Problem:
Set up a definite integral to find the area of an ellipse with axis lengths $a$ and $b$. Use a trigonometric substitution to find a formula for the area. What happens if $a=b$? Does this agree with a Geometry formula for a circle? Explain.
$$\frac {x^2}{a^2} + \frac{y^2}{b^{2}} = 1$$
where $a$ & $b$ are positive constants.
$B$ = Area of the first quadrant of the ellipse.
Total area = $4B$
\begin{align}
& y^{2} = \left(\frac{b^{2}}{a^{2}}\right)({a^{2}-x^{2}}) \\
\implies & y = \frac{b}{a}\sqrt{a^{2}-x^{2}}
\end{align}
Hence,
\begin{align}
B & = \int_0^a{\frac{b}{a}\sqrt{a^{2}-x^{2}}}\;dx = \frac{b}{a}\int_0^a{\sqrt{a^{2}-x^{2}}}\;dx
\end{align}
To remove $\sqrt{\quad}$, make a trig sub.
$$1 – \sin^{2} \Theta = \cos^{2}\Theta$$
$$x = a\,\sin\Theta$$
\begin{align}
B & = \frac{b}{a}\int_0^a{\sqrt{a^{2}-a^{2}\sin^{2}\Theta}\;dx} = \frac{b}{a}\int_0^a{\sqrt{a^{2}(\cos^{2}\Theta)}}\;dx\\
& = \frac{b}{a}\int_0^a{\sqrt{a^{2}}\,\sqrt{\cos^{2}\Theta}}\;dx = \frac{b}{a}\int_0^a{a\,\cos\Theta}\;dx
\end{align}
\begin{align}
\frac{dx}{dΘ} = (a \, \sin\Theta)' \implies dx = a \, \cos \Theta \, d\Theta
\end{align}
Therefore,
\begin{align}
B & = \frac{b}{a}\int_0^a{a \, \cos\Theta(a \, \cos \Theta \, d \Theta)} = \frac{a^{2} \, b}{a}\int_0^a{\cos^{2}\Theta}\;d\Theta = (a \, b)\int_0^a{\cos^{2}\Theta}\;d\Theta
\end{align}
I am a little lost up until this point and the formula doesn't seem to be going in the direction it needs to so that it will become the area of an ellipse. I feel like I made a mistake somewhere along the way. Any help is much appreciated!
Best Answer
Hint...You just need to change the limits to $0$ and $\frac {\pi}{2}$ and use the identity $\cos^2\theta=\frac 12(1+\cos2\theta)$ and you will be finished