- #1

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First, using rectangular coordinates,

1/2s=S{[1+(f'(x))^2]^(1/2)}dx taken from x=-a to x=a

Since, y^2=b/a(a^2-x^2)

2y*y'=-2bx/a

y'=-bx/(ay)

[f'(x)]^2=(x^2)/(a^2-x^2)

At this point, I'm already uncomfortable because b is no longer in the equation, and clearly the circumference should depend on both a and b.

Next, using parametrics, I have

s=S[(bcosx)^2+(asinx)^2]^(1/2)dx from x=0 to x=2pi

This integral shows more promise for finding the answer. I expect the answer to be C=pi(a+b) simply because this would reduce to C=(2pi)r for the case when a=b. I've tried manipulating the second integral in every way possible to fit in trig substitution but it just won't work. It doesn't look like integration by parts will help. Of course, there's always the possiblity that these integrals do not give the circumference of an ellipse at all. Even so, it would be satisfying to find an answer.

Can someone give me a hint?