Question

Compute the scalar line integral \( \int_C fds \), where \( f \) abd \( C \) are as indicated.

\( f(x,y)=x+y;C \) is the perimeter of the square with vertices (0,0),(1,0),(1,1) and (0,1).
\( \)

Answer 1

solution 

we have \( f(x,y)=x+y;C \) is the perimeter of the square with vertices (0,0),(1,0),(1,1) and (0,1).

Let A=(0,0) B=(1,0),C=(1,1) and D=(0,1) and \( \ \begin{cases} x=x_0+at & \quad \\\ y=y_0+bt & \quad \ \end{cases} \ \)

Let \( C_1,C_2,C_3,C_4 \) be line segment of AB,BC,CD,DA, respectively,

For \( C_1=>r_1(t)=(t,0)=>r_1'(t)=(1,0)=>||r_1'(t)||=1,t\in [0,1] \)For \( C_2=>r_2(t)=(1,t)=>r_2'(t)=(0,1)=>||r_2'(t)||=1,t\in [0,1] \)For \( C_3=>r_3(t)=(1,1-t)=>r_3'(t)=(0,-1)=>||r_3'(t)||=1,t\in [0,1] \)For \( C_4=>r_4(t)=(1-t,0)=>r_4'(t)=(-1,0)=>||r_4'(t)||=1,t\in [0,1] \)\( =>\int_Cfds=\int_{C_1}fds+\int_{C_2}fds+\int_{C_3}fds+\int_{C_4}fds \)

                    \( =\int_0^1tdt+\int_0^1(1+t)dt+\int_0^1(2-t)dt+\int_0^1(1-t)dt=4 \)

answer 

Therefore, scalar line integral \( \int_Cfds=4. \)