Find the area enclosed by the ellipse

Question

I am given the formula ${x^2/a^2}+{y^2/b^2}=1$ and a diagram with an ellipse that is symmetrical about the $x$ and $y$ axes.

I able to get the area formula of the ellipse which is $4 \int_0^a (b/a) \sqrt{a^2-x^2}\ dx $

Then substituting $x=a\sin(\theta)$ and $dx=a\cos(\theta)\,d\theta$ to get:

$A=4\int^a_0(b/a)(a\cos\theta)(a\cos\theta)\,d\theta$

Then: $A=4\int^a_0ba(\cos^2\theta)\,d\theta= 4ab \int(1/2)(1+ \cos(2\theta) d\theta$

$(a=\pi/2)$

The total area should equal $\pi ab$

What I am confused on is how do you get the limit of $a$ and how do you go from $\frac ba$ to $(ab)$

Answer 1

Second, you have an $a$ before each of your $\cos \theta$'s, so that multiplies $\frac{a}{b}$ by $a^2$, giving $ab$.

First, you're using $a$ for two different things. The reason your second $a$ is $\frac{\pi}{2}$ is that when you change variables to $\theta$ the range for $\theta$ is $0$ to $\frac{\pi}{2}.$ It draws $1/4$ of the ellipse.

Third, please read the tutorial on MathJax.

Answer 2

Your notation is confusing, but you are right when you assert that the area is$$4\int_0^a\frac ba\sqrt{a^2-x^2}\,\mathrm dx$$and also when you say that we can compute this integral by doing $x=a\sin\theta$ and $\mathrm dx=a\cos\theta\,\mathrm d\theta$, thereby getting$$4\int_0^{\frac\pi2}\frac baa\cos\theta \,a\cos\theta\,\mathrm d\theta=4\int_0^\frac{\pi}2ab\cos^2\theta\,\mathrm d\theta.$$Since $\int_0^\frac\pi2\cos^2\theta\,\mathrm d\theta=\frac\pi4$, you get the number ($\pi ab$) that you're after.

Answer 3

We may use the "known" area of the circle, and the substitution $X=x/a$, $Y=y/a$ to pass linearly from the area of the (full) ellipse to the area of the (full) circle or radius one. The factor, the determinant of the map is $ab$.

Alternatively we can compute in full length using the substitution $x = a\,r\cos t$, $y=b\, r\sin t$, then formally

$dx= a\, \cos t\, dr-a\, r\sin t\, dt$,

$dy= b\, \sin t\, dr+b\, r\cos t\, dt$,

$dx\, dy = dx\wedge dy = ab\, r\, dr\wedge dt$,

and the integral to be computed is $$ \int_{r\in[0,1]}\int_{t\in[0,2\pi]} ab\, r\, dr\wedge dt\ , $$ which is easily computed.

Answer 4

The standard form of ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ The curve is symmetric about both the $x$ and $y$ axes.

We need to find the area in the first quadrant and multiply the result by $4$

$$\mbox{Area}=4\int_{0}^{a}ydx=4\int_{0}^{a}\sqrt{b^2\left(1-\dfrac{x^2}{a^2}\right)}dx=4\int_{0}^{a}\dfrac ba\sqrt{a^2-x^2}dx$$ Put $x=a\sin\theta,dx=a\cos\theta d\theta$

When $x=a,\theta=\dfrac{\pi}{2}$

When $x=0,\theta=0$

$$\mbox{Area}=\dfrac{4b}{a}\int_{0}^{\dfrac{\pi}{2}}\sqrt{a^2-a^2\sin^2\theta}(a\cos\theta d\theta)=4ab\int_{0}^{\dfrac{\pi}{2}}\cos^2\theta d\theta=4ab\int_{0}^{\dfrac{\pi}{2}}\dfrac{1+\cos2\theta}{2}d\theta$$ $$=2ab\int_{0}^{\dfrac{\pi}{2}}(1+\cos2\theta)d\theta=2ab\left(\frac{\pi}{2}\right)=\pi ab$$