Evaluate the line integral
β«π(π₯, π¦, π§) Γ β π ,
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].Β
"\\int {u(x,y,z)dr = } \\int\\limits_C {{y^2}} dx + xdy + zdz"
"y = {e^x} \\Rightarrow dy = {e^x}dx"
"z = {e^x} \\Rightarrow dz = {e^x}dx"
Then
"\\int\\limits_C {{y^2}} dx + xdy + zdz = \\int\\limits_0^1 {{e^{2x}}dx + x{e^x}} dx + {e^x} \\cdot {e^x}dx = \\int\\limits_0^1 {2{e^{2x}}dx + } \\int\\limits_0^1 {x{e^x}} dx = \\left. {{e^{2x}}} \\right|_0^1 + \\int\\limits_0^1 {x{e^x}} dx = {e^2} - {e^0} + \\int\\limits_0^1 {x{e^x}} dx"
Since
"\\int\\limits_0^1 {x{e^x}} dx = \\left| {\\begin{matrix}\n{u = x}&{dv = {e^x}dx}\\\\\n{du = dx}&{v = {e^x}}\n\\end{matrix}} \\right| = \\left. {x{e^x}} \\right|_0^1 - \\int\\limits_0^1 {{e^x}dx = } \\left. {x{e^x}} \\right|_0^1 - \\left. {{e^x}} \\right|_0^1 = e - 0 - e + {e^0} = 1"
Then
"\\int\\limits_C {{y^2}} dx + xdy + zdz = {e^2} - {e^0} + \\int\\limits_0^1 {x{e^x}} dx = {e^2} - 1 + 1 = {e^2}"
Answer: "{e^2}"
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