Just to emphasize the point and to show you another technique often used to solve path integrals, here is another example in which an object descends along a helical path - like the shape of a spring. As a real-life analogy, maybe this represents some object traveling down a spiral ramp in a parking garage. [br][br]The easiest way to solve a path integral with a complicated path like this is to use parametric equations. The parametric equations for a helical path that starts at some height h look like this:[br][math]x(t)=r \cos(\omega t),[/math] [math]y(t)=r\sin(\omega t),[/math] [math]z(t)=h-t.[/math][br][br]This parametric function is plotted below, with the green arrows representing downward, uniform gravity everywhere which we assume must exist in this problem. If you want to see it in 3D and have red/cyan 3D glasses, click on the cube shape in the menu and then on what looks like glasses in the sub-menu. Play with the other controls and see what they do. You can also drag the plot with a mouse, and if you do so dynamically you can induce a spin.

We need to write the infinitesmal path vector [math]\vec{ds}[/math] in terms of dx, dy and dz. So we differentiate each of the above parametric expressions with respect to t. Keep in mind that t here is NOT time. It is just the commonly used parameter. After differentiating the parametric equations above, we find that[br] [br][center][math] [br]dx=-r\omega\sin(\omega t)dt, \\[br]dy=r\omega\cos(\omega t)dt, \\[br]dz=-dt. [/math][/center] [br]This means we can write [math]\vec{ds}=-r\omega\sin(\omega t)dt\hat{i}+r\omega\cos(\omega t)dt\hat{j}-dt\hat{k}.[/math] It is now in a form that can be entered into the path integral along with the gravitational force which will now be in the negative z direction: [math]\vec{F}_g=-mg\hat{k}.[/math] The work integral becomes: [br][center][math]W=\int_{t_i}^{t_f} (0\hat{i}+0\hat{j}-mg\hat{k})\cdot(-r\omega\sin(\omega t)\,dt\hat{i}+r\omega\cos(\omega t)\,dt\hat{j}-dt\hat{k}).[/math][/center] [br]This gives [math]W=\int_{t_i}^{t_f} mgdt=mg(t_f-t_i).[/math] We must be careful here to find the correct limits on the parametric variable t. If, for instance, we start at a height of 20m above the ground and descend to 0m, we look at the z(t) equation and see that z(0)=20m, when h=20m (since it was defined as initial height) and when t=0. Once t=20 we find z(20)=20-20=0m. So our limits on the parameter t are from 0 to 20. That means our integral becomes W=20mg=200J if m=1kg, or the same as just dropping a ball straight downward! [br][br]This is another example of the [b]path-independence[/b] that arises in gravitational fields. It just goes to show that gravity is a [b]conservative field[/b]. These two terms go hand-in-hand.