Intereting Posts

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normalized subgroup by another subgroup
Prime Appearances in Fibonacci Number Factorizations

I’m finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green’s theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating

$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$

- Prove that $\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots.$
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- Evaluate $\int_0^{\frac{\pi}{2}}\frac{x^2}{1+\cos^2 x}dx$

So we should be able to find the area of the ellipse by

$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$

$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$

$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$

$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$

$$=\pi\frac{a^2+b^2}{2}.$$

First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.

What am I doing wrong?

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HINT:

Putting $x=r\cos\theta,y=r\sin\theta$

$$\frac {x^2}{a^2}+\frac{y^2}{b^2}=1,$$

$$r^2=\frac{a^2b^2}{b^2\cos^2\theta+a^2\sin^2\theta}=b^2\frac{\sec^2\theta}{\frac{b^2}{a^2}+\tan^2\theta}$$

Your question has been answered, so now we look at how to find the area, using your parametrization, which is a perfectly good one.

The area is the integral of $|y\,dx|$ (or alternately of $|x\,dy|$. over the appropriate interval.

We have $y=b\sin\theta$ and $dx=-a\sin\theta\,d\theta$. So the area is

$$\int_0^{2\pi} |-ab\sin^2\theta|\,d\theta.$$

Using $\sin^2\theta=\frac{1-\cos 2\theta}{2}$, we find that the area is

$$\int_0^{2\pi} ab\frac{1-\cos 2\theta}{2}\,d\theta.$$

This is indeed $\pi ab$.

Here you go – this person even made your mistake, then someone else corrected it.

There are already a lot of good answers here, so I’m adding this one

primarily to dazzle people w/ my Mathematica diagram-creating skills.

As noted previously,

$x(t)=a \cos (t)$

$y(t)=b \sin (t)$

**does** parametrize an ellipse, but t is **not** the central angle. What

is the relation between t and the central angle?:

Since y is b*Sin[t] and x is a*Cos[t], we have:

$\tan (\theta )=\frac{b \sin (t)}{a \cos (t)}$

or

$\tan (\theta )=\frac{b \tan (t)}{a}$

Solving for t, we have:

$t(\theta )=\tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

We now reparametrize using theta:

$x(\theta )=a \cos (t(\theta ))$

$y(\theta )=b \sin (t(\theta ))$

which ultimately simplifies to:

$x(\theta)=\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$

$y(\theta)=\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$

Note that, under the new parametrization, $y(\theta)/x(\theta) =

tan(\theta)$ as desired.

To compute area, we need $r^2$ which is $x^2+y^2$, or:

$r(\theta )^2 = (\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2+

(\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2$

(note that we could take the square root to get r, but we don’t really need it)

The above ultimately simplifies to:

$r(\theta)^2 =

\frac{1}{\frac{\cos ^2(\theta )}{a^2}+\frac{\sin ^2(\theta )}{b^2}}$

Now, we can integrate $r^2/2$ to find the area:

$A(\theta) = (\int_0^\theta

\frac{1}{\frac{\cos ^2(x )}{a^2}+\frac{\sin ^2(x )}{b^2}} \, dx)/2$

which yields:

$A(\theta) = \frac{1}{2} a b \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

good for $0\leq \theta <\frac{\pi }{2}$

Interestingly, it doesn’t work for $\theta =\frac{\pi }{2}$ so we

can’t test the obvious case without using a limit:

$\lim_{\theta \to \frac{\pi }{2}} \, \frac{1}{2} a b \tan

^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

which gives us $a*b*Pi/4$ as expected.

This is another way to do it when one know the area of a circle: Consider the area of a circle with radius 1 in coordinates $(\xi, \eta)$ this is:

$$

\int d\xi d \eta = \pi

$$

now if you define new coordinates in your ellipse equation $\xi = \frac{x}{a}, \quad \eta= \frac{y}{b}$ you obtain a circle of radius one: $\xi^2 + \eta^2 =1$

The area of the ellipse you want is $ \int dx dy = ab \int d\xi d\eta = \pi ab$.

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