# Integral along a curve

Given a curve $C$, parametrized by $\gamma (t)$ in the interval $[a, b]$, and a function $f$ defined in this interval, then:

If $f$ is a scalar field (so, if $f (x, y, z)$ belongs to the real $l$). Then: $$\displaystyle \int_C f \cdot dL= \int_a^b f(\gamma(t)) \ ||\gamma '(t)|| dt$$

That is to say, the integral of $f$ along the curve $C$ is the integral between $a$ and $b$ of the function composed with the parametrization, multiplied by the norm of the derivative of the parametrization.

## Procedure

1. Take the parametrization of the surface $C$, compute its derivative and the norm of the derivative.
2. Substitute $x$, $y$ and $z$ by $x (t)$, $y (t)$ and $z (t)$ in the function $f$, in accordance with the given parametrization.
3. Compute the resultant integral.

Compute the integral of the function $f(x, y, z)=z$ along the curve $\gamma (t) = (t\cos t, t \sin t , t)$ , which is a parametrization of a spiral, with a similar shape to a "tornado".

1. $\displaystyle \gamma '(t)=(\cos t-t \sin t, \sin t + \cos t, 1 )$, $\displaystyle ||\gamma '(t)||=\sqrt{(\cos t- t \sin t)^2+(\sin t +t \cos t)^2+1^2 }=\sqrt{2+t^2}$

2. $\displaystyle f(\gamma(t))=t$

3. $\displaystyle \int_0^{2\pi}f(\gamma(t)) ||\gamma '(t)|| \ dt=\int_0^{2\pi} t\cdot \sqrt{2+t^2} \ dt= \frac{1}{3}\Big[(2+t^2)^\frac{3}{2}\Big]_0^{2\pi}$

If F is a vector field $F(x,y,z)=(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$. Then, $$\displaystyle \int_CF \cdot dL=\int_a^bF(\gamma (t)) \cdot \gamma'(t) \ dt$$

that is to say, the integral of $f$ along the curve $C$ is the integral between $a$ and $b$ of the function composed with the parametrization, in doing the scalar product with the derivative of the parametrización.

## Procedure

1. Take the parametrization of the surface $C$ and compute its derivative.
2. Substitute $x$, $y$ and $z$ by $x (t)$, $y(t)$ and $z(t)$ in the function $f$, in accordance with the given parametrization.
3. Calculate the scalar product of the results of steps 1 and 2.
4. Compute the resultant integral.

Compute the integral of $F(x,y,z)=(-x^2,-3 \cdot x \cdot y+z,y)$ along the parametrized curve by $\gamma (t)=(\cos t, \sin t, 2)$, $t \in [0,2\pi]$

1) $\gamma '(t)=(-\sin t, \cos t ,0)$

2) $F(\gamma(t))=(-\cos^2t, -3\sin t \cdot \cos t+2, \sin t)$

3) $F(\gamma(t))\cdot \gamma'(t)=(-\cos^2t, -3\cdot \sin t \cdot \cos t +2, \sin t)\cdot(-\sin t, \cos t, 0)=$

$=\sin t \cdot cos^2t-3\cdot \sin t \cdot cos^2 t=2 \cdot cos t- 2 \cdot cos^2t \cdot \sin t$

4) $\displaystyle \int_0^{2\pi} 2\cdot \cos t -2 \cdot \cos^2 t \cdot \sin t \ dt=\Big[-\frac{2}{3}cos^3 t+2 \sin t\Big]_0^{2\pi}=\frac{2}{3}+0-\Big(-\frac{2}{3}\Big)-0=\frac{4}{3}$