# Thread: Line integrals and a Force vector

1. ## 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...

2. ## Re: Line integrals and a Force vector

The first integral is easily calculated without the parameterization

$\begin{gathered}y = \sqrt {4 - {x^2}} ~~ \Rightarrow ~~dy = \frac{{ - x\,dx}}{{\sqrt {4 - {x^2}} }} \hfill \\[5pt] I_1= \int\limits_{L_1}(2 + x^2)\,dx + 3xy\,dy= \int\limits_2^0\!\left( {2 + {x^2} + 3x\sqrt {4 - {x^2}} \frac{{ - x}}{{\sqrt {4 - {x^2}} }}} \right)\!dx = 2\int\limits_2^0 (1-x^2)\,dx= \ldots = \frac{4}{3} \hfill \\ \end{gathered}$

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

$\begin{gathered}\frac{x-0}{-1-0} = \frac{y-2}{0-2}~~ \Leftrightarrow~~ y = 2x + 2~~ \Rightarrow~~ dy = 2\,dx \hfill \\ I_2= \int\limits_{L_2}(2 + {x^2})\,dx + 3xy\,dy = \int\limits_0^{ - 1} {(2 + x^2 + 3x(2x + 2)2)\,dx} } = \int\limits_0^{ - 1}(2 + 12x + 13x^2)\,dx= \ldots = - \frac{1}{3} \hfill \\\end{gathered}$

So, $I=I_1+I_2=\frac{4}{3}+\left(-\frac{1}{3}\right)=1$.

3. ## Re: Line integrals and a Force vector

Ah! I see now.
thanks DeMath!