<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] Here is my proof if it is any use to anyone.
The equation of an ellipse is given by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Rearrange and write this ( the equation of an ellipse ) in terms of y: $$y=b\sqrt{1-\frac{x^2}{a^2}}$$
Lets find the area of one quarter of the ellipse and multiple that by 4 to get the area of the entire ellipse. We will integrate from $x=0$ to $x=a$. So we are looking for: $$4\int_0^a y.dx$$
Which is the same as ... $eq1$: $$4\int_0^a b\sqrt{1-\frac{x^2}{a^2}}.dx$$
Let: $$\sin u=\frac{x}{a}$$
Rearrange to get: $$x=a\sin u$$
Differentiate to get: $$\frac{dx}{du}=a\cos u$$
Therefore: $$dx=a\cos u.du$$
Sub this back into equation $eq1$.
But we also need to find the range of $u$ values.
Using $\sin u = \frac{x}{a}$ we can sub in our original limits of $x=0$ and $x=a$ to get the new limits in terms of $u$ of $u=0$ and $u=\frac{\pi}{2}$.
Subbing this back into $eq1$ as well we get $eq2$: $$4\int_0^\frac{\pi}{2} b\sqrt{1-(\sin u)^2}.a\cos u.du$$
From trigonometry we know that: $$\sin^2\theta + \cos^2\theta = 1$$
Therefore $$\sqrt{1-\sin^2u}.\cos u = \sqrt{\cos^2u}.\cos u = \cos^2u$$
Subbing this back into $eq2$ we get: $$4\int_0^\frac{\pi}{2} ab\cos^2u.du$$
Which is the same as: $$4ab\int_0^\frac{\pi}{2} \cos^2u.du$$
But again from trigonometry we know that: $$\cos^2\theta=\frac{1}{2}(1+\cos2\theta)$$
Subbing gives us: $$4ab\int_0^\frac{\pi}{2} \frac{1}{2}(1+\cos2u).du$$
Which is the same as: $$2ab\int_0^\frac{\pi}{2} (1+\cos2u).du$$
Now we can finally integrate to get: $$2ab[u+\frac{\sin2u}{2}]_0^\frac{\pi}{2}$$ 
Which goes to: $$2ab[(\frac{\pi}{2}+0)-(0+0)]$$
Which finally goes to: $$\pi ab$$
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