Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, where c is given by r(t), −1 ≤ t ≤ 1.

With[tex]\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k[/tex]we have[tex]\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt[/tex]The vector field evaluated over this parameterization is[tex]\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k[/tex]so the line integral is[tex]\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt[/tex][tex]=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4[/tex]