Intergation problem

Can you post what you did / the given answer?
from a back of the envelope sketch it should be ~1?

Can you post what you did / the given answer?
from a back of the envelope sketch it should be ~1?

out of curiosity how did you find the area to be 1 by sketching?

its fine i got the answer in the end i went wrong by just a silly error. when integrating -24sin^3t i split it inot sin^2t and sint then used trig identity to get it to 1-cos^2(t) x sint and all i did worng from here was foregot to times by -1.

Your right it was just shy of 1 but the exact form i got it in was some nasty surd 16-(26/3)(root3)

its fine i got the answer in the end i went wrong by just a silly error. when integrating -24sin^3t i split it inot sin^2t and sint then used trig identity to get it to 1-cos^2(t) x sint and all i did worng from here was foregot to times by -1.

Your right it was just shy of 1 but the exact form i got it in was some nasty surd 16-(26/3)(root3)

Must admit, Ive not worked it through parametrically (neither do I want to), but in x-y coords youre simply integrating
4 - x^2/9
from 3sqrt(3) to 6 (for the curved section) which aint hard. Similarly the intersection point is (3sqrt(3),1) and the grad of the tangent is -2sqrt(3)/3 whch gives the area of the triangle.

It's not very pretty, which is why I wasn't volunteering an answer earlier until someone else had done it first The important thing with the parametrics is to remember that in going from left-to-right, t goes from pi/6 to 0, so if you get that wrong there's a negative contribution coming in. I did think of doing the cartesian equivalent to check my answer, but ran out of energy / time and other things took priority

I hadnt done either originally, just a quick sketch, but the parametric description simply obfuscates things here.
Its a few lines to do it using the parabolic x-y representation.