Evaluate the line integral β«π(π₯, π¦, π§) Γ β π πΆ , where π(π₯, π¦, π§) = (π¦ 2 , π₯, π§) and the curve πͺ is described by π = π¦ = π π₯ with π₯ β [0,1].
Since,z=exdz=exdxAnd,y=exdy=exdxTherefore,dr=(dx,dy,dz)=(dx,exdx,exdx)=(1,ex,ex)dxThen,β«u.dr=β«(y2,x,z).(1,ex,ex)dx=β«(e2x,x,ex).(1,ex,ex)dx=β«01(e2x+xex+e2x)dx=β«01(xex+2e2x)dx=[xexβex+e2x]01=e2Since,\\ z=e^x\\ dz=e^xdx\\ And,\\ y=e^x\\ dy=e^xdx\\ Therefore,\\ dr=(dx,dy,dz)=(dx,e^xdx,e^xdx)=(1,e^x,e^x)dx\\ Then,\\ \int u.dr\\ =\int(y^2,x,z).(1,e^x,e^x)dx\\ =\int(e^{2x},x,e^x).(1,e^x,e^x)dx\\ =\int_0^1(e^{2x}+xe^x+e^{2x})dx\\ =\int_0^1(xe^x+2e^{2x})dx\\ =[xe^x-e^x+e^{2x}]_0^1\\ =e^2Since,z=exdz=exdxAnd,y=exdy=exdxTherefore,dr=(dx,dy,dz)=(dx,exdx,exdx)=(1,ex,ex)dxThen,β«u.dr=β«(y2,x,z).(1,ex,ex)dx=β«(e2x,x,ex).(1,ex,ex)dx=β«01β(e2x+xex+e2x)dx=β«01β(xex+2e2x)dx=[xexβex+e2x]01β=e2
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