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, .
Hey everyone,
I have to evaluate F=<2+x^{2},3xy> where a curve is formed by moving along x^{2} + y^{2} = 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)^{2}sin(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, 6t^{2}-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...