Line Integral Evaluation

TheJason

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\(\displaystyle r(t) = \) 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 \(\displaystyle | = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}\)

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

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

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

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

Your Turn

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 \(\displaystyle 1, 1, 1,\).
 
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