Answer:
Yes, you are correct!!!
The result is [tex]\int\limits^{1}_{0} {4t+1dt} \,=3\sqrt{11}[/tex]
Step-by-step explanation:
The given line integral is
[tex]\int\limits_C {x+y} \, ds[/tex], where C is the line segment from [tex](0,1,1)[/tex] to [tex](3,2,2)[/tex].
First we need to get the direction vector,
[tex]d=\:<\:3,2,2\:>-\:<\:0,1,1\:>[/tex]
[tex]d=\:<\:3,1,1\:>[/tex]
The parametric equation now becomes,
[tex]r(t)=\:<\:0,1,1\:>+\:t<\:3,1,1\:>[/tex].
[tex]\Rightarrow r(t)=3ti+(1+t)j+(1+t)k[/tex].
We parameterize the curve with the equations,
[tex]x=3t[/tex] and [tex]y=t+1[/tex]
[tex]ds=|r'(t)|dt[/tex]
[tex]\Rightarrow ds=|3i+j+k|dt[/tex]
[tex]\Rightarrow ds=\sqrt{3^2+1^2+1^2}dt[/tex]
[tex]\Rightarrow ds=\sqrt{11}dt[/tex].
The line integral now becomes,
[tex]\int\limits^{1}_{0} [{3t+(t+1)]\sqrt{11}dt} \,[/tex]
[tex]\sqrt{11}\int\limits^{1}_{0} {4t+1dt} \,=3\sqrt{11}[/tex]
