: just as a circular sector of angle $\theta$ on the unit circle has area $\theta/2$, the hyperbolic angle $t$ is twice the area of the corresponding hyperbolic sector. That is what the "argument" $t$ in $\cosh t$ and $\sinh t$ actually measures — an area, not an arc length.

How to apply it

1. Recognize the curve $x=\cosh t,\,y=\sinh t$ as the unit hyperbola and the region as the sector bounded by the $x$-axis, the arc, and the ray to the endpoint.
2. Plug $x',y'$ into $A=\tfrac12\int (x\,y'-y\,x')\,dt$; the integrand becomes the constant $1$.
3. Read off $A=\tfrac12(t_1-t_0)$. For a sector starting at the vertex $(1,0)$, simply $A=t_1/2$.

Mini-example

Find the area of the hyperbolic sector from $t=0$ to $t=3$ (origin to $(\cosh 3,\sinh 3)$, closed by the $x$-axis).

$A=\frac{1}{2}\int_{0}^{3}\bigl(\cosh^{2}t-\sinh^{2}t\bigr)\,dt=\frac{1}{2}\int_{0}^{3}1\,dt=\frac{3}{2}.$

No evaluation of $\cosh 3$ is ever needed — the parametrization does all the work.

Exam trick: Whenever you see $x=\cosh t,\,y=\sinh t$ (or $x=a\cosh t,\,y=b\sinh t$) and a "sector from the origin" area, do not integrate $y\,dx$ by parts. Use $A=\tfrac12\int(x\,y'-y\,x')\,dt$: the identity $\cosh^2 t-\sinh^2 t=1$ collapses the integrand to a constant and the answer is just $\tfrac12(t_1-t_0)$ — times $ab$ if the semi-axes are $a,b$.