Evaluate the line integral ∫𝒖(𝑥, 𝑦, 𝑧) × ⅆ𝒓 𝐶 , where 𝒖(𝑥, 𝑦, 𝑧) = (𝑦 2 , 𝑥, 𝑧) and the curve 𝑪 is described by 𝒛 = 𝑦 = 𝑒 𝑥 with 𝑥 ∈ [0,1].
Since,z=exdz=exdxAnd,y=exdy=exdxTherefore,dr=(dx,dy,dz)=(dx,exdx,exdx)=(1,ex,ex)dxThen,∫u.dr=∫(y2,x,z).(1,ex,ex)dx=∫(e2x,x,ex).(1,ex,ex)dx=∫01(e2x+xex+e2x)dx=∫01(xex+2e2x)dx=[xex−ex+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,∫u.dr=∫(y2,x,z).(1,ex,ex)dx=∫(e2x,x,ex).(1,ex,ex)dx=∫01(e2x+xex+e2x)dx=∫01(xex+2e2x)dx=[xex−ex+e2x]01=e2
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