Evaluate the line integral of →{v}=xyi+y^{2}zj+e^{2z}k over the curve C whose parametric representation is given by x=t^{2},y=2t,z=t,0<t<1.
Given that
v=xyi+y2zj+e2zkx=t2,y=2t,z=t,0<t<1.{v}=xyi+y^{2}zj+e^{2z}k\\ x=t^{2},y=2t,z=t,0<t<1.\\v=xyi+y2zj+e2zkx=t2,y=2t,z=t,0<t<1.
Now the line integral is
∫CVdrv(t)=xyi+y2zj+e2zk =2t3i+4t3j+e2tkdr=(2t,2,1)v(t)dr=2t(2t3)+2(4t3)+1(e2t) =4t4+8t3+e2t∴∫CV.dr=∫01(4t4+8t3+e2t)dt(45t5+84t4+12e2t)012310+12e2\int_C Vdr\\ {v(t)}=xyi+y^{2}zj+e^{2z}k\\ \space \space \space \space \space \space \space = 2t^3i+4t^3j +e^{2t}k\\ dr = (2t,2,1)\\ {v(t)}dr=2t(2t^3)+2(4t^3)+1(e^{2t})\\ \space \space \space \space \space \space \space \space \space \space \space \space = 4t^4+8t^3+e^{2t}\\ \therefore \int_C V.dr = \int_0^1(4t^4+8t^3+e^{2t})dt\\ (\frac{4}{5}t^5+\frac{8}{4}t^4+\frac{1}{2}e^{2t})_0^1\\ \frac{23}{10}+\frac{1}{2}e^2∫CVdrv(t)=xyi+y2zj+e2zk =2t3i+4t3j+e2tkdr=(2t,2,1)v(t)dr=2t(2t3)+2(4t3)+1(e2t) =4t4+8t3+e2t∴∫CV.dr=∫01(4t4+8t3+e2t)dt(54t5+48t4+21e2t)011023+21e2
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