# Thread: Area bounded by the ellipse

1. ## Area bounded by the ellipse

Show that the area bounded by the ellipse
$
\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1
$

is given by $A = \pi a b$

So I take the area in the first quadrant (with boundaries from 0 to a) and multiply it by 4

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

$
A = 4b(\frac {x}{2a} \sqrt {1 - \frac {x^2}{a^2}} + \frac {1}{2} sin^{-1} \frac {x}{a})
$

from 0 to a

However, as the final result I get $A = \pi b$ and I am missing that "a".

Any ideas would be greatly appreciated!

2. Hello, hasanbalkan!

Show that the area bounded by the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} \:= \:1$
. . is given by: . $A \:=\: \pi a b$
I would solve for $y$ like this . . .

Multiply by $a^2b^2\!:\;\;b^2x^2 + a^2y^2 \:=\:a^2b^2\quad\Rightarrow\quad a^2y^2 \:=\:a^2b^2-b^2x^2$

. . $a^2y^2 \:=\:b^2(a^2-x^2) \quad\Rightarrow\quad y^2 \:=\:\frac{b^2}{a^2}(a^2-x^2) \quad\Rightarrow\quad y \:=\:\pm\frac{b}{a}\sqrt{a^2-x^2}$

Then: . $A \;=\;4\cdot\frac{b}{a}\int^a_0\left(a^2-x^2\right)^{\frac{1}{2}}\,dx$

Let: $x \:=\:a\sin\theta\quad\Rightarrow\quad dx \:=\:a\cos\theta\,d\theta$

Substitute: . $A \;=\;\frac{4b}{a}\int a\cos\theta(a\cos\theta\,d\theta) \;=\;4ab\int\cos^2\theta\,d\theta$

. . $= \;2ab\int(1 + \cos2\theta)\,d\theta \;=\;2ab\left(\theta + \frac{1}{2}\sin2\theta\right) \;=\;2ab(\theta + \sin\theta\cos\theta)$

Back-substitute:. . $A \;=\;2ab\left(\sin^{-1}\frac{x}{a} + \frac{x\sqrt{a^2-x^2}}{a^2}\right)\,\bigg]^a_0$

Evaluate: . $A \;=\;2ab\left[\sin^{-1}(1) + 0\right] - 2ab\left[\sin^{-1}(0) + 0\right] \;=\;2ab\left(\frac{\pi}{2}\right) \;=\;\boxed{\pi ab}$