Answer to Question #269465 in Calculus for Margreta

Question

Evaluate the line integral

∫𝒖(𝑥, 𝑦, 𝑧) × ⅆ𝒓 ,

where 𝒖(𝑥, 𝑦, 𝑧) = (𝑦^2 , 𝑥, 𝑧) and the curve 𝑪 is described by 𝒛 = 𝑦 = 𝑒 𝑥 with 𝑥 ∈ [0,1]. 

Answer 1

$\int {u(x,y,z)dr = } \int\limits_C {{y^2}} dx + xdy + zdz$

$y = {e^x} \Rightarrow dy = {e^x}dx$

$z = {e^x} \Rightarrow dz = {e^x}dx$

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

$\int\limits_0^1 {x{e^x}} dx = \left| {\begin{matrix} {u = x}&{dv = {e^x}dx}\\ {du = dx}&{v = {e^x}} \end{matrix}} \right| = \left. {x{e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}dx = } \left. {x{e^x}} \right|_0^1 - \left. {{e^x}} \right|_0^1 = e - 0 - e + {e^0} = 1$

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: ${e^2}$