Hello,
I'm trying to prove the area of ellipse formula via integration and I have gotten close but I've got a negative sign on my final answer which I know shouldn't be there.
I reaaranged the equation of ellipse to get the top half of the ellipse as;
y=ab(a2−x2)
Giving the area of the top half as;
ab∫−aaa2−x2dx
I didn't know how to approach this integral at first so I tried one round of IBP and got:
xa2−x2+∫a2−x2x2dx
I realised that a trig sub would be better got the remaining integral down to the following by letting x = asinu
(a2∫(sin2θ)dθ
Then once integrated and subbing x back in I got
ab(xa2−x2+a2(21arcsin(ax)−41sin(2arcsin(ax))))
But then when evaluating the inner bracket from -a to a I got
A=−2abπ
However I have realised now that if I used arcsin(-1)=-pi/2 instead of arcsin(-1) = 3pi/2 I would have got a positive answer. But shouldn't these two angles give the same result?
I'm trying to prove the area of ellipse formula via integration and I have gotten close but I've got a negative sign on my final answer which I know shouldn't be there.
I reaaranged the equation of ellipse to get the top half of the ellipse as;
y=ab(a2−x2)
Giving the area of the top half as;
ab∫−aaa2−x2dx
I didn't know how to approach this integral at first so I tried one round of IBP and got:
xa2−x2+∫a2−x2x2dx
I realised that a trig sub would be better got the remaining integral down to the following by letting x = asinu
(a2∫(sin2θ)dθ
Then once integrated and subbing x back in I got
ab(xa2−x2+a2(21arcsin(ax)−41sin(2arcsin(ax))))
But then when evaluating the inner bracket from -a to a I got
A=−2abπ
However I have realised now that if I used arcsin(-1)=-pi/2 instead of arcsin(-1) = 3pi/2 I would have got a positive answer. But shouldn't these two angles give the same result?
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