# Finding limits of a line integral with vector fields

A quick question about limits on a line integral involving vector fields.

1. Evaluate the line integral $\int_CF\cdot\mathrm ds$ where $$F(x,y)=(e^x\sin y+3y,e^x\cos y+2x-2y)$$ and $C$ is the ellipse $4x^2+y^2=4$ choosing the counterclockwise direction. (2 points)

I know that the parametrization of this curve is the following

\begin{align} r(t) &= [\cos(t), 2 \sin(t)]\\ r'(t) &= [-\sin(t), 2 \cos(t)] \end{align}

and we have our $F(r(t)) = F(x(t), y(t))$ $$F(r(t)) = e^{\cos(t)}\sin(2\sin(t))+6\sin(t), e^{\cos(t)}\cos(2\sin(t)) +2\cos(t)-4\sin(t)$$

and so by brute force we have the formula for the line integral $$\int_?^? F(r(t)) \cdot r'(t) \,\textrm{d}t$$

What would my limits be in this case? A wild guess would be 0 to $2\pi$

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Hint: Holy crow, $F(r(t))\cdot r'(t)$ looks like a hideous function to integrate! Use a handy-dandy theorem involving a way to write a line integral as something else (that seems completely unrelated to line integrals). – user5137 Nov 9 '11 at 7:08
yeah but we havn't gotten to stokes or greens theorems yet which is what I suspect I will use to make these easier – Tyler Hilton Nov 9 '11 at 7:12

The easiest way to determine the limits of a line integral is simply to look at the function $r(t)$. Since your curve $C$ is an ellipse, your wild guess is right, because $r(t)$ runs over $C$ only once when $r(t)$ runs from $0$ to $2 \pi$. The limits do not depend on the vector field $F$, they must only be such that $r(t)$ runs over $C$ once when $t$ goes over the chosen domain of $r$. What I mean by "once" is that for instance had we chosen $0$ and $4 \pi$ for the limits, you would run over $C$ twice.
I just needed confirmation. Thanks. I got $-2\pi$ as my answer. Sound correct? – Tyler Hilton Nov 9 '11 at 7:15
I'll have a wild guess and believe that $F(x,y) = (e^x \sin y + 3y,e^x \cos y + 2x - 2y)$? – Patrick Da Silva Nov 9 '11 at 7:25