Evaluate the line integral of the given vector field The formula I'm using is $\int_C \mathbf F \bullet d\mathbf r$
The given vector field is
$$F(x,y) = \frac{x}{(x^2 + y^2)^{3/2}}\vec i +\frac{y}{(x^2 + y^2)^{3/2}} \vec j$$
and the curve C given parametrically by
$$r(t)=e^{2t}\cos(3t)\vec i + e^{2t}\sin(3t)\vec j,\space 0\leq t\leq 2\pi$$
$$r^{'}(t)= -3e^{2t}\sin(3t)+2e^{2t}\cos(3t)\vec i + 2e^{2t}\sin(3t)+3e^{2t}\cos(3t)\vec j$$
The integral will then be
$$\int_0^{2\pi}\left<\frac{e^{2t}\cos(3t)}{(e^{2t}\cos(3t))^2 + (e^{2t}\sin(3t))^2)^{3/2}}+ \frac{e^{2t}\sin(3t)}{(e^{2t}\cos(3t))^2 + (e^{2t}\sin(3t))^2)^{3/2}}\right> \bullet \left<-3e^{2t}\sin(3t)+2e^{2t}\cos(3t), \space2e^{2t}\sin(3t)+3e^{2t}\cos(3t)\right>dt$$
After working that whole thing out I got
$$\int_0^{2π} \frac{2}{e^{2t}}dt=1-\frac{1}{e^{4\pi}}$$
Am I correct?
 A: $F(x,y)$ is a scalar valued function.
We don't use an inner product, we use the magnitude of $dr$
$\int F(x(t),y(t))\|r'(t)\|\ dt$
$r'(t) = e^{2t} \langle 2\cos 3t - 3\sin 3t, 2\sin 3t + 3\cos 3t\rangle$
$\|r'(t)\| = \sqrt{2^2 + 3^2} e^{2t}$
$\int_0^{2\pi} \frac {e^{2t}(\sin t + \cos t)}{e^{3t}} e^{2t}\sqrt 13 \ dt\\
\sqrt {13} \int_0^{2\pi} e^{t}(\sin t + \cos t) \ dt\\
\sqrt {13} (\frac {e^t(\cos 3t + 3\sin 3t)}{10} + \frac {e^t(\sin 3t - 3\cos 3t)}{10})|_0^{2\pi}\\
\frac {\sqrt {13}\ e^t}{10}(4\sin t - 2\cos t)|_0^{2\pi}\\
\frac {\sqrt {13} (1-e^{2\pi})}{5}\\
$
A: Your working is correct. However I am not sure if you are aware that for conservative vector field, there is a much easier way for you to get to the answer. The line integral will only depend on the end points and not on the chosen path. Here is details for this problem -
$F(x,y) = \frac{x}{(x^2 + y^2)^{3/2}}\vec i +\frac{y}{(x^2 + y^2)^{3/2}} \vec j$ is a conservative vector field.
Please note the potential function for the vector field is
$f(x,y) = -\frac{1}{\sqrt{x^2 + y^2}} + c \,$ where $c$ is a constant.
Just to confirm, you can see  $\frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j \,$ is your vector field.
So $\int_C \vec{F}  \cdot{d\vec{r}} = \int_C \nabla {f} \cdot {d\vec{r}} = f(\vec{r}(2\pi)) - f(\vec{r}(0))$.
Now $\vec{r}(0) = (1, 0) \,$ and $\vec{r}(2\pi) = (e^{4\pi}, 0) \,$
So our line integral is simply $ = f(e^{4\pi}, 0) - f(1,0) = \displaystyle 1 - \frac{1}{e^{4\pi}}$.
