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Find the work done in moving a particle in a force field given by $\bar{F}=3xy\ \hat{i}-5z\hat{j}+10x\hat{k}$ along the curve x=t2+1, y=2t2, z=t3 from t=1 to t=2
$\\[2ex] r=xi+yj+zk \displaystyle \\[2ex] dr=dxi+dyj+dzk\\[2ex] dr=2t\ dt\ i+4t\ dt\ j+3t^2\ dt\ k \displaystyle \\[2ex] \bar{F}=\ 3xyi-5zj+10xk=3\ \left(t^2+1\right)\left(2t^2\right)i-5t^3j+10\left(t^2+1\right)k \displaystyle \\[2ex] \bar{F}\ .dr\ \ =6t^2\left(t^2+1\right)2t\ dt-20t^4dt+30t^2\left(t^2+1\right)dt \displaystyle \\[2ex] \bar{F}=\left(12t^5+10\ t^4+12t^3+30t^2\ \right)dt\ \displaystyle$
$\\[2ex] Work\ done=\ \int_c\bar{F}.dr\\[2ex] \displaystyle =\int_1^2\left(12t^5+10\ t^4+12t^3+30t^2\ \right)dt \displaystyle \\[2ex] ={\left[2t^6+2t^5+3t^4+10t^3\right]}_1^2\ \displaystyle \\[2ex] =\left[2\times{}\ 2^6+2\times{}2^5+3\times{}2^4+110\times{}2^3\right]-\left[2+2+3+10\right] \displaystyle \\[2ex] =303 \displaystyle \\[2ex] Hence\ the\ work\ done\ by\ the\ \bar{F}=303\ units.\ \displaystyle$