Question : Find the area bounded by $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and above the x-axis
By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis $\displaystyle a,b$ , i.e. $\displaystyle \frac{\pi ab}{2}$, but of course you need to prove this with integral calculus, so:
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Longrightarrow a^2y^2=b^2(a^2-x^2)\Longrightarrow y=b\sqrt{1-\left(\frac{x}{a}\right)^2}$ , so you need the integral $\displaystyle \int_{-a}^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx=2b\int_0^a\sqrt{1-\left(\frac{x}{a}\right)^2}dx$ , again by symmetry, this time wrt the y-axis.
You may want to try the trigonometric substitution $\displaystyle x=a\sin t\,,\,0\le t\le \frac{\pi}{2}$ , to make things easier.
Tonio
Taking $\displaystyle x \ = \ a sint$ as $\displaystyle 0 \le t < \frac{ \pi}{2}$
= $\displaystyle 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \left( \frac{a \ sint}{a} \right)^2}$
= $\displaystyle 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sint )^2}$
= $\displaystyle 2b \int_{0}^{\frac{\pi}{2}} 1 - ( sint )^2$
= $\displaystyle 2b \int_{0}^{\frac{\pi}{2}} 1 - sint $
= $\displaystyle b \pi$...............Is this corect???
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.
you forgot a few things
x=asinz
dx=acosz dz
$\displaystyle 2b \int_{0}^{\frac{\pi}{2}} \sqrt{1 - ( sinz )^2}acoszdz$
$\displaystyle 2b \int_{0}^{\frac{\pi}{2}} \sqrt{(cosz)^2}acoszdz$
$\displaystyle 2b \int_{0}^{\frac{\pi}{2}} cosz a coszdz$
$\displaystyle 2b \int_{0}^{\frac{\pi}{2}} a(cosz)^2$
$\displaystyle 2ab \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2z}{2}$
$\displaystyle ab [\int_{0}^{\frac{\pi}{2}} (1)+\int_{0}^{\frac{\pi}{2}}cos2z]$
$\displaystyle ab [\frac{\pi}{2}+\frac{1}{2}(sin(\frac{2\pi}{2})-0)]$
$\displaystyle ab [\frac{\pi}{2}]=\frac{ab\pi}{2}$