# Line Integral Evaluation

#### TheJason

Staff member
Moderator
$$\displaystyle r(t) =$$ ti + tj + tk, $$\displaystyle (a \leq t \leq b)$$

$$\displaystyle \int_{C}f(x,y,z) \,ds = \int_{a}^{b} ,f(g(t),h(t),k(t))\,\,|v(t)| \,dt$$

Integrate $$\displaystyle f(x,y,z) = x - 6y^{2} + z\,\,\,$$ over the line segment $$\displaystyle C$$ joining the origin of $$\displaystyle 0,0,0$$ to the point $$\displaystyle 1, 1, 1,$$.

OK, the most simple (smooth) parameterization would be $$\displaystyle r(t) = |$$ ti + tj + tk,$$\displaystyle |\,(0 \leq t \leq 1)$$

Now, looking at the numbers going into $$\displaystyle v(t)| =$$| i + j + k | - we would make the calculation: $$\displaystyle (x_{2},y_{2},z_{2}) - (x_{1},y_{1},z_{1}) = (1,1,1) - (0,0,0) = (1,1,1)$$

Now, $$\displaystyle v(t)| = |$$ i + j + k $$\displaystyle | = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}$$

$$\displaystyle \int_{C}f(x,y,z) \,ds = \int_{0}^{1} ,f(t,t,t)\,\sqrt{3} \,dt$$

$$\displaystyle = \int_{0}^{1} (t - 6t^{2} + t)(\sqrt{3})\,dt$$

$$\displaystyle = [(\dfrac{t^{2}}{2} - \dfrac{6t^{3}}{3} + \dfrac{t^{2}}{2} ) - (\dfrac{t^{2}}{2} - \dfrac{6t^{3}}{3} + \dfrac{t^{2}}{2} ) ]\sqrt{3}$$

$$\displaystyle = [(\dfrac{(1)^{2}}{2} - \dfrac{6(1)^{3}}{3} + \dfrac{(1)^{2}}{2} ) - (\dfrac{(0)^{2}}{2} - \dfrac{6(0)^{3}}{3} + \dfrac{(0)^{2}}{2} ) ]\sqrt{3} = 2\,\sqrt{3}$$

Integrate $$\displaystyle f(x,y,z) = x - 5y^{3} + 7z\,\,\,$$ over the line segment $$\displaystyle C$$ joining the origin of $$\displaystyle 0,0,0$$ to the point $$\displaystyle 1, 1, 1,$$.