HallsofIvy said:You are making the substitution x= a sin\theta but then your integral has both x and \theta. That's not right.
However, I would advise using the parametric equations x= a sin(\theta), y= b cos(\theta) rather than that complicated equation.
genericusrnme said:your handwriting is incredibly neat, good show!
emc92 said:I'm not sure I understand what you mean by the parametric equations.. how does that fit into what I already have?
Dick said:You were doing alright until after you drew the triangle. But you want to get rid of all of the x's in the thing you are integrating. And you never used dx=a cos(θ) dθ, probably because you weren't writing the dx in the integration. You need that.
emc92 said:oh darn! i did forget. okie well, now that I have dθ in there and i changed x^2, i still have a really ugly equation.. what should i do next?
emc92 said:i completely reworked it, and it looks so much better now! lol.
now i have integral from 0 to a of sqrt(1- (b^2/a^2)cos(θ))
emc92 said:right. k = 1-(b^2/a^2). did i cancel sin^2(θ) instead of cos(θ)?
emc92 said:i've completely messed up. i don't know what to use for substitutions.. i always end up in the same place. and i don't know where dx = a cos(θ) dθ fits into all of this.
sorry I'm a lot confused!
emc92 said:but if k = 1 - (b^2/a^2) and under the radical says 1-ksin^2(θ), shouldn't the end result under the radical, when expanded, be 1 - (sin(θ))^2- (b^2/a^2)(sin(θ))^2?
and where does the 4 outside the integral come from?