Compute $\int_Yyz \ dS.$

Question

Compute the work done by the vectorfield $F(x,y)=[x^2,x^2]$ along the curve $x^2-y^2=1$ from the point $(\sqrt{5},2)$ to $(\sqrt{10},-3).$

I have two things I need to get answered:

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1) Using the parameterization $\vec{r}(x,y)=\vec{r}(\cosh(t),\sinh(t))$ I get that

$$W=\int_\gamma \vec{F} \cdot\frac{d\vec{r}}{dt} \ dt =\int_{\sinh^{-1}(2)}^{\sinh^{-1}(-3)} \cosh^3(t) + \cosh^2(t)\sinh(t) \ dt = \frac{5(2\sqrt{10}-\sqrt{5}-10)}{3}.$$

I used WolframAlpha to evaluate that integral, how is this done by hand?

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2) We have not covered the hyperbolic functions in this course and we are not supposed to use them to solve this problem (hence my difficulties with the above integral). Is there any other way to parameterize the curve without using hyperbolics?

Answer 1

It is easy to integrate $\cosh^2(t)\sinh(t)$ since $\frac{d}{dt}\cosh(t)=\sinh(t)$, so $$\int\cosh^2(t)\sinh(t)\,dt=\frac{\cosh^3(t)}{3}+C$$ To integrate $\cosh^3(t)$, note that $\cosh^2(t)=1+\sinh^2(t)$, so $\cosh^3(t)=\cosh(t)+\sinh^2(t)\cosh(t)$.

Now $$\int \cosh(t)\,dt=\sinh(t)+C$$ and $$\int \cosh(t)\sinh^2(t)\,dt=\frac{\sinh^3(t)}{3}+C,$$ since $\frac{d}{dt}\sinh(t)=\cosh(t)$.

To parameterise without hyperbolics, use $(x,y)=(\sec t,\tan t)$.

Answer 2

The curve $\gamma$ going from ${\bf p}=(\sqrt{5},2)$ to ${\bf q}=(\sqrt{10},-3)$ is an arc of a hyperbola. It can be parametrized using $y$ as parameter, i.e., viewing $\gamma$ as a graph $x=\sqrt{1+y^2}$. I'm introducing nevertheless a $t$ here to make things clearer: $$-\gamma:\quad t\mapsto \bigl(x(t),y(t)\bigr):=\bigl(\sqrt{1+t^2}, \ t)\qquad(-3\leq t\leq2)\ .$$ Note that the chosen parametrization has interchanged the initial and endpoint of $\gamma$; therefore the minus sign. (You can see it this way: In this parametrization we always have $y'(t)=1>0$, whereas on the given $\gamma$ the tangent vector along $\gamma$ always points downwards.) We then have to compute $$-\int_{-3}^2 \bigl(x^2(t)x'(t)+ x^2(t)y'(t)\bigr)\>dt=-{1\over3}\int_{-3}^2\bigl(x^3(t)\bigr)'\>dt-\int_{-3}^2(1+t^2)\>dt\ .$$ Here the first integral on the RHS is for free, and the second is simple.