# Find $\int_{L}\overrightarrow{F} \cdot d\overrightarrow{r}$

Question $\def\vec{\overrightarrow}$Let$$\vec{F} = \left(3+2xy\right)\hat{i}+\left(x^{2}-3y^{2}\right)\hat{j}$$ and let $L$ be the curve$$\vec{r}=e^{t}\sin t \hat{i}+e^{t}\cos t \hat{j}.$$ Then find $\int_{L}\vec{F} \cdot d\vec{r}$.

MY Approach I know that i need to parameterize$$\vec{F} = \left(3+2xy\right)\hat{i}+\left(x^{2}-3y^{2}\right)\hat{j},$$ but i don't know How TO DO THAT.

x= $e^{t}sint$ , y=$e^{t}cost$

$\Longrightarrow$$x^{2}+y^{2}=e^{2t}$ .L is boundary of a circle

I have done questions where curve is f$\left(x,y\right)$and vector field is also function of $x$ and $y$.

• You don't need parameterization.
– Rab
Feb 9, 2018 at 8:53
• @RabMakh Should I try to use greens theorem Feb 9, 2018 at 8:58
• Its a gradient field, so calculate the value of $f$ at the final and initial points alone Feb 9, 2018 at 9:19
• @MohanSharma, Green's Theorem works for closed loops. The curve isn't closed in this case. Feb 21, 2019 at 8:58

Since the curl of $F$ is zero, $F$ must be a gradient field, that means there is a function $\phi$ which satisfies $F=-\nabla \phi$. From the equations \begin{align} 3+2xy=-\partial_x\phi\\ x^2-3y^2=-\partial_y\phi \end{align} we can calculate $\phi(x,y)=-3x-x^2y+y^3$. Finally it's just \begin{align} \int_L F\cdot \mathrm{d}r=\int_L -\nabla\phi \cdot\mathrm{d}r=[-\phi]^{r(\pi)}_{r(0)} \end{align}
• It's also possible to write $F=\nabla \phi$. Furthermore $\phi$ is not unique. You can add any constant to $\phi$. Dec 18, 2021 at 9:11