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