Question : Find the area bounded by and above the x-axis
By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis , i.e. , but of course you need to prove this with integral calculus, so:
, so you need the integral , again by symmetry, this time wrt the y-axis.
You may want to try the trigonometric substitution , to make things easier.
your set up should look like: 2 integral (from 0 to a) of sqrt(b^2 - (b^2)(x^2)/a^2) dx.
you do not need a trig substitution, simply factor out a b^2/a^2 term inside the square root and you will get sqrt[(b^2/a^2) (a^2 - x^2)]. sqrt(b^2/a^2) becomes b/a so you can factor that constant out of the integral. so now the integral should look like 2b/a integral (from 0 to a) of sqrt(a^2 - x^2) dx. now look at the integral, and geometrically it just means the area of a circle of radius "a" and since the limit is from 0 to a, the integral only wants 1/4 of the whole circle. so integral (from 0 to a) of sqrt(a^2 - x^2) dx is just (pi)(a^2)/4 since it's 1/4 of the area of the whole circle. then with the constant outside of the integral earlier, you get (2b/a)((pi)(a^2)/4)) = (pi)(a)(b)/2 and since the area of an ellipse is pi(a)(b), your result makes sense because you only wanted the top half of the ellipse.