Make parametric equation for closed space

Question

I have the following problem.

Let $a$ be a positive number. A curve $K$ in the $(x,z)$-plane is given by the parametric equation:

$$r(u)=(2 \mbox{sinh(u)},\mbox{cosh(u)}), u \in[0,a]$$

Let $A$ be a bounded, closed, and connected area in the $(x,z)$-plane.

The boundary of $A$ consists of $K$, a straight line from $(0,0)$ to $(0,1)$, and a straight line from $(0,0)$ to $r(a)$.

Make a parametric equation for $A$. What is the area of $A$?

I'm having great difficulty with making a parametric equation for $A$ because it's boundary consists of three separate lines.

I can make three separate parametric equations for this problem, but I don't think that will help me with the next part, to find the area of $A$.

Can you solve this problem using three separate parametric equations, or do you have to make one single parametric, and if so how can you do it?

Answer 1

Since $A$ is a region of the plane rather than a curve, you’ll need two parameters. One fairly obvious possibility is $$A(u,v) = v r(u), (u,v)\in[0,a]\times[0,1].$$ If you hold $u$ constant, varying $v$ traces the line segment from the origin to $r(u)$.

Using this parameterization, the area of $A$ is given by the integral \begin{align}\iint_A dA &= \int_0^a \int_0^1{\partial(x,y)\over\partial(u,v)}\,dv\,du \\ &= \int_0^a \int_0^1 \begin{vmatrix}2v\cosh u & 2\sinh u \\ v\sinh u&\cosh u\end{vmatrix}\,dv\,du \\ &= \int_0^a \int_0^1 2v\,dv\,du \\ &= a.\end{align}

This value can be verified by computing the area using Green’s theorem with the integrand $\frac12(x\,dy-y\,dx) = \frac12\begin{vmatrix}x&y\\x'&y'\end{vmatrix}\,dt$. The integral vanishes along the straight line segments bounding $A$ (verify this!), leaving $$\frac12\int_a^0 \begin{vmatrix}2\sinh t & \cosh t \\ 2\cosh t & \sinh t\end{vmatrix}\,dt = \int_0^a dt = a.$$

Answer 2

I think that the point of this problem is to write the curve as a function $f(x)$.

You can rewrite the curve $K$ as the image of the function $$f(x)=\sqrt{1+\dfrac{x^2}{4}}$$ To do this, put $(2\sinh u,\cosh u) = (x,f(x))$ and sobstitute these equalities in the relation $\cosh^2 u - \sinh^2 u = 1$.

Now the domain is: $$ A=\left\{(x,z) \ \Big| \ \frac{\cosh a}{2\sinh a}x\leq z \leq f(x), \ 0\leq x\leq 2\sinh a\right\} $$ So: \begin{multline} Area(A)=\int_{0}^{2\sinh a} \left(\int_{\frac{\cosh a}{2\sinh a}x}^{f(x)}dy\right) \ dx=\\ \int_{0}^{2\sinh a}\left(\sqrt{1+\dfrac{x^2}{4}} - \frac{\cosh a}{2\sinh a}x\right) \ dx=\\ \int_{0}^{2\sinh a}\sqrt{1+\dfrac{x^2}{4}} \ dx - \frac{\cosh a}{2\sinh a}\left[ \frac{x^2}{2}\right]_0^{2\sinh a} =\\ \int_{0}^{2\sinh a}\sqrt{1+\dfrac{x^2}{4}} \ dx - \sinh a\cosh a = \quad (x=2\sinh t)\\ \int_{0}^{a}\sqrt{1+\dfrac{4\sinh^2 t}{4}} \ 2\cosh t \ dt - \sinh a\cosh a = \\ \int_{0}^{a}2\cosh^2 t \ dt - \sinh a\cosh a = \\ \left[t + \sinh t\cosh t\right]_0^a - \sinh a\cosh a =\\ a + \sinh a\cosh a - \sinh a\cosh a = a \end{multline}