first, the formula for the length of the parametric curve is not the formula for area. Not even close.
I guess if worse comes to worse, I guess I can always convert the parametric equation into Cartesian and then integrate from there.
actually, thats pretty much what you are supposed to do.
The formula for area under a curve is generally
∫ y dx from x = a to x = b
for parametric curves, we do the same thing, only we replace y with y(t) and dx with x'(t) dt.
So suppose our parametric function is y = t^2 x = ln(t)
we have then that dx = 1/t dt and y = t^2, so we can substitute to get:
∫ y dx = ∫ (t^2)1/t dt = ∫ t dt
the only important thing here, is that if you are integrating from x = a to x = b, then you have to find what values of t correspond to these values of x. Here, we have x = ln(t), so if x = a, then we'd have that t = e^a. Likewise, for x = b, we'd have t = e^b.
So your final integral is: ∫ t dt from t = e^a to e^b
Note, however, that they will probably give you the limits of integration in terms of t, rather than x, which simplifies things greatly.
Also remember that sometimes we like to find the area between a graph and the y axis, this way:
∫ x dy. If you were to do this parametrically, you'd do the same this as above, only you'd have to differentiate the y function instead.
LAST bit of advice, you need to be careful about what sort of parametric curve you are playing with because they can loop back over themselves. If this sort of thing happens, you will (a) probably be given a graph, and (b) might need to break up the integration over a few inteverals. For instance, if the graph looks like a circle, you would only find the area under the lower half of the circle.
Anyway, hope I'm not dumping too much info on you at once, these are just things you should be wary of. They like to trick you! They TRY to trick you!
