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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!

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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

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The only mistake you've made so far was that when you substitued $\Theta$ for $x$, you didn't substitute the limits of integration. Since $x$ is going from $0$ to $a$, $\Theta$ has to go from $0$ to $\frac\pi2$. Once you fix that, your answer will indeed come out to the area of an ellipse.

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The integral is pretty trivial. Spoiler:

$$\int \sqrt{a^2-t^2}\ dt = \frac{1}{2} \left(t \sqrt{a^2-t^2}+a^2 \tan ^{-1}\left(\frac{t}{\sqrt{a^2-t^2}}\right)\right)$$

This can be obtained by a simple substitution $t = a\sin(z)$, $dt = a\cos(z)\ dz$

Which turns the integral into

$$a^2\int \cos^2(z)\ dz$$

Which is trivial and straightforward.

Remember that $z = \arcsin\left(\frac{t}{a}\right)$

The final result will be thence:

$$\int_0^a \sqrt{a^2-x^2}\ dx = \frac{1}{4} \pi a^2$$

Now, adjust with the constants.

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