Curve integral of $f(x,y,z) = z$ over $\gamma(t) = ( \sqrt{2}t, e^t, e^{-t} )$

Question

What is the curve integral

$$\int_\gamma f ds$$

of $$f(x,y,z) = z$$ over $$\gamma(t) = (e^t, e^{-t}, \sqrt{2}t)$$

Answer 1

The parametric equations of the curve are $\gamma(t)=(e^t, e^{-t}, \sqrt{2} t)$ and therefore the velocity is $\dot \gamma(t) = (e^t,- e^{-t},\sqrt{2})$; the speed $|\dot \gamma(t)|=\sqrt{2+e^{2t}+e^{-2t}}=\sqrt{2(1+\cosh(2t))}$. This expression is simplified by noting that $\cosh(2t)=2 \cosh^2t-1$, so $|\dot \gamma(t)|=2\cosh t$, since $\cosh$ is always positive.

Now, if $t\in[a,b]$, $$ \int_\gamma f ds \equiv \int_{a}^b f(\gamma(t))|\dot \gamma(t)|dt, $$ and, since $f(\gamma(t))=z(t)=\sqrt{2}t$, $$ \int_\gamma f ds = 2\sqrt{2}\int_a^b t \cosh t\, dt = {\sqrt 2}\left(\int_a^b t e^t dt+ \int_a^b t e^{-t} dt\right). $$ These integrals can be evaluated using integration by parts: $$ \int_a^b t e^t dt = (te^t)\Big|_a^b - \int_a^b e^t dt = (be^b-a e^a)-(e^b-e^a) $$ $$ \int_a^b t e^{-t} dt = (-te^{-t})\Big|_a^b + \int_a^b e^{-t} dt = (-be^{-b}+a^{-a}) - (e^{-b}-e^{-a}) $$ putting everything back together $$ \int_\gamma f ds=2\sqrt{2} \left(b \sinh b- a \sinh a - \cosh b + \cosh a \right). $$

Answer 2

Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be a scalar field and $\gamma:I\to\mathbb{R}^{n}$ be a parametrization of a curve $\Gamma$ where $I$ denotes an interval $[a,b]$ ($a<b$). We have:

$$\int_{\gamma}f\text{d}s=\int_{a}^{b}f(\gamma(t))\Vert\gamma'(t)\Vert\text{d}t$$

Here, we have $f:\mathbb{R}^{3}\to\mathbb{R}:(x,y,z)\mapsto z$ and $\gamma(t)=(e^{t},e^{-t},\sqrt{2}t)$, so that

$$\Vert\gamma'(t)\Vert=\left\Vert\left(e^{t},-e^{-t},\sqrt{2}\right)\right\Vert=\sqrt{e^{2t}+e^{-2t}+2}=\sqrt{2(1+\cosh(2t))}$$

so that

$$\begin{align}\int_{a}^{b}f(\gamma(t))\Vert\gamma'(t)\Vert\text{d}t &=\int_{a}^{b}\sqrt{2}t\sqrt{2(1+\cosh(2t))}\text{d}t\\ \end{align}$$ Now, see that

$$1+\cosh(2t)=\frac{2+e^{2t}+e^{-2t}}{2}=\frac{(e^{t}+e^{-t})^{2}}{2}=2\left(\frac{e^{t}+e^{-t}}{2}\right)^{2}=2(\cosh(t))^{2}$$

so that the integral becomes:

$$\begin{align}\int_{a}^{b}f(\gamma(t))\Vert\gamma'(t)\Vert\text{d}t &=\int_{a}^{b}\sqrt{2}t\sqrt{2(1+\cosh(2t))}\text{d}t\\ &=\sqrt{2}\int_{a}^{b}t\sqrt{4(\cosh(t))^{2}}\text{dt}\\ &=2\sqrt{2}\int_{a}^{b}t\cosh(t)dt \end{align}$$

Now, integration by part yields:

$$\begin{align}\int_{a}^{b}f(\gamma(t))\Vert\gamma'(t)\Vert\text{d}t &=2\sqrt{2}\int_{a}^{b}t\cosh(t)dt\\ &=2\sqrt{2}\left([t\sinh(t)]_{a}^{b}-\int_{a}^{b}\sinh(t)\text{d}t\right)\\ &=2\sqrt{2}\left(b\sinh(b)-a\sinh(a)-\cosh(b)+\cosh(a)\right)\\ \end{align}$$