How Do You Calculate This Complex Line Integral?

[aruwin]

I have no idea how to even start with this problem. I know the basics but this one just gets complicated. Please guide me!


Find the line integral:
∫C {(-x^2 + y^2)dx + xydy}
When 0≤t≤1 for the curved line C, x(t)=t, y(t)=t^2
and when 1≤t≤2, x(t)= 2 - t , y(t) = 2-t.
Use x(t) and y(t) and C={(x(t),y(t))|0≤t≤2}
Help!

 

[DonAntonio]

I have no idea how to even start with this problem. I know the basics but this one just gets complicated. Please guide me!


Find the line integral:
∫C {(-x^2 + y^2)dx + xydy}
When 0≤t≤1 for the curved line C, x(t)=t, y(t)=t^2
and when 1≤t≤2, x(t)= 2 - t , y(t) = 2-t.
Use x(t) and y(t) and C={(x(t),y(t))|0≤t≤2}
Help!

x(t)=t\,\,,\,y(t)=t^2\Longrightarrow x'(t)=1\,\,,\,y'(t)=2t\,\,,\,\,t\in [0,1]\Longrightarrow \oint_{C_1}(-x^2+y^2)dx+xydy=\int_0^1\left[(-t^2+t^4)\,dt+t^3(2tdt)\right]=
=\int_0^1(3t^4-t^2)dt=\frac{3}{5}-\frac{1}{3}=\frac{4}{15}
and now you do the other part of the curve, with x(t)=2-t\,\,,\,y(t)=2-t\,\,,x'(t)=y'(t)=-1\,\,,\,t\in [1,2], following

the argument as above.

DonAntonio

 

[aruwin]

x(t)=t\,\,,\,y(t)=t^2\Longrightarrow x'(t)=1\,\,,\,y'(t)=2t\,\,,\,\,t\in [0,1]\Longrightarrow \oint_{C_1}(-x^2+y^2)dx+xydy=\int_0^1\left[(-t^2+t^4)\,dt+t^3(2tdt)\right]=
=\int_0^1(3t^4-t^2)dt=\frac{3}{5}-\frac{1}{3}=\frac{4}{15}
and now you do the other part of the curve, with x(t)=2-t\,\,,\,y(t)=2-t\,\,,x'(t)=y'(t)=-1\,\,,\,t\in [1,2], following

the argument as above.

DonAntonio

Oh, got it! Well, one important question, how do you know if it's a scalar or vector? Because if it's scalar then we must find the magnitude of the derivatives. In this case, you didn't find the magnitude which means it's a vector but how do you know it is a vector?

 

[HallsofIvy]

You start by knowing the difference between a "scalar" and a "vector". Since there was no mention of a vector in this problem, do you have a specific problem where that question might arise?

"Because if it's scalar then we must find the magnitude of the derivatives."
I've never heard of such at thing. The problem here was entirely a "scalar" problem and there is no finding the magnitudes of anything.