Thread: Line integrals and a Force vector

Hey everyone,
I have to evaluate F=<2+x²,3xy> where a curve is formed by moving along x² + y² = 4 from (2,0) to (0,2)
then along the line segment from (0,2) to (-1,0)

So far I've parameterized the first curve as r(t)=<2cost(t), 2sin(t)>, taking the derivative of that to get my tangent vector.
Then I parameterized F(r(t)) to get F(r(t))=<2+ 4cos(t)^2, 12cos(t)sin(t)>
Dotted my new F and tangent vector to get 16cos(t)²sin(t) - 4sin(t) and took the integral from 0 to pi/2
For this I got: -28/3, wolfram got 4/3

For the second curve I set r(t) = < -t, 2-2t> from 0 to 1 and got the tangent vector as <-1,-2>
My new F is F(r(t))=<2+t^2, 6t²-6t>
I dotted those to get 12t - 13t^2 - 2 and took the integral from 0 to 1
For this I got -1/3 and wolfram got -1/3

Adding these I got -29/3 and wolfram got 1, neither of these is what it's supposed to be, I guess the answer is -17/3 but I can't find how to get that anyway!
All help is greatly appreciated! Really really confused...

The first integral is easily calculated without the parameterization



To calculate the integral along the second line, first find the equation of the line (straight line)



So, .

Ah! I see now.
thanks DeMath!