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Dear all,
I have a question regarding the computation of the area of an ellipse. The parametric form of the ellipse with axes a and b is
$$x(t) = a\cos{(t)}, \ \ \ y(t) = b\sin{(t)} $$
Using this to evaluate the area of the ellipse, usually one takes one halve or one quarter of the ellipse and uses symmetry. E.g.,
$$\frac{1}{4} \text{area} = \int_{x=0}^{x=a} y(x) dx = -ab \int_{t=\frac{\pi}{2}}^{t = 0} \sin^2{(t)} dt = \frac{\pi ab}{4} $$
My question is as follows: if one does the calculation as follows,
$$\text{area} = ab \int_{t=0}^{t = 2\pi} \sin^2{(t)} dt = \pi ab $$
one obtains the right answer. However, one integrates from one point on the ellips to the very same point, and the integrand is also not positive for half of the integral, so I can see that integration from t =0 to t = 2 pi is not allowed. Is it then a coincidence that one still obtains the right answer?
I have a question regarding the computation of the area of an ellipse. The parametric form of the ellipse with axes a and b is
$$x(t) = a\cos{(t)}, \ \ \ y(t) = b\sin{(t)} $$
Using this to evaluate the area of the ellipse, usually one takes one halve or one quarter of the ellipse and uses symmetry. E.g.,
$$\frac{1}{4} \text{area} = \int_{x=0}^{x=a} y(x) dx = -ab \int_{t=\frac{\pi}{2}}^{t = 0} \sin^2{(t)} dt = \frac{\pi ab}{4} $$
My question is as follows: if one does the calculation as follows,
$$\text{area} = ab \int_{t=0}^{t = 2\pi} \sin^2{(t)} dt = \pi ab $$
one obtains the right answer. However, one integrates from one point on the ellips to the very same point, and the integrand is also not positive for half of the integral, so I can see that integration from t =0 to t = 2 pi is not allowed. Is it then a coincidence that one still obtains the right answer?