Answer to Question #269465 in Calculus for Margreta

Question #269465

Evaluate the line integral

βˆ«π’–(π‘₯, 𝑦, 𝑧) Γ— ⅆ𝒓 ,

where 𝒖(π‘₯, 𝑦, 𝑧) = (𝑦^2 , π‘₯, 𝑧) and the curve π‘ͺ is described by 𝒛 = 𝑦 = 𝑒 π‘₯ with π‘₯ ∈ [0,1].Β 



1
Expert's answer
2021-11-24T17:22:21-0500

"\\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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