Question

Find an expression as a definite integral for the length of an ellipse x²/a²+y²/b²= 1 by using the parametrization x=a sinφ, y= b cosφ and letting

                        $k=e=\sqrt{((a^{2}-b^{2})/a^{2})}$ you should get $\int_{0}^{u}\sqrt{1-k^{2}sin^{2}\phi }d\phi$

Answer 1

We have to find a define integral for the length of an ellipse x^2/a^2 + y^2/b^2 = 1 where x = a sin phi, y = b cos phi dx/d phi = a cos phi dy/d phi = -b sin phi Therefore length of ellipse is L = integral^25_0 squareroot (dx/d phi)62 + (dy/d phi)^2 d phi = integral^25_0 squareroot a^2 cos 2 phi + b^2 sin^2 d phi = integral^2 pi_0 squareroot a^2 (1 - sin^2 phi) + b^2 sin^2 phi d phi = 4 integral^pi/2_0 squareroot a^2 - a^2 sin^2 phi + b^2 sin^2 phi d phi = 4 integral^pi/2 squareroot a^2 - (a^2 - b^2) sin^2 phi d phi = 4a integral^pi/2 squareroot 1 - (a^2 - b^2/a^2) sin^2 phi d phi = 4a integral^pi/2_0 squareroot 1 - u^2 sin^2 phi d phi (Because K middot e = squareroot a^2 - b^2/a^2) Therefore L = 4a integral^pi/e_0 squareroot 1 - x^2 sin^2