VC4.11

Along the given curve C,we have $r=xi+yj+zk=2t^2i+tj+t^3k\,$

Also, $A=(2t+3)i+2t^5j+(t^4-2t^2)k\,$ [On putting the values of x,y,z]

From the first equation, $\frac{dr}{dt}=4ti+j+3t^2k\,$

Therefore, $\int_{C}A\cdot\,dr=\int_{C}[A\cdot{\frac{dr}{dt}}]\,dt\,$

= $\int_{0}^{1}[(2t+3)i+2t^5j+(t^4-2t^2)k]\cdot(4ti+j+3t^2k)\,dt\,$

= $\int_{0}^{1}[4t(2t+3)+2t^5+3t^2(t^4-2t^2)dt\,$

= $\int_{0}^{1}(8t^2+12t+2t^5+3t^6-6t^4)\,dt\,$

= $[\frac{8t^3}{3}+6t^2+\frac{t^6}{3}+\frac{3t^7}{7}-\frac{6t^5}{5}]_{0}^{1}\,$

= $\frac{8}{3}+6+\frac{1}{3}+\frac{3}{7}-\frac{6}{5}=\frac{864}{105}=\frac{288}{35}\,$