<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] 
In order to find the the area inside the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can use the transformation $(x,y)\rightarrow(\frac{bx}{a},y)$ to change the ellipse into a circle. Since the lengths in the $x$-direction are changed by a factor $b/a$, and the lengths in the $y$-direction remain the same, the area is changed by a factor $b/a$. Thus $$\text{Area of circle} = \frac{b}{a}\times \text{Area of
ellipse},$$
which gives the area of the ellipse as $(a/b\times\pi b^2)$, that is $\pi ab$.
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