Line integral convert to polar coordinates

[Scott77]

Homework Statement

I need to find the work done by the force field:
$$\vec{F}=(5x-8y\sqrt{x^2+y^2})\vec{i}+(4x+10y\sqrt{x^2+y^2})\vec{j}+z\vec{k}$$
moving a particle from a to b along a path given by:
$$\vec{r}=\frac{1}{2}\cos(t)\vec{i}+\frac{1}{2}\sin(t)\vec{j}+4\arctan(t)\vec{k}$$

The Attempt at a Solution

So I set up my line integral:
$$\vec{F}(\vec{r}(t))=(\frac{5}{2}\cos(t)-2\sqrt{2}\sin(t))\vec{i}+(2\cos(t)+\frac{5\sqrt{2}}{2}\sin(t))\vec{j}+(4\arctan(t))\vec{k}$$
$$\vec{r'}(t)=\left(-\frac{1}{2}\sin(t)\right)\vec{i}+\left(\frac{1}{2}\cos(t)\right)\vec{j}+\left(\frac{4}{t^2+1}\right)\vec{k}$$
$$\int_0^{1}\left(-\frac{5+5\sqrt{2}}{4}\sin(t)\cos(t)+\sqrt{2}\sin^2(t)+\cos^2(t)+\frac{16\arctan(t)}{t^2+1}\right)\; \text{d}t=5.86436$$

I have left a lot of steps out, it gets messy! Could this problem be solved by reducing the integral to polar coordinates?

 

[SteamKing]

Homework Statement

I need to find the work done by the force field:
$$\vec{F}=(5x-8y\sqrt{x^2+y^2})\vec{i}+(4x+10y\sqrt{x^2+y^2})\vec{j}+z\vec{k}$$
moving a particle from a to b along a path given by:
$$\vec{r}=\frac{1}{2}\cos(t)\vec{i}+\frac{1}{2}\sin(t)\vec{j}+4\arctan(t)\vec{k}$$

The Attempt at a Solution

So I set up my line integral:
$$\vec{F}(\vec{r}(t))=(\frac{5}{2}\cos(t)-2\sqrt{2}\sin(t))\vec{i}+(2\cos(t)+\frac{5\sqrt{2}}{2}\sin(t))\vec{j}+(4\arctan(t))\vec{k}$$
$$\vec{r'}(t)=\left(-\frac{1}{2}\sin(t)\right)\vec{i}+\left(\frac{1}{2}\cos(t)\right)\vec{j}+\left(\frac{4}{t^2+1}\right)\vec{k}$$
$$\int_0^{1}\left(-\frac{5+5\sqrt{2}}{4}\sin(t)\cos(t)+\sqrt{2}\sin^2(t)+\cos^2(t)+\frac{16\arctan(t)}{t^2+1}\right)\; \text{d}t=5.86436$$

I have left a lot of steps out, it gets messy! Could this problem be solved by reducing the integral to polar coordinates?

What makes you think you didn't use polar coordinates to get your result?