1. ## [SOLVED] Line integral

I must evaluate $\displaystyle I=\int_C F dr$ where $\displaystyle C$ is described by $\displaystyle r(t)=(t^2,t^3)$, $\displaystyle 0\leq t\leq 1$. $\displaystyle F(x,y)=(e^x,xy)$.
My attempt : $\displaystyle |r'(t)|=\sqrt{4t^2+9t^4}$.
$\displaystyle I=\int_0^1 (e^{t^{2}}+t^5)\sqrt{4t^2+9t^4}dt$. I don't think I've made it right... Seems hard to solve.

2. Originally Posted by arbolis
I must evaluate $\displaystyle I=\int_C F dr$ where $\displaystyle C$ is described by $\displaystyle r(t)=(t^2,t^3)$, $\displaystyle 0\leq t\leq 1$. $\displaystyle F(x,y)=(e^x,xy)$.
My attempt : $\displaystyle |r'(t)|=\sqrt{4t^2+9t^4}$.
$\displaystyle I=\int_0^1 (e^{t^{2}}+t^5)\sqrt{4t^2+9t^4}dt$. I don't think I've made it right... Seems hard to solve.
i believe you want $\displaystyle \oint_C F \cdot dr$, as in the dot product of the vector field $\displaystyle F$ and the vector $\displaystyle dr$

3. F(x,y) is a vector field, not a function.

4. Originally Posted by Random Variable
F(x,y) is a vector field, not a function.
Ah ok, then I should read how to proceed.

Originally Posted by Jhevon
i believe you want $\displaystyle \oint_C F \cdot dr$, as in the dot product of the vector field $\displaystyle F$ and the vector $\displaystyle dr$
I think you're right although the exercise didn't precise any $\displaystyle \oint$ but an $\displaystyle \int_C$.

5. Originally Posted by arbolis
Ah ok, then I should read how to proceed.

I think you're right although the exercise didn't precise any $\displaystyle \oint$ but an $\displaystyle \int_C$.
yes, i misused the symbol here. textbooks might not put that integral sign unless they want to specify it is a closed curve/contour, but i like the symbol

6. $\displaystyle \int_{C} F \cdot dr = \int_{a}^{b} \Big(F(r(t)) \cdot r'(t)\ \Big)dt$

$\displaystyle \int_{0}^{1} \Big((e^{t^2}, t^{5}) \cdot (2t, 3t^2)\Big)dt$

$\displaystyle \int_{0}^{1} (2te^{t^{2}} + 3t^{7}) \ dt$