<s>[INST] Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx$$
I can't get the answer $\pi ab$
 [/INST] from the equation of the ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
$$  y = \sqrt{\frac{a^2b^2 - x^2b^2}{a^2}}  = \frac{b}{a}\sqrt{a^2-x^2}$$
since proceding with change of variable in integration is, using x = a*cos(Θ)::
$$ 
\\
\sqrt{a^2 - x^2} = \sqrt{a^2-a^2cos^2\theta} =asin\theta
\\
\int_{}^{}\sqrt{a^2 - x^2} = a\int_{}^{}sin\theta \
dx \
= a^2\int_{}^{}sin^2\theta d\theta = a^2\int_{}^{}\frac{1 - cos 2\theta}{2} d\theta = \frac{a^2\theta}{2}-\frac{a^2sin^22\theta}{2}+C = \frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x\sqrt{1-\frac{x^2}a^{}}}{a})+C
$$
so the area in the first quadrant is:
$$ \frac{b}{a} \int_{0}^{a} \sqrt{a^2-x^2} dx$$
that is:
$$ \frac{b}{a}((\frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x \sqrt{1-\frac{x^2}{a^2}}}{a}))|^a_0) = \frac{ab \pi}{4}$$
so for the area of the entire ellipse:
$$ \frac{ab\pi}{4} 4  = ab\pi$$
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